Myers’ Math 140 Notes Table of Contentsdale/140/Math140LectureNotes.pdf · curves, just because...

Myers’ Math 140 Notes

Table of Contents

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SYLLABUS

COMMON ERRORS

Study these before doing the Take-home quiz.

SAMPLE GATEWAY EXAMS (5)These are practice exams for the Gateway Exams. To get a “C” or better, you

must pass some Gateway exam with a score of at least 28 (7 problems out of 8).

PRACTICE EXAMS AND FORMULA REVIEWS 1-5Before each exam, go over the corresponding practice exam thoroughly. You

must be able to complete it within 50 minutes. Each exam problem has a workedexample on a practice quiz sheet. Also take the Online exams.

LECTURES 1-30These are almost verbatim copies of the lectures.

Lecture videos are on the web.

HOMEWORK AND IN-CLASS WORKSHEETS 1-27Fill in these worksheets and hand them in. Do not turn in scratch paper.

RECOMMENDED PROBLEMS AND WORKED EXAMPLES

These recommended problems are for practice only.

Print the syllabus from the class website: www.math.hawaii.edu/140

Precalculus Syllabus, Math 140

Class website. math.hawaii.edu/140 or Google myers 140.

Phone. 956-4672 (PSB 318), (Myers home: 536-1285, ask for "Prof. Myers" not "Dr. Myers" which is my wife, an M.D.)

Email address. [email protected] Include your TA's name in all emails

Office hours, PSB 318. See class website, math.hawaii.edu/140 If you have to wait, do so on a chair inside the office. Waiting outside is always futile.

Text. Stewart, Precalculus. Register online for WebAssign with the code in your textbook. Buy the required worksheet packet and iClicker from the bookstore. To register your iClicker, go to the iClickers link.

Prerequisites. Two years of high school algebra, one year of high school geometry or trigonometry, passage of the placement exam.

Grade. Grade score = 10*(four chapter exams + 200 point written final) + homework +WebAssign homework + classwork + (1/3)* (iClicker+preexams+gateway). Letter grades are assigned according to rank (curve - not percentage) on the total grade score. Letter grades are not averaged. To get a C or better, you must pass a gateway exam with a score of 56/64 or better. Curve for an average class (to within 3%): 10% A's 30% A&B's 40% C's 30% D&F's 10% F's. You are ranked against students in the two sections taught by your lab session instructor. Students in sections taught by the other lab instructors are on a separate curve with separate grade lines. After each midterm, grades are posted in the middle of the class website page. Along with your grades are the class grade curves. Do not read your scores horizontally across the class curves, just because you are 10th from the top in the curve for one exam does not mean you are the 10th from the top in the other curves.

Grading. I handle matters involving doctor-signed excuses and grading of exams, WebAssign homework, and iClicker quizzes. The teaching assistants handle all matters relating to written homework and classwork.

iClicker quizzes. The MW lecture periods are devoted to iClicker quizzes. Talking is not allowed. Prepare for these quizzes with the lectures in the worksheet packet or the online lecture videos on the class website. To preview a quiz, go to the class website math.hawaii.edu/140, select LecturesHW, click "quiz" after the lecture.

Homework. Homework assignments: Practice on odd numbered problems, they have answers in the back of the book. Homework problems are for credit, not practice. Homework worksheets are due at the beginning of the lab section. For homework due-dates, click "Home" above then "Full

http://www.math.hawaii.edu/%7Edale/140/index.html

https://www.webassign.net/login.html

http://math.hawaii.edu/%7Edale/140/iClicker-registration.html

http://www.math.hawaii.edu/%7Edale/140/index.html

http://math.hawaii.edu/%7Edale/140/lectures.html

Calendar". Homework and written exams (Exam 3, 4 and the final): Simplify your answers: unacceptable: (b-a)/(a-b), y-7 = x-2, -6/9, 0/2 acceptable: -1, y = x + 5, -2/3, 0 Use reduced fractions instead of decimals or mixed fractions. E.g., 3/2, not 1.5 or 1 1/2. Fractions of fractions such as (1-1/x)/(1+x) are not acceptable. Simplify to a single fraction. Give exact answers instead of decimals. E.g. pi instead of 3.14. Circle your answers. All homework assignments must be written on the worksheets provided in the lecture note packet or on line in the "LecturesHw" link. Write all your work in the space provided. Do not attach scratch paper to either homework or written exams. Your lab session instructor grades your homework and the first page of the written exams. I grade the last. Extra credit (5 points) for being the first to find an error in the notes, in a WebAssign problem or on exams. chk: If you see "chk=8" on a homework assignment, it means the checksum of your answer is 8. The checksum is the sum of all the digits. For -13/12, chk=1+3+1+2=7. For x-2+3x-1/2, chk=2+3+1+2=8. Web homework: WebAssign homework is due 48 hours after the corresponding lecture unless specified otherwise. Web homework is not an alternative to the written homework, both are required. If you have an incorrect answer, you get two additional tries for that problem. To get a second or third try, click anywhere in the question, then click "New Randomization".

Late credit. Excuses require a doctor-signed medical excuse or a University-approved excuse for off-island athletic events. Make up missed exams and assignments within 48 hours the last excused day. Late homework assignments: If only one lab session overdue, 90%. Else if only one week overdue 80%. Otherwise 70% late credit. If you turn in Hw 11 when 12, 13, 14 are due, you get 90%, 80%, 70% respectively. If you have a dated, doctor-signed medical excuse for a missed homework or exam, your due date for this missed exam is one period after the last excused date. Homework submitted 1, 2, 3, ... periods after this due date will get 90%, 80%, 70% as above. The last day of instruction is the last day to submit late homework. Late web assignments: 85% if late. WebAssign does not accept assignments which are more than 14 days late. To submit a late web assignment, you request an automatic extension and accept their terms. If you have a doctor-signed excuse I can remove either this 85% late penalty or extend the due date (but at most 14 days). Late iClicker quizzes and classwork: You can't make up late iClicker quizzes or classwork. If you have a doctor-signed medical excuse, show it to your TA who will give you your group's grade for the missed day and show the excuse to me and I will average in your missed iClicker exam. I will also average in scores for iClicker quizzes missed due to taking a midterm during a lecture period.

Class work. In-class group work must be done with a group of between three to six students. No credit otherwise. To get credit, you work with your group until everyone has finished or until the end of the class period. Help those in your group who need it; get help if you need it. Everyone in the group gets the lowest score in the group.

Exams. Exams are closed book. No calculators. Use only the provided scratch paper. They have three pages, put your name on each page. Do not use erasable ink. Show your work and circle your answers. Practice exams are in your worksheet packet. Try to do the Practice-quiz problems in the time indicated. Practice-quiz problem numbers correspond to practice exam problem numbers. Test problems are similar the practice exam and pre-exam problems and the more difficult homework problems. Pre exams: For each exam, Exam1, Exam2, Exam3, Exam4, Final, there is a pre exam, PreExam1, PreExam2, PreExam3, PreExam4, PreExam5 which you do at home and submit before the beginning of the exam week. I suggest taking the no-credit practice exams before taking the pre-exams. For the exams, you only get one try. For the pre-exams, you get three tries as with Web homework. You can take the no-credit practice exams 50 times. To get the dates for the exams and preexams, click "Home" above then "Full Calendar". Gateway exams: Gateway exams are 25 minute WebAssign exams held in the computer lab in PSB 208. See GatewaySchedule for the available exam times. Passing is 56/64 or better. Until the last day to withdraw, you must take them at least once every week until you pass or run out of your 20 tries. You must pass a gateway exam before the last day to withdraw in order to get a grade of C or better. This does not imply the converse, that passing the gateway will get you a C or better. If you don't pass any gateway exam, the best grade you can get is C- even if your grade other than the gateway is an A. But passing the Gateway, won't necessarily mean that you pass the course, an F is still possible. Final exam: The final exam is a cumulative 2-hour exam, 12:00-2:00, Wednesday of finals week.

Late registration. If you register late, read “FirstWeekDuties” on the class website. Any missed assignments must be made up within one week.

http://math.hawaii.edu/%7Edale/140/gateways.html

http://math.hawaii.edu/%7Edale/140/gateway-schedule.pdf

Exact answers. On exams and homework assignments such as Hw 12, 14, 18 you are asked to give exact answers rather

than decimal answers. This means writing π and not its nonexact decimal approximation 3.14. This means writing 2instead of a finite decimal approximation given by a calculator. The exact answer for the area of a circle with radius 2is , the decimal approximation is The exact answer for the solution of is .�22 = 4� 4(3.14) = 12.56. x2 − 5 = 0 x = ! 5The decimal approximation to two places is . !2.24

Take-home quiz. Note that . In general and are not and and2 = 12 + 12 ! 1 + 1 = 2 a2 + b2 a2 − b2 a + b a − b

can not be simplified further. However is true for positive a, b. a2b2 = ab

In a fraction, if you can factor out common factor from top and bottom, you can cancel. Otherwise you can’t. Hence

. However no cancellation is possible for . 2a−2b

2b =2(a−b)

2b =a−b

b2a−b

2a ,2a−b

b

Note that . Secondly . 123

83 =(4$3)3

(4$2)3 =4333

4323 =33

23

a5−a3

a3 =a3(a2−1)

a3 = a2 − 1

Finally, . 2737−26

26 =26(2$37−1)

26 = 2 $ 37 − 1

Write with fewer symbols: . −a−b

2b−2a =b−a

2(b−a) =12

, find and . and .f(x) = x + 1 f (f (x)) f (x + y) f (f (x)) = f (x) + 1 = x + 1 + 1 f (x + y) = (x + y) + 1 = x + y + 1

Gateway problems. All Gateway exams have a problem which asks you to complete the square for a formula. The mostcommon mistake is to try to make the formula into an equation and then complete the square for the equation. E.g.,when asked to complete the square of many try to make this an equation . Formulas andx2 − x + 1 x2 − x + 1 = 0equations are two different things.

The most common error for factorization problems is to list the roots rather than give the factorization. For theproblem “Factor ”, the correct answer is “ ” not “ ”.x2 − 4 x2 − 4 = (x − 2)(x + 2) x = −2, x = 2

Hw 0, Prob. 3, 4. Rewrite without | |’s: for .|x + 1| + 4|x + 3| x < −3 is either or . |x + 1| (x + 1) −(x + 1)

To determine which, determine if is or . x + 1 m 0 < 0If , then . x + 1 m 0 |x + 1| = (x + 1)

If , then . x + 1 < 0 |x + 1| = −(x + 1)

Hw 0, Prob. 6, 7. Write as an interval: {x: |x−4| < 4}. Write as a union of two intervals: {x: |x+5| > 2}. A inequality such as is equivalent to < |x + 6| < 2

which in turn reduces to (subtract 6)−2 < x + 6 < 2 which gives one interval −8 < x < −4 (−8, −4)

can also be written .−8 < x < −4 −8 < x and x < −4A inequality such as is equivalent to> |x + 6| > 2

which is reduces to (subtract 6) x + 6 < −2 or 2 < x + 6 which gives two intervals x < −8 or −4 < x (−∞, −8) 4 (−4,∞)

A inequality involves an “or” instead of an “and” and two intervals instead of one. > or m

Hw 1, Prob. 2. Solve 3x + 1 = 4When solving equations with radicals, get the radical by itself on one side and everything else on the other. After solving the equations one must substitute the answers back into the original equation to determine if they arevalid.

Math 140 Common ErrorsCompiled by Myers and Andrew Conner

1

Hw 1, Prob 4. . x+4

2x−5 [ 0

The most common mistake is to multiply both sides of by , to get . But things aren’t thisx+4

2x−5 [ 0 2x − 5 x + 4 [ 0

simple. Since we don’t know whether is positive or negative, we don’t know whether the new inequality is 2x − 5 or the reverse . Don’t multiply an inequality by a term such as unless you know if it is positivex + 4 [ 0 x + 4 m 0 2x − 5

or negative. Use the key number method given in Lecture 1 and see the worked examples for Homework 1.

Similarly, you can’t simplify by multiplying both sides by x. First get the 2 on the left, , thenx −3x [ 2 x −

3x − 2 [ 0

put the left side over a common denominator, , then factor, , then find the key numbersx2−2x−3

x [ 0(x−3)(x+1)

x [ 0where the fraction is 0 (where the numerator is 0) and where the fraction is undefined (where the denominator is 0).

Hw 2, Prob 2. .x2 + y2 − 10x + 2y + 17 = 0To solve ,x2 + y2 − 4x − 6y − 12 = 0first group the x’s together and the y’s together and get the constant on the right.

(x2 − 4x ) + (y2 − 6y ) = 12Now, complete the square by adding the square of one half the coefficient of or .x y

Whatever is added to the left must also be added to the right side of the equation.

(x2 − 4x + (−42 )2) + (y2 − 6y + (

−62 )2) = 12 + (

−42 )2 + (

−62 )2

(x2 − 4x + 4) + (y − 6y + 9) = 12 + 4 + 9(x − 2)2 + (y − 3)2 = 25Now write the right side as a square.

.(x − 2)2 + (y − 3)2 = 52

The center is and the radius is .(a, b) = (2, 3) r = 5

Hw 2, Prob 9. Find the slope of the line through and .(−3, 0) (4, 9)

When finding the slope of the line through , the most common error was invert the slope formula,(x1, y1) and (x2, y2)

i.e. use instead of the correct or equally correct . Make sure that the order of them =x1−x2

y1−y2 m =y1−y2

x1−x2 m =y2−y1

x2−x1

coordinates on top matches the order of the coordinates on the bottom, e.g. would be wrong.m =y1−y2

x2−x1

Hw 3, Prob 3, 4.Factor and find all roots x3 + 7x2 + 11x + 5 = 0. − 1 is one root.Likewise for x3 + 8x2 − 3x − 24 = 0. − 8 is one root.First error: not giving both the roots and the factorization.Your answer should be of the form:Factorization: (x + 3)(x + 2)(x − 1)

Roots: .x = −3, −2, 1Second error: not knowing how to get started. Notice the theorem of Lecture 3: a is a root iff is a factor. (x − a)

Secondly, if is a factor, then you get the other factor using long division.(x − a)

Hence if 11 is a root, then is one factor. Divide the polynomial by . The resulting quotient is the other(x − 11) (x − 11)

factor. This quotient might have to be factored even further.

Hw 3, Prob 6. When calculating a domain, you must exclude all points which makes the denominator of a fraction 0.

Hw 3, Prob 11. Note that is not multiplication. is multiplication. means replace each x in by f(g(x)) f(x)g(x) f(g(x)) f(x)

. Hence if , then , , , , , g(x) f(x) = 2x + 3 f(t) = 2t + 3 f(8) = 2(8) + 3 f(1x ) = 2 1

x + 31

f(x) =1

2x+3 f(x2) = 2x2 + 3

, , .f(y + 1) = 2(y + 1) + 3 f(x + h) = 2(x + h) + 3 f(f(x)) = 2f(x) + 3 = 2(2x + 3) + 3If , then , in particular, .g(x) = x2 g(x + h) = (x + h)2 g(x + h) ! x2 + h2


2

Hw 4, Quiz 4. When listing intervals of increase of decrease, separate the intervals with a comma rather than taking theunion as is done when listing a function’s domain. Write when giving a domain but write [0, 1) 4 (2,∞) [0, 1), (2,∞)

when listing two intervals of increase. Secondly, when listing an interval of increase or decrease, include the endpoint(i.e., use if the function is defined at the endpoint; exclude the endpoint (i.e., use ) if the function is undefined.[, ] (, )

Hence if are two intervals of increase for a function, then the function is defined at 0 but undefined at 1[0, 1), (2,∞)

and 2.A common error is to wrongly assume that for any function f, . For multiplication by a number cf(a + b) = f(a) + f(b)

it is true that . But for functions, it is usually the case that .c(a + b) = ca + cb f(a + b)!f(a) + f(b)

Almost always, we have: , , , , , a + b ! a + b |a + b| ! |a| + |b| (a + b)n ! an + bn xa+b ! xa + xb 1a+b !

1a +

1b

log(a + b) ! log a + log b, sin(a + b) ! sin a + sin b, cos(a + b) ! cos a + cos b, tan(a + b) ! tan a +, tan b.

Likewise, it is usually the case that and , ... .a − b ! a − b |a − b| ! |a| − |b|

Hw 5, Prob 2(k). Do problem 2(k) on page 137 carefully; it is often missed. You will see variants in quizzes and exams.

Quiz 5, Prob 9, 14b. Problems 9 and 14(b) were often missed. Getting was the most common error in Problem 9.g(x + h)

If , then , in particular, (see Hw 3, Prob 34).g(x) = x2 g(x + h) = (x + h)2g(x + h) ! x2 + h2

If , then the correct answer is . Note that , ,g(x) = 2 −3x g(x + h) = 2 −

3x+h g(x + h) ! 2 +

3x + h g(x + h) ! 2 +

3x +

2h

and .g(x + h) ! (2 +3x ) + 2 +

3h

Quiz 5, Prob. 14b. The most common mistake was to solve the problem with the following incorrect sequence of steps. The problem lies with the last step, instead of replacing x with x +1, it incorrectly replaces -xf(x) d f(−x) d f(−x + 1)

with -x +1. The recommended solution is to rewrite as . The correct sequence is obtained byf(−x + 1) f(−(x − 1))

working backward from to and then to . This results in the sequence . Notef(−(x − 1)) f(−x) f(x) f(x) d f(−x) d f(−(x − 1))

that this last step involves replacing x by x−1 which shifts the graph right one unit.

Practice Exam 1, Prob. 14b. As with the preceding quiz problem, the majority of students can’t graph . Theyf(1 − x) − 1fail to rewrite the problem as and fail to find the correct sequence which is f(−x + 1) − 1

. Start with the value work backward one step at a time to andf(x) d f(x + 1) d f(−x + 1) d f(−x + 1) − 1 f(−x + 1) f(x)

work forward to the desired value (see Quiz 5, Prob. 14b). f(−x + 1) − 1

Practice Exam 1, Prob. 12. Graphing piecewise defined functions is easy but many miss this problem.

Practice Exam 1, Prob. 16. Most get this problem but fail similar problems such as (1) if , then find orf(x) = 1 − x2 f (f (x))

(2) if then find (see quiz 5 prob. 9, 14b). g(x) = 1 − x2g(x+h)−g(x)

h

Quiz 10. Many were able to find the vertical and horizontal asymptotes and x and y-intercepts but yet could not draw thegraph. The best remedy is to plot more points until you see the shape of the graph.

Quiz 11, Prob. 9. Most mistakes here were due to confusion about the rules of exponent arithmetic — when you shouldadd, subtract or multiply exponents. Here practice makes perfect, do the recommended exercises to developcompetence with exponents.

Practice Exam 2, Prob. 1(b). To solve for y, regard x as a constant and use the quadratic formula. Gettingy2 − 2y = x

everything on the left gives Note that this is where . The quadraticy2 − 2y − x = 0. ay2 + by + c = 0 a = 1, b = −2, c = −x

formula then gives . y =−b! b2−4ac

2a =2! 4+4x

2 = 1 + 1 + x


3

Practice Exam 2, Prob. 4. Similar problems on Exam 2 often require knowing formulas for volume and surface area.Know all the facts on the inside of the front cover of the text except Heron’s formula and the circular cone formulas.

Practice Exam 2, Prob. 5. You need to know that for similar triangles, ratios of corresponding sides are equal. In theanswer’s picture the sides “6” and “s” of the small triangle correspond to the sides “10” and “x + s” of the largertriangle which is similar to the smaller one.

Hw 16. While you can calculate the trigonometric functions for the following angles using reference angles, they occuroften enough that it is best be able to picture them graphically on the unit circle. Know the x and y coordinates of eachpoint.

π/2

π/3

π/4

π/6

−π/6

−π/4

−π/3

3π/4

π=−π

5π/6

2π/3

7π/6=−5π/6

5π/4=−3π/4

4π/3=−2π/33π/2=−π/2

0=2π

Here are the coordinates for the unit circle points for the first-quadrant angles:

, , , . The x coordinate is the cosine and theP(0) = (1, 0) P(�/6) = ( 3 /2, 1/2) P(�/4) = (1/ 2 , 1/ 2 ) P(�/3) = (1/2, 3 /2)

y coordinate is the sine of the angle. The coordinates and hence sines and cosines of the other angles are the same astheir first quadrant reference angles except for a change of sign. Points below the x-axis have a negative y coordinate.Points left of the y-axis have a negative x coordinate.

Hw 16. Don’t continue writing sin or cos after calculating sin or cos. is correct but sin(−�/4) = − sin(�/4) = 1/ 2

is wrong. sin(−�/4) = − sin(�/4) = sin(1/ 2 )

Hw 17. Common error when writing proofs: don’t write “iff ” or between formulas; don’t write “=” betweeng

equations. Write “=” between formulas; write “iff” or between equations.g

Practice Exam 3, Prob. 12, 13. The most common error (with a 1 point loss) was failure to insert the needed “iff”s betweenequalities or wrongly inserting “=”s between equalities. Some students made the opposite mistake of placing “iff”s

between terms. E.g. “ ” is correct but “ ” is incorrect — you should place “=” not “iff”2x + 4 = 2(x + 2) 2x + 4 iff 2(x + 2)

between two numbers. One the other hand “ ” is correct but “ ” is incorrect — you should2x = 4 iff x = 2 2x = 4 = x = 2place “iff” not “=” between two equations.

Exam 4 errors. When graphing tan and cot, remember that the period is π. The other trig functions have period 2π.


4

While most students remember the graphs of sin, cos, and tan. Most fail to graph csc, sec or cot. These are obtainedfrom sin, cos, and tan respectively by taking reciprocals. This was done for csc and sec in the lecture notes and in thehomework. The same technique will give the graph of cot.

When calculating phase shifts, remember that it is the number which is subtracted from x. Hence the phase shift of

is π and the graph of sin is shifted π units to the right. The phase shift of is -1 since sin(x − �) tan(�(x + 1))

and this corresponds to shifting the graph of one unit to the left. tan(�(x + 1)) = tan(�(x − (−1))) tan(�x)

Practice Exam 4, Prob. 4. Many try to apply the addition formula to and . This makes the problemsin(x + y) cos(x + y)

more complicated. One should note that the entire expression is the right-hand side ofsin(x + y) cos x − cos(x + y) sin x

the a subtraction formula for sin: . Then simplifies tosin((x + y) − x) = sin(x + y) cos x − cos(x + y) sin x sin((x + y) − x)

. In general, many students have a hard time remembering the sum and product rule identities. There is moresin x

memorization required here than in high school math classes but much less to remember than in an organic chemistryclass.

Practice Exam 4, Prob. 10. Many who were able to solve this equation were not able to solve the2 cos2x − 5 cos x = −2related quiz-type equation . The first step here would be to replace by and make this2 sin2x + 5 cos x = 4 sin2x 1 − cos2x

a problem involving only cosines. Many also were unable to solve . Since there is no squares anothercos x + sin x = 0

tactic is needed. Divide by to get which is which is . Then solve this. cos x 1 +sin xcos x = 0 1 + tan x = 0 tan x = −1

Practice Exam 4, Prob. 11a, 11b. Most mistakes here involved ignoring what the restricted ranges are for the inverse trigfunctions. For any value x, must be an angle in the upper two quadrants, i.e., an angle in . must becos−1(x) [0,�] sin−1(x)

an angle in the two right hand quadrants, i.e., an angle in . [−�/2,�/2]


5

1. Expand and then simplify with no parentheses in the final answer.2s 5t2 3st

4s2 17st 25t2

2. Factor .81c3 90c2 25c

c9c 52

3. Use inequalities to write without absolute value signs.|z 1| 0

z 1

4. Let State the domain of and compute .gd 1/d . g g1/16

domain: . d 0 g1/16 2

5. Complete the square for .3c2 3c 7

3c 12 2 31

4

6. Simplify .y1/2 1y 2 y y1/3

y1/2 2y y1/6

7. Solve for n: .25/3

47/3 2n

n 3

8. Find all real numbers b such that .b b

b 0, 1

Sample Gateway Exam 1You must get at least 7 out of 8 problems on some Gateway Exam in order to get a C or better in Math 140. No calculators.

1. Simplify .52827

27

58 2 1

2. Expand .x 1x 2x 3

x3 6x2 11x 6

3. Find all real numbers b such that .9b 12 5b

no solutions

4. Factor .8E3 27

2E 34E2 6E 9

5. Solve for a: .36

95/2 a33

3 3

6. Let . Compute and simplify it.Yj 4 j 32 Y12

3

7. Complete the square for .2L2 4L 7

2L 12 9

8. Use inequalities to write without absolute value signs.|z 3| 3

z 6 or z 0


1. Factor .x2 9x 8

x 1x 8

2. Express the following inequalities by using one pair of absolute value signs: .3 t or t 0

|t 32 | 3

2

3. Simplify for 7a4

a5 a 0

7a

4. Solve for c: 1254/3253 c5

c 25

5. Let . Compute and simplify it.Lz z1/3 Lw 13

w 1

6. Complete the square for .2b2 7b 12

2b 74

2 478

7. Find all real numbers a such that .5a2 4a 8a 1

no solution

8. Expand and then simplify with no parentheses in the final answer.a 3b c ac 2b

ab 5b 3c


1. Complete the square for .6F2 3F 11

6F 14

2 858

2. Simplify .5621520

520

5 621 1

3. Let . State the domain of g and compute .gx 13 x g 1

27

domain: . x 0 g 127 3

4. Find all real numbers d such that .6d2 d2 7d 2

d 25 , 1

5. Expand .x 1x 2x 3

x3 6x2 11x 6

6. Use inequalities to write without absolute value signs.|t 2| 12

t 32 or 5

2 t

7. Factor .x2 5x 6

x 2x 3

8. Solve for s: .242/332/3 8s1

s 13


1. Use inequalities to write without absolute value signs.|t 72 | 3

2

5 t 2

2. Let . What is the domain of g and what is .gx x/x g2

domain: . x 0 g2 1

3. Find all real numbers E such that .17E2 25E 10 E3 E2 E 1 E3

E 34

4. Factor 25b3 30b2 9b.

b5b 32

5. Simplify .3 8b6

2b2

6. Solve for d: .63/2

2 3 d1/2

d 27

7. Complete the square for .20q2 5q 12

20q 18

2 18716

8. Expand and then simplify with no parentheses in the final answer.a 32a 2 5a 1

a3 4a2 8a 23


Solving quadratic equations`Find all real numbers x such that .9x 12 5

xFirst rewrite this with everything on the left, 0 on the

right.9x 12 5

x9x 12 5

x 09x2 12x 5 0First try factoring. In this case, it doesn’t work, so use

the quadratic formula .x b b24ac2a

x 12 122495

29 12 144180

18 undefinedSince 144-180 is clearly negative, we can’t take the

square root, hence there are no solutions. Don’t use theimaginary number i here, the problem asks for realnumbers.

Answer: no solution.

`Find all real numbers b such that .b bSquare to get rid of the square root, then put everything

on the left.b b

b b2

b b2 0We multiply by -1 and put the highestb2 b 0

power first to make things look more familiar.Now factor.bb 1 0Set each factor to zero separately.

Answer: b 0, 1

`Find all real numbers d such that .6d2 d2 7d 2First rewrite this with everything on the left, 0 on the

right.5d2 7d 2 0Now try to factor.

. 5d 2d 1 0Set each factor to zero separately.

Answer: d 25 , 1

Completing the square`Complete the square for .2x2 7x 12

Warning: completing the square for the formula and completing the square for the equation 2x2 7x 12

are two different problems. The2x2 7x 12 0common mistake is to assume the formula is an equation.For the equational version of completing the square, seethe circle equation section of Lecture 2. Completing thesquare for formulas is different. With an equation, youcan add a number to both sides. When you add somethingto a formula, you must subtract it off somewhere else inthe formula.

2x2 7x 12First factor the coefficient of x2 out of the first two

terms. Leave the constant alone. Factoring a 2 out of is7xthe same as dividing by 2. 7x2x2 7x 122x2 7

2 x 12To complete the square of a factor , divide thex2 ax

coefficient of x by two, then square. This gives . Add a2 2

this to the factor in parentheses to get which is the perfect square . x2 ax a

2 2 x a2 2

In , a = . x2 72 x 7

2Dividing by 2 and squaring gives: . 7

2 74 49

16Thus becomes2x2 7

2 x 12

2x2 72 x 49

16 12 2 4916

Since we added , we compensate by subtracting it4916

from the end of the formula. Where did the 2 come from?Since the that we added was multiplied by 2 (the 2 at the49

16beginning of the formula), the we subtract must also be49

16multiplied by 2. Failure to do this multiplication is acommon error.

Hint. Cancellation almost always occurs at this step.Don’t multiply 2x49; cancel 2 off the 16. You should get

rather than .498

9816

2x2 72 x 49

16 12 498

2x 74 2 128

8 498

2x 74 2 9649

8

2x 74 2 47

8

Answer: 2x 74 2 47

8

Be careful when adding fractions; partial credit isn’tallowed on gateway exams.

Worked examples of gateway exam problemsYou must get at least 7 out of 8 problems on some Gateway Exam in order to get a C or better in Math 140. No calculators.

Inequalities and absolute valuesYou can replace absolute value inequalities with a pair

of algebraic inequalities. iff a = 0|a| 0 iff |a| 2 a 2 or 2 a iff 0|a| 2 2 a 2

Ùse inequalities to write without absolute|z 1| 0value signs.

iff iff |z 1| 0 z 1 0 z 1Answer: z 1

Ùse inequalities to write without absolute|z 3| 3value signs.

iff |z 3| 3 z 3 3 or 3 z 3 iff z 6 or 0 z

Answer: z 6 or z 0

Ùse inequalities to write without absolute|t 72 | 3

2value signs.

iff |t 72 | 3

2 32 t 7

2 32

iff 32

72 t 3

2 72

iff 5 t 2Answer: 5 t 2

Èxpress the following inequalities by using one pair ofabsolute value signs:

.t 1 or 4 t

Draw a picture of the set; find the midpoint between theendpoints; find the distance between the midpoint and theendpoints.

The distance between a and b is . |a b|The midpoint between a and b is .a b/2

41 5/2

3/23/2

Picture:

The midpoint is (1+4)/2 = 5/2. The distance between the midpoint and the endpoints is

5/2 - 1 = 3/2. t is in the set iff the distance between t and 5/2 is greater than 3/2.iff Answer: |t 5

2 | 32

Domain problems

`Let State the domain of and compute gd 1/d . g. g1/16

You can’t divide by 0, you can’t take the square root ofa negative.

Hence we must have d=0 and we must have 1/d > 0.That latter means d > 0. This plus givesd 0

Answer: domain: . d 0

g1/16 1/1/16 16 4 2

Answer: .g1/16 2

`Let . What is the domain of g and what is gx x/x.g2

You can’t divide by 0. Hence we must have x=0.Answer: domain: .x 0Note: Domains must be calculated without

simplifications. If we simplified to 1, we wouldx/xwrongly conclude that the domain was (-5,5).

g2 2/2 1

Answer: .g2 1


Exponential equation: variable inexponent

Rewrite the equation so that both sides are powers ofthe same base . Then equate the exponents bp bq p qand solve for the variable.

`Solve for n: .25/3

47/3 2n

Since 4=22, everything can be written as a power of 2.25/3

47/3 2n

25/3

227/3 2n

25/3

214/3 2n

You don’t have to do this much detail.25/3214/3 2n

If you can do this in your head, great.2 5/314/3 2n

29/3 2n

23 2n

3 n

Answer: n 3

`Solve for x: .242/332/3 8x1

First simplify.242/332/3 8x1

3 82/332/3 8x1

since 32/382/332/3 8x1 32/332/3 32/32/3 30 1Now equate exponents.82/3 8x1

2/3 x 1x 2/3 1x 1 2/3 1/3

Answer: x 13

Exponential equation: variable in base

`Solve for x: .63/2

2 3 x1/2

To get x, square both sides. 63/2

2 3 2 x1/22

63/22

2 6 x

x 63

21/26

x 63

23 233

23 2333

23 33 27Answer: x 27

`Solve for x: .36

95/2 x33

To get x, take the 33rd root of both sides.x33 36

95/2

x331/33 36

95/2 1/33

x331/33 36

325/2 1/33

x 36

35 1/33 36 351/33

x 3111/33 311/33 31/3

Answer: 3 3


Math 140 Practice Exam 1 /100Exam 1. Lectures 1-6. Use provided scratch paper. No calculators.Understanding is not enough, you must develop enough proficiency

to complete this exam in 50 minutes. Tutoring available in Bilger Annex 209.

1(5). Rewrite without 's: for x i[3,6].|x 3| 2|2 x|

2(5). Write as a union of two intervals: {x: |x+2| > 1}.

3(4). Find all real solutions x. Write Q if there are none. 3x 4 x

4(16). Solve for x. Write your answer in interval notation.(a) One interval

2xx3 1

(b) A union of two intervals1 1x 2

2x

5(6). Find the circle :x2 2x y2 7(a) Find the center and radius.

(b) Find the x and y-intercepts.

6(4). Find the equation for the line through (-1,-1) and(2,3).

7(5). Find the domain of f (x) = . Union of two1

1 x1

intervals.

8(6). Factor and find all roots given that -1 is one root.x3 x2 3x 3 0

9(4). Simplify to a single polynomial:g(x) = x3. [g(x+h)-g(x)]/h =

10(3). For the function h pictured,

h

-2-10123

0 1 2 3 4 5 6 7 8 9 10

for which values of x does h(x)=0?

11(14). For the domains and ranges, use interval notation,[1,5), or set notation, {1,3,5}. The answer may be none.

interval(s) of decrease

interval(s) of increase

min value

max value

turning point

range

domain

lx1/xfunction

12(6). Graph f (x) where

fx

1/x if x 1

x1/x

if 1 x 1if x 1

13(4). Graph .1 x 1

14(6). The graph of f (x) is a line segment from (0,2) to(2,0). (a) Graph y = -f (x+1)+1. (b) Graph y = f (1-x)-1.

15(4). f(x) = (1/x)+1, g(x) = (1/x)-1, find and simplify(fog)(x).

16(2). G(-4) = -3, G(-3) = 1, G(-1) = 2, G(1) = 5. F(-3)= -4, F(1)=-3, F(2)= -1, F(5)=1. Find (GoGoF)(1).

17(6). Write each function as a composition f (g(x)) of twosimpler functions f (x) and g(x).(a) . (b) .Gx x4 x2 1 Fx 1

x x

Answers1. 3x - 7

2. (-5,-3]u[-1,5)

3. x = 4, (not x = -1, it doesn't work in the original equation.)

4a. (-3,3) 4b. 2,1 0, 2

5. (a) center = (1,0), radius = , 2 2(b) x-intercept: , y-intercept: 1 2 2 7

6. y 43 x 1

3

7. [-1,0)u(0,5)

8. roots .x 3 x 3 x 1 3 ,1, 3

9. 3x2 + 3xh + h2

10. x = 1, 9

11. See homework problem Hw 4:7

12.

-11

1

-1

13. 14(a). 14(b).

-1 -11 1

1 1

-1 -1

15. 1/(1/x)-1)+1 = 1/(1-x) .

16. G(G(F(1))) = G(G(-3)) = G(1) = 5

17. (a) G(x) = f (g(x)) where g(x) = x2 and f (x) = x2+x-1. (b) F(x) = f (g(x)) where g(x) = and f (x) = 1/x + x.x

All exams will consist solely of homework type problems.

All numeric answers must be exact, no decimals, no mixed

fractions. E.g., not, or 1.5 or 1.414... .32 , 2 1 1

2

ABSOLUTE VALUE |x| = x if x > 0. |x| = -x if x < 0.

DISTANCE. = the distance between a and b. |a − b|

DEFINITION. The domain of a function (real values, realvariables) is the set of numbers for which it is defined.

} Use ( )’s with <, with >, around +5, and where f(x) is

undefined. } Use [ ]’s when f(x) is defined and the inequality is < or > .

(a,b) is the point on the plane whose x-coordinate is a andy-coordinate is b.

DEFINITION. The graph of an equation is the set of all pointswhich satisfy the equation.

DISTANCE FORMULA. The distance d between points (x1, y1) and

(x2, y2) is d = (x1 − x2)2 + (y1 − y2)2

CIRCLE EQUATION. The circle of radius r with center (a,b) has

equation: . (x − a)2 + (y − b)2 = r2

COMPLETING THE SQUARE. To make a perfect square, addx2 + ax

. . (a2 )2 x2 + ax + (

a2 )2 = (x +

a2 )2

FACTS. For positive a, . For all a, .a = ( a )2 a2 = |a|

DEFINTION. An x-intercept is the x-coordinate of a point wherethe graph crosses the x-axis. A y-intercept is they-coordinate of a point where the graph crosses the y-axis.

DEFINITION. The slope of a line is the ratio of the vertical(height) change in y over a horizontal change in x.

THEOREM.} The slope of the line through (x1,y1) and (x2,y2):

m =y2−y1

x2−x1

} The equation of the line through (x1,y1) with slope m:

y − y1 = m(x − x1)

} The equation of the line with slope m and y-intercept b:

y = mx + b} The slope of a horizontal line is 0.

} The slope of a vertical line is undefined.

} The equation of the horizontal line through (a,b) is y=b.

} The equation of the vertical line through (a,b) is x=a.

LINE EQUATION FORMAT. Line equations must be in the one ofthese four forms:

y = mx+b, y = mx, y = b, x = a.

THEOREM. If , a > 0 x2 − a = (x − a )(x + a )

DIVISION LAW. If p(x)/d(x) has quotient q(x) and remainder r(x)then p(x)

d(x) = q(x) +r(x)

d(x) and thus p(x) = d(x)q(x) + r(x)

X-INTERCEPT. a is a root or zero of p(x) iff p(a)=0.

THEOREM. a is a root of p(x) iff (x-a) is a factor of p(x).

DEFINITION. For sets A and B, a function from A to B is a rule

which assigns a value in B to each x in A. The domainf(x)

of f is A; the range of f is the set of all possible values . f(x)

DEFINITION. The graph of a function f is the set of all points(x,y) with y = f (x). The height y = the function value f (x).

FACT. A curve is the graph of a function iff no vertical lineintersects it more than once.

DEFINITION. f increases on an interval if the value f (x)increases as x moves from left to right in the interval. f decreases if f (x) decreases as x moves from the left to theright. c is a turning point of f if f increases on one side of cand decreases on the other. f (a) = c is a maximum value iff c is > all other values. f (a) = c is a minimum value iff c is < all other values.

Know these graphs. 1, x, x2, x3, 1/x.

1/x2, , |x|, .x 1 − x2

THEOREM. Translating a graph f (x) or reflecting it across anaxis, changes the function as follows:

f (-x)-f (x)f (x−1)f (x+1)f (x)−1f (x)+1

reflect iny-axis

reflect inx-axis*right 1*left 1down 1 up 1 unit

* Horizontal changes are the opposite of what one wouldexpect.

DEFINITION. For functions f and g, define f +g, f-g, f.g, f /g by (f+g)(x) = f(x)+g(x), Note: (f+g)(x) is not (f+g).(x)(f−g)(x) = f(x)-g(x), The first is function application.(fg)(x)= f(x)g(x), The second is multiplication.(f/g)(x) = f(x)/g(x).

DEFINITION. For functions f and g, define fοg, the composition

of f and g, by(fοg)(x) = f(g(x))

f is the outer function; g is the inner function.

Review of Theorems and Formulas for Exam 1

Math 140 Practice Exam 2 /100Exam 2. Lectures 7-12. No calculators. Understanding is not

enough; you must be proficient enough to complete the exam in 50minutes.

Know the area and volume formulas for triangles, rectangles,circles, boxes, cans (including curved surface area).

1(12). (a) , find . f1x x3

x31 f x (b) , find . fx x2 2x for x 1 f 1x

2(4). The given lines are the main diagonal and the graphof . Graph if it exists. If not, write “not 1-1”.f x f 1x

3(10). Graph and on the graph mark the vertex and giveboth coordinates. Also mark the intercepts.

y 3x2 6x

In 4 and 5, Picture: Draw the picture. On the picture indicate your variables. Given: Write the equations which relate the variables. Answer: Solve for the wanted quantities.

4(12). A square is inscribed inside a circle. Express thearea A of the circle as a function of the width x of thesquare.

5(12). A 6’ man stands x feet away from a 10’ high lamp. Express the length s of his shadow as a function of x.

6(22). Graph. Mark the intercepts and the asymptotes.

(a) . (b) .y 2x x2

x 12 y x 12

2x x2

7(2). Estimate as a power of 10. 230

8(4). Solve for t: 32t3 3

9(4). Simplify to a rational number.(a) (b) log21/ 2 lne e

10(8). Graph. Mark the intercepts and asymptotes and listthe domain and range. y 1 ln2 x

11(10). Find the exact answer. No decimals.(a) Solve for w: . Use an appropriate logarithm.32w5 10

(b) Solve using natural logarithms: .3x 2x1

Answers

1a. 1b. f x 3x

1 x f 1x 1 1 x

2. 3.

(-1,3)

-2 0

4. Picture:

x

xr

r

Given: A r2 x2 x2 2r2

Answer: A x2

2

5. Picture:

10 6

x s

Given: 10x s 6

sAnswer: s 3x

2

6a. 6b.

02

x=1 vert asym

y=-1 hor asym

1y=-1 h.a.

x=2 v.a.x=0 v.a.

7. 109

8. t = 7/4

9(a). -½ (b). 3/2

10. Hor. asymp.: none. Vert. asymp.: .x 2

11a. w = log3(10)/10 11b. x ln2/ ln3/2

`Know volume, area, and circumference formulas (theyare inside the front cover of the text).

All exams will consist solely of homework type problems. All numeric answers must be exact, no decimals, no mixed


2

DEFINITION. f -1, the inverse of f , is the function, if any, such that, f (f -1(x)) = x when f -1(x) is defined and f -1(f (x)) = x when f (x) is defined.

THEOREM. The graph of y = f -1(x) is the reflection of the graph of y = f(x) across the major diagonal y = x.

DEFINITION. f is 1-1 (“one-to-one”) iff x = y implies f (x) = f (y).

THEOREM. f has an inverse iff f is 1-1 iff no horizontal lineintersects its graph more than once.

DEFINITION. A quadratic function is a degree-2 polynomial

with a=0. y = ax2 + bx + c

The graph is a parabola.} If a>0, the horns point up.

} If a<0, the horns point down.

} If |a|>1, the parabola is narrower than y = x2.

} If |a|<1, the parabola is wider than y = x2.

THEOREM. Every quadratic function may be written in the form:

y = a(x − x0)2 + y0

where (x0, y0) is the vertex (nose) of the parabola. For an expanded polynomial with axn the term of highestaxn + ... + c

degree: axn is the leading term, a is the leading coefficient, and nis the degree. The y-intercept is the constant term c.

y=x y=-x

y=xy=x

y=x 2 2

43 3

y=-x

} For large x (near +5), graph looks like the leading term axn.

} As x goes to 5, y goes to +5 if a>0, to -5 if a<0.

} Graphs of odd degree go to +5 in one direction, -5 in the other,

like y = x3, y = -x3. } Graphs of even degree either go to +5 in both directions or to -5

in both directions, like y=x2, y=-x2.} At roots of degree 1, the graph crosses x-axis like y=x or y=-x.

} At roots of odd degree n>1, the graph crosses the x-axis like y=x3

or y=-x3

} At roots of even degree, the graph touches but doesn't cross the

x-axis, like y=x2 or y=-x2

DEFINTION. A ratio of two polynomial functions is a rational function.

It is reduced if the top and bottom have no common factors.In the graphs below, x = 0 (the y-axis) is the vertical asymptote, y =

0 (the x-axis) is the horizontal asymptote.

y=1/x

y= -1/x y=-1/x

y=1/x

y= -1/x

y=1/x 2

2 3

3

y=0

y=0

x=0

x=0

x=0

y=0

x=0

y=0

x=0

y=0

y=0

x=0vert

vert

hor

hor

For odd degree vertical asymptotes, one side goes to +5, the other to-5.

For even degree vertical asymptotes, both sides go to +5 or both goto -5.

For a reduced rational function: } x-intercepts (roots) occur where the top is 0.

If the root has degree n, the x-intercept looks like that of y = xn or y= -xn .} If the bottom is 0 at a, then x=a is a vertical asymptote.

If the factor has degree n, the vertical asymptote looks like that of y=1/xn or y = −1/xn .} As xƒ+5, the graph resembles the graph of the leading term

which is either a constant b or of the form a/xn or axn. (1) If a constant b, then y = b is a horizontal asymptote.

(2) If it is a/xn, then y = 0 is a horizontal asymptote. (3) If it is axn, there is no horizontal asymptote.

DEFINITION. An exponential function is of the form y = bx with thebase b>0. b0 = 1, b1 = b, b2 = b.b, ... b-n = 1/bn

b1/n = , the nth root of b bp/q = bp.(1/q) = (bp)1/q = (b1/q)p n b

EXPONENT RULES Examples

(bn)m = bnm (52)3 = 56

bnbm = bn+m 5254 = 56

, bn

bm = bn−m 53

57 = 5−4 =154

57

53 = 54

(ab)n = anbn 2535 = 65

(ab )n =

an

bn (23 )5 =

25

35

PROPERTY. If b = 1, . Hence is 1-1. bx = by g x = y bx

means a is approximately equal to b.a l b

Fact: , i.e., 210 is approximately equal to 103. 210 l 103

The graph of y = 1x is the horizontal line y = 1. Otherwise, the graph of y = bx } has y-intercept 1 but no x-intercept,

} it goes to 5 in one direction,

} it has the horizontal asymptote y = 0 in the other.

For b >1, the graph of bx is like the graph of 2x as below.

For 0 <b <1, the graph is like (½)x.

y=2x

xy=(1/2)

1

y=0 hor asymp

DEFINITION. ex is the unique exponential function bx whose the tangent

at (0,1) has slope 1. Fact: e l 2.7

DEFINITION. , the log of x to the base b, is the inverse of thelogb(x)

exponential function bx.

ln(x), the natural logarithm, = =the inverse of ex.loge(x)

Inverses act in opposite directions and inverses cancel. τ

iff . y = ln(x) iff .y = logb(x) by = x ey = x

. ln(ex) = x. eln(x)=x.logb(bx) = x. b logb(x) = x

If the exponent base is b instead of e, ln and bx don’t completelycancel: the bottom left equation becomes:

. The exponent comes down to the outside.ln(bx) = x ln b

FACT. e0 = 1ˆ 0 = ln(1). e1 = e ˆ 1 = ln(e)

Know area, circumference, volume formulas for triangles,

rectangles, boxes and cans. See inside front cover of the text.


Math 140 Practice Exam 3 /100Exam 3. Lect. 13-18. No calculators. Use the provided scratch paper.1(2). Combine into a single logarithm: 2 ln 1

x 3 lnx 12(3). Write as a sum / difference / multiple of the simplest

possible logarithms. logb

3 xx1

3(6). Find the one valid solution x. lnx lnx 1 ln2

4(6). Solve for x. ln2x 1 lnx 1

5(4). Express in terms of natural logarithms: log1014

6(18). Initially, 200 bacteria are present in a colony. Eighthours later there are 500.

(a) Determine the growth constant k.

(b) What is the population two hours after the start?

(c) How long did it take for the population to double?

7(12). The half-life of a radioactive substance is 9 years.Initially a sample has 20 grams. How many grams remain after 8 years?

8(2). Convert 300o to radian measure using p.

9(2). Find the radian measure of an angle which interceptsa 10 inch arc on a circle of radius 36 inches.

10(4). A point rotates around a circle of radius 3 feet at 20revolutions/minute.

(a) Find its angular speed. Exact radian answer using p.

(b) Find its linear speed. Exact answer using p.

11(10). (a) cos q = -4/5 and p < q < 3p/2. Find sin q.

(b) The side opposite angle of a right triangle is 6 andthe hypotenuse is x2, find . tan2 and tan 2

(c) The two legs of a right triangle are 2 and 3. What isthe cosine of the smallest angle? Hint: the smallest angle isopposite the smallest side.

12(4). Prove: tan2A 1 sec2A

13(8). Prove: cos2 sin2 1tan21tan2

14(6).(a) Find cot(11p/3). Give the exact answer.

(b) Rewrite with a reference angle in [0, p/2].cos 165

Don't give the decimal or exact answer.

15(7). Simplify to 3 symbols: cot2tsec2t1sec2ttan2t1

16(6). Simplify to 5 or 7 symbols: tantcost cost

Answers

1. lnx13

x2

2. 13 logbx logbx 1

3. x = 2

4. x = e/(e-2)

5. ln(14)/ln(10)

6.(a) k 1/8 ln5/2

(b) N(2) = 200e1/4 ln5/2

(c) 8 ln2/ ln5/2

7. N(8) = 20e8/9 ln1/2

8. 5p/3

9. 5/18 radians

10. (a) 40p radians/minute

(b) 120p ft/min

11.(a) -3/5

(b) , 6/ x4 36 x4 36 /6

(c) 3/ 13

12. iff sin2Acos2A 1 1

cos2A iff sin2A cos2A 1 iff true.

13. 1tan21tan2

1sin2/ cos21sin2/ cos2

cos2sin2cos2sin2

cos2sin21 cos2 sin2

14.(a) -1/l3

(b) cos 5

15. 1/2

16. 2 sin t

All exams will consist solely of homework type problems. All numeric answers must be exact, no decimals, no mixed


2Most properties of logarithms are the same as for exponentiation but

with

( _ ) n;x1exchanged with

n( _ )-+0the symbols

bnx bxnlogbxn = n logbx

bxy bx/bylogbx/y = logbx - logby

bxy bxbylogbxy = logbx + logby

b1 blogbb = 1

b0 1logb1 = 0

EXPONENT PROPERTIESLOG PROPERTIES

When b = e, logb(b) = 1 becomes: ln(e) = 1

CHANGE OF BASE FORMULA. .logax logbx

logbaBASE e FORM LEMMA. Every exponential function can be written in the

form

. Nt N0ekt

N0 = N(0) is the initial amount. k, the coefficient of t, is the growth constant. If k > 0, N(t) measures exponential growth. If k < 0, N(t) measures exponential decay.

FACT: Every a > 0 can be written as a power of e:

. Thus 2 = .a elna eln2

DEFINITION. Suppose the vertex of an angle is at the center of a circleof radius r. Let s be the length of the arc the angle intercepts on thecircle. Then

q = s/r is the radian measure of the angle. For unit circles, the radius r =

1 and radian measure equals arc length: q = s.Clockwise angles ?are negative. CONVERSION FORMULAS. 180o = p radians. Thus

1o = p/180 radians and 1 radian = 180/p degrees. DEFINITION. If an object travels a distance d in time t, its linear speed

is d/t. If an object rotates through an angle q in time t, its rotational

speed is w = q/t. THEOREM. If a point rotates around a circle of radius r with rotational

speed w, then its linear speed is wr. The unit circle has radius one and center (0,0). There are four

quadrants I, II, III, IV as pictured. An angle is in standard position if its vertex is at the origin (0,0) and

its initial side is on the positive x-axis. The other side of the angleis the terminal side.

(x,y)

qI

II

III IV

x

DEFINITION. Given an angle of radian measure q in standard position,let (x,y) be the coordinates of the intersection of the terminal sidewith the unit circle.

1q y = sinq

x=cosq

(x,y)

The trigonometric functions of q are:

cot q = cos q / sin qtan q = sin q / cos q

sec q = 1/cos qcos q = x

csc q = 1/ sin qsin q = y

FACTS. Know the sin and cos (and hence tan, cot, sec and csc) of: 0, p/6, p/4, p/3, p/2.

0

p/6=30

p/4=45p/3=60

p/2=90oo

o

o

o

1

(0,1)

(1,0)

l

l l

( 3/2,1/2)

(1/ 2, 1/ 2)

(1/ 2, 3/ 2) l

sin(0) = 0 cos(0) = 1 tan(0) = 0

sin(p/6) = 1/2 cos(p/6) = tan(p/6) = 1/l33 /2sin(p/4) = cos(p/4) = * tan(p/4) = 11/ 2 1/ 2sin(p/3) = l3/2 cos(p/3) = 1/2 tan(p/3) = l3sin(p/2) = 1 cos(p/2) = 0 tan(p/2) = undef.

* is also ok. 2 /2DEFINITION. For any angle in standard position (vertex at the origin,

initial side on the positive x-axis), its reference angle is the acutepositive angle (in [0,p/2]) between the x-axis and the angle’sterminal side.

THEOREM. The sin and cos of q equals the sin and cos of its referenceangle except for a possible change of sign.

NOTATION. sin q means sin(q); sin2 q means (sin q)2.THEOREM. Since sin q and cos q are the legs of a right triangle of

hypotenuse 1, sin2 q + cos2 q = 1 RECALL. sin2 q+cos2 q = 1

tan q = sin q/cos q cot q = cos q/sin qsec q = 1/cos q csc q = 1/sin q

DEFINITION. An identity is an equation which is always true.THEOREM. For any right triangle with an acute angle q:

sin q=side opposite/hypotenuse* csc q=hypotenuse/side oppositecos q=side adjacent/hypotenuse sec q=hypotenuse/side adjacenttan q=side opposite/adjacent cot q=side adjacent/opposite

THEOREM. sin(p/2 - q) = cos q sin(90o - q) = cos q cos(p/2 - q) = sin q cos(90o - q) = sin q

DEFINITION. For any function f: f is even iff f (-x) = f (x). f is odd iff f(-x) = -f (x).

MINUS THEOREM. sin(-t) = -sin(t), cos(-t) = cos(t), tan(-t) = -tan(t).DEFINITION. A function f is periodic with period p iff f(x+p) = f(x)

for all x and p is the smallest such positive number. 2p THEOREM. sin(t+2p) = sin(t), cos(t+2p) = cos(t), sin(t+2kp) =

sin(t), cos(t+2kp) = cos(t), k an integer. p THEOREM. sin(t+p) = -sin(t), cos(t+p) = -cos(t), tan(t+p) = tan(t).PYTHAGOREAN IDENTITIES. sin2t + cos2t = 1, tan2t +1 = sec2t, cot2t +1 =

csc2t.


Math 140 Practice Exam 4 Exam 4. Lectures 19-24. No calculators.Write angles in exact radians, no decimals, no degrees.1(10). For the graph below, find the amplitude, find the

period, and find an equation of the form y =+ Asin(Bx)or y = +Acos(Bx).

(3,2)

2(14). Graph .y 2 sinx

3(9). Graph over one period. List they tanx/2 x-intercepts and the vertical asymptotes which occur inthis period.

4(9). Graph . Draw the asymptotes as dottedy secx lines. Be prepared to graph cot and csc.

5(6). Simplify . sinx y cos x cosx y sinx

6(6). Simplify . tan/5tan/30

1tan/5 tan/30

7(12). . x sin/2, 2

(a) Find sin2

(b) Find sin/2

(c) Find tan/2

8(5). Write as a sum or difference of trig functions.

cosx 1 sinx 19(5). Find all solutions to . cos 3

2 0Two sets of solutions.

10(10). Find all solutions of . 2 sin2x 5cos x 4Two sets of solutions.

11(14). Find the exact value (no decimals) of (a)(3) (b)(3) cos1 3 /2 tan1 3

(c)(4) (d)(4) sin1sin 57 tanarcsin 3

5

Answers 1. amplitude = 2, period = 6, y 2 cos

3 x

2.

1 2

2

3

3. x-inter: 2p, vert. asymp: x = p, x = 3p.

3pp0

2p

4.

p/2-p/2 3p/2

5. siny

6. 1/ 3

7. (a) 4x 1 4x2

(b) 1 14x2

2

(c) 2x1 14x2

8. 12 sin2 sin2x

9. q = 5/6 2n, 5/6 2n

10. x 3 2n, x

3 2n

11. (a) 5p/6 (b) -p/3 (c)2p/7 (d) ¾

All exams will consist solely of homework type problems.

All numeric answers must be exact, no decimals, no mixed


2

Graph of sin(x).

π/2 π 3π/2 2π−π/2

1

−1

sin(x)

y=x

x-intercept: x = ..., -2π, -π, 0, ππππ, 2ππππ, 3π, ... .

Max value: 1 at x = ..., -3π/2, ππππ/2, 5π/2, ...

Min value: -1 at x = ..., -5π/2, -π/2, 3ππππ/2, ...

Period: 2π Amplitude: 1 (see definition below)

At x = 0, the line y = x is tangent to the graph of sin(x).

Graph of cos(x). Shift the above graph left π/2 units.

DEFINITION. For any periodic function f (x), the amplitude

of f is half the distance between the min value and themax value of f. Like periods, amplitudes are always positive.

THEOREM. For y = Asin(Bx) or y = Acos(Bx) with B > 0,} period = 2π/B,

} amplitude = |A|. Reflect the graph around the x-axis if A<0.

TRIGONOMETRIC ADDITION FORMULAS. sin(s ! t) = sin s cos t ! cos s sin t

cos(s + t) = cos s cos t − sin s sin t

cos(s − t) = cos s cos t + sin s sin t

tan(s + t) =tan s+tan t

1−tan s tan t

tan(s − t) =tan s−tan t

1+tan s tan t

DOUBLE-ANGLE FORMULAS. sin 2� = 2 sin � cos �cos 2� = cos2� − sin2

�

tan 2� =2 tan �

1−tan2�

HALF-ANGLE FORMULAS.

+ if θ/2 is in quadrant I or II.sin�

2 = !1−cos�

2

+ if θ/2 is in quadrant I or IV.cos�

2 = !1+cos�

2

tan�

2 =sin �

1+cos�

PRODUCT-TO-SUM FORMULA. sin x sin y =

12

[cos(x − y) − cos(x + y)]

cos x cos y =12[cos(x − y) + cos(x + y)]

sin x cos y =12[sin(x − y) + sin(x + y)]

DEFINITION. sin -1, cos

-1, tan -1 are the inverses of the above

restrictions of sin, cos, and tan. Thus

sin -1(x) = the θι[-π/2,π/2] such that sin(θ) = x,

cos -1(x) = the θι[0,π] such that cos(θ) = x,

tan -1(x) = the θι(-π/2,π/2) such that tan(θ) = x.

sin-1(x) and cos-1(x) have domain [-1,1], for x2[-1,1], theyare undefined. tan

-1(x) has domain (-5,5).

NOTATION. sin -1(x), cos

-1(x) and tan -1(x) are also written:

arcsin(x), arccos(x), and arctan(x).

Since inverses undo each other, we have

tan(tan -1

x) = x for any xtan-1(tan x) = x if xι(-π/2,π/2)

cos(cos-1 x) = x if xι[-1,1]cos

-1(cos x) = x if xι[0, π]

sin(sin-1 x) = x if xι[-1, 1]sin-1(sin x) = x if xι[-π/2, π/2]

THEOREM. For xι[-1,1],

cos(sin−1x) = 1 − x2

sin(cos−1x) = 1 − x2

In a triangle with hypotenuse 1 and side x, = angle opposite x,sin−1x

= angle adjacent to x.cos−1x

Note that sin−1x + cos−1x = �/2 (= 90o)


Math 140 Practice Exam 5

The final covers everything ∞ Lectures 1-30. No Gateway problems.

This practice exam covers the material since the last exam. About a

third of the exam will be on this material; the other exam problems

will be on the previous material (see previous practice exams). No

decimal answers; all answers must be exact. No calculators. If you must wear a baseball cap, wear it backwards.

1(6). Find the area (exact value) of a triangle with sides a = 2, b = 3, and an included angle ςC = 25o.

2(10). An antenna sits on top of a 1000 ft building. From a

point level with the base of the building, the buildinghas elevation 20o and the antenna has elevation 25o.How high is the top of the antenna above the ground?

A

ca

bC

B

Give exact answers, not decimal answers.

3(8). a = 3, c = 2, ςC = 60o, solve for all b.

4(9). a = 3, ςC = 60o, ςA = 50o, solve for b.

5(12). a = 3, b = 2, ςB = 30o, solve for ςC.

6(13). Find the perimeter of a pentagon inscribed in a unitcircle (exact answer using radians rather than degrees).

7(3). Convert the given polar coordinates to rectangularcoordinates:

(a) (2,�/4)

(b) (−1, 2�)

(c) Convert the given rectangular coordinates to polarcoordinates: (�,�)

8(6). (a) Convert the polar equation to a rectangularequation: r2 = cos 2�

(b) Convert the rectangular equation to a polar equation:

x2 − y2 = 1

9(14). Find all axes and focus or focal points and draw thegraph of 4y2 − x2 + 2x − 2 = 0

10(14). Find all axes and focus or focal points and drawthe graph of 4y2 + x2 − 2x = 0

Answers

1. .3 sin 25o

2. .1000 tan 25o

tan 20o ft

3. No solutions.

4. . b =3 sin 70o

sin 50o

5. 150o − sin

-1(¾) or sin -1(¾) − 30o.

6. or 5 . 5 2 − 2 cos 2�5 (sin

2�5 )/(sin

3�10 )

7.(a) , (b) , (c) . ( 2 , 2 ) (−1, 0) (� 2 ,�/4)

8.(a) . (x2 + y2)2 = x2 − y2

(b) .r2(cos2� − sin2�) = 1

9. 4y2 − x2 + 2x = 2 4y2 − (x − 2x) = 2

4y2 − (x2 − 2x + 1) = 1

vertical hyperbolay2

(1/2)2 −(x−1)2

12 = 1

a = 1/2, b = 1, c = a2 + b2 = 1/4 + 1 = 5 /2Shift: (0,¯ ¯ -½)(0,½¯ ¯ ¯ ¯ ¯ ¯ ) right 1, get major axis: (1,¯ ¯ -½)(1,½¯ ¯ ¯ ¯ ¯ ¯ )Shifting (-1,0)(1,0¯ ¯ ¯ ¯ ¯ ¯ ¯ ) right 1, get minor axis: (0,0)(2,0¯ ¯ ¯ ¯ ¯ ¯ ¯ )Shifting x = -1 up and right 1 gives directrix: x = 0.

Shift (0, + ) right 1 get foci: . 5

2 (1,− 5

2 ), (1,5

2 )

Now graph (see below left). The foci are ≈ (1,-1.11) and (1,1.11).

10. 4y2 + (x2 − 2x + 1) = 1, 4y2 + (x − 1)2 = 1

horizontal ellipse y2

1/4 + (x − 1)2 = 1,(x−1)2

12 +y2

(1/2)2 = 1

≈ .87 c = 1 −14 =

3

2

Shift: right 1 unit. Shift (-1,0)(1,0¯ ¯ ¯ ¯ ¯ ¯ ¯ ) right 1, get major axis: (0,0)(2,0¯ ¯ ¯ ¯ ¯ ¯ ¯ ). Shift (0,¯ ¯ ½)(0,½¯ ¯ ¯ ¯ ¯ ) right 1, get minor axis: (1,¯ ¯ -½)(1,½¯ ¯ ¯ ¯ ¯ ¯ ).

Shift (+ 0) right 1 get foci: . 3

2 , (1 −3

2 , 0), (1 +3

2 , 0)

Now graph (below right). The foci are ≈ (-.13,0) and (1.87,0).

(0,0)(0,0) (2,0)(2,0)

(-.13,0) (1.87,0)

(1,-1.11)

(1,1.11)

(1,1/2)(1,1/2)

(1,-1/2)(1,-1/2)

THEOREM. The area of a triangle with sides a and b and includedangle q is ½absin(q).

STRAIGHT ANGLE SUM FACT. The sum of a triangle’s 3 angles is astraight angle.

VA +VB +VC = p (=180o)CONVENTION. Assume side a is opposite angle A, side b is opposite

angle B and side c is opposite angle C.

SINE LAWS. sin Aa sin B

b sin Cc

asin A b

sin B csin C

COSINE LAWS. a2 b2 c2 2bccosAb2 a2 c2 2accosBc2 a2 b2 2abcosC

Suppose two sides and a nonincluded angle are known. When solving for the third side using the cosine law, you may get an

answer of the form . Then there is: s t} no solution if t <0 or both , s t 0} two solutions if t >0 and both , s t 0} one solution otherwise. When solving for a second angle using the sine law, you may get an

answer of the form q = sin-1t. There is: } no solution if t >1, } two solutions if t<1 and the given angle is adjacent to the larger

of the two sides (if one angle is q, the other p-q), } one solution otherwise.

DEFINITION. The polar coordinates (r, q) of a point P are its distance r(radius) from the origin and the angle q between the positivex-axis and the line from (0,0) to P.

The usual (x, y) are the rectangular coordinates.THEOREM. If a point has rectangular coordinates (x, y) and polar

coordinates (r, q), then: and x, y r cos, r sin

*. r, x2 y2 , arctan yx

* Since tan is periodic in p, q and arctan(y/x) may differ by amultiple of p. Add p if (x,y) is in quadrant II or III.

NEGATIVE r. is the point on the opposite side of ther, r,origin (0,0) as . r,

PARABOLASVERTICAL PARABOLA THEOREM. For p=0, the graph of is thex2 ky

vertical parabola with focus (0,p) and directrix y = -p where p =k/4. The axis is the y-axis; the vertex is (0,0).

HORIZONTAL PARABOLA THEOREM. For p=0, the graph of is ay2 kxhorizontal parabola with focus (p,0) and directrix x = -p where p =k/4. The axis is the x-axis, the vertex =(0,0).

(p,0) = focus

x=-pdirectrix

4px=y2

axis(0,0) vertex

focal-width point

ELLIPSES

focifoci

minor axis

major axis

vertices

center

THEOREM. Let a be half the major axis’s length, b be half the minoraxis’s length, and c be half the distance between the foci. Then

. t .a2 b2 c2 c a2 b2

THEOREM. For a>b>0, the graphs of and x2

a2 y2

b2 1 y2

a2 x2

b2 1are ellipses. The center is (0,0). The major axis has length 2a; theminor axis has length 2b; the distance between foci is 2c where

.c a2 b2

} For , the ellipse is horizontal, the foci are (-c,0)x2

a2 y2

b2 1and (c,0), the major axis is the line segment (-a,0)(¯ ¯ a,0)¯¯, the minoraxis is (0,¯¯-b)(0¯¯ ,b).

} For , the ellipse is vertical, the foci are (0,-c) andy2

a2 x2

b2 1(0,c), the major axis is the line segment (0,-¯ ¯ a)(0¯¯ ,a), the minor axisis (-b,0)(¯ ¯ b,0)¯¯.

HYPERBOLAS

--------

minor axis

-----------major axis

----verte

xvertex---- ---focal axis---focal pointfocal

point----

----asy

mptote

center


THEOREM. The graphs of and arex2

a2 y2

b2 1 y2

a2 x2

b2 1hyperbolas (a must be below the positive square). The center is(0,0). The major axis has length 2a; the minor axis has length 2b;and the distance between focal points is 2c where . c a2 b2

} For , the hyperbola is horizontal, x2

a2 y2

b2 1 the foci are (-c,0) and (c,0), the major axis is (-a,0)(¯ ¯ a,0), the minor axis is (0,¯ -b)(0,¯ ¯ b).

} For , the hyperbola is vertical, y2

a2 x2

b2 1 the foci are (0,-c) and (0,c), the major axis is (0,-¯¯a)(0,¯ ¯ a), the minor axis is (-b,0)(¯ ¯ b,0).

The final covers everything. These are the Theorems and Formulas not listed in earlier reviews .

Practice on the Hw 0 recommended-problem worksheet or the recommended textbook problems listed above.

1=1+1 has 5 symbols, (-5,0] has 6 symbols, has 8 symbols, has 6 symbols, has 5 symbols, has 3 symbols. 12 x 3 x2 2x 13

12 0 xDo not turn in scratch paper; put your work in the space provided.You get extra credit if you are the first to find an error in the lecture note packet sold at the Bookstore.Problems marked with a superscript E have notes in the note-packet section titled “Errors”.

1(1). Is the inequality true or false? p2 < 12

2(1). Draw the interval on the number line: (-4, 0] chk=4

3E(1). Rewrite without | |’s: |x-3| + |x-4| for x > 4Answer has 4 symbols, chk=9

4E(1). Rewrite without | |’s: |x+1| + 4|x+3| for x < -36 symbols, chk=9

5(1). Write using | |’s: The distance between x and 1 is less than ½.9 symbols, chk=4

6E(1). Write as an interval: {x: |x-4| < 4}5 symbols, chk=8

7E(1). Write as a union of two intervals: {x: |x+5| > 2}14 symbols, chk=10

Math 140 Hw 0 Name _________________________________ Score ____/ 7 Recommended. §1.7 84: 64-76 Odd ones have answers in the back of the text. Evens will be on computer homework.

See if you can do these "practice quiz" problems on your own in the time limit given. They will be covered in the lab sessions. Problems on the exams will be very similar to these problems and problems on the practice exams. You should be able to complete each quiz in the time listed after the score. You need this level of proficiency to complete an exam on time.“(5)” after the problem number means the corresponding final exam problem will probably have 5 points.

1(5). Rewrite without ’s: for xi[4, 6].2|4 x| |x 7|

x 4, 6 x 4 4 x 0 |4 x| 4 x x 4x 4, 6 x 6 x 7 6 7 0 |7 x| x 7 7 x2|4 x| |x 7| 2x 4 7 x 2x 8 7 x Answer: 3x 15

2(4). Write in interval notation: .x : |x 2| 3

|x 2| 3 3 x 2 3 5 x 1Answer: x 5, 1

3(5). Write in interval notation: .x : |x 1| 2

|x 1| 2 x 1 2 or 2 x 1 x 1 or 3 x Answer: x ,1 3,

Math 140 Discussion Lecture 0 Exam problems points/time 14pts,10mins

Rewrite each expression in a form that does not containabsolute values.

A. |x 3| |x 4| where x 3.

B. |x 3| |x 4| where 3 x 4.

C. |x 1| 4|x 3| where 52 x 3

2 .

Rewrite each statement using absolute values.D. The distance between x and 4 is 8.

E. The distance between x and 1 is ½.

F. The distance between x and 1 is at least ½.

G. The distance between y and -4 is less than 1.

H. The number y is less than three units from the origin.

Write the following in interval notation.I. |x| 4.

J. |x| 1.

K. |x 5| 3.

L. |x 3| 4.

M. |x 13 | 3

2 .

N. |x 5| 2.

AnswersA. 2x 7.B. 1.C. 3x 11.D. |x 4| 8.E. |x 1| 1

2 .F. |x 1| 1

2 .G. |y 4| 1.H. |y| 3.I. 4, 4.J. ,1 1,.K. .2,8L. 1, 7.M. 11

6 , 76 .

N. , 3 7,.

Math 140 Hw 0 Recommended problems, don't turn this in.

Rewrite each expression in a form that does not contain

absolute values.

C. |x + 1| + 4|x + 3| where −52 < x < −

32 .

First we need the following facts from Lecture 1:|w| = w if w> 0. Example, |3| = 3.|w| = -w if w < 0. Example, |-3| = -(-3) = 3. In -w, the minus sign cancels the minus sign in w.To get rid of the absolute value sign, you mustdetermine whether w is positive or negative.

Is positive or negative?|x + 1| ˆ . x < −

32 x + 1 < −

32 + 1 = −

12 < 0

τ is negative. x + 1 < 0Ητ .|x + 1| = −(x + 1)

Is positive or negative?|x + 3| ˆ ˆ −

52 < x −

52 + 3 < x + 3 0 <

12 < x + 3

τ is positive.x + 3τ |x + 3| = x + 3τ |x + 1| + 4|x + 3| = −(x + 1) + 4(x + 3)

−x − 1 + 4x + 12 = 3x + 11Answer: 3x + 11

Each of problems 51 - 61 can be solved by v using key numbers,

v restating the absolute value as a distance, or

v restating the problem in terms of inequalities.

Use any method you wish.

Write the following in interval notation.

L. |x − 3| [ 4.We’ll solve this one using key numbers. First solve the

equality. The solutions to the equality are “keynumbers” which divide the line into "key intervals".Plug in numbers from each interval to determine whichintervals satisfy the inequality.First solve the equality .|x − 3| = 4In general, iff w = a or w = -a. |w| = a

For example, iff w = 3 or w = -3.|w| = 3τ iff or |x − 3| = 4 x − 3 = 4 x − 3 = −4iff or .x = 7 x = −1The key numbers: -1, 7. The key intervals: (-5, -1], [-1,7], [7,5).To find out which intervals satisfy the inequality, pick anumber from each interval and see if it makes theinequality true.

is false for x = -2 ι (-5,-1] 3|x − 3| [ 4 is true for x = 0 ι [-1,7] ,|x − 3| [ 4 is false for x = 10 ι [7,5) 3|x − 3| [ 4

τAnswer: iff x ι [-1, 7].|x − 3| [ 4

Note that we use “[ ]” rather than “( )” since theinequality < means the endpoints are included. If the

inequality have been <, the endpoints would beexcluded and the answer would have been (-1, 7)instead of [-1,7].

M. |x +13 | <

32 .

We’ll solve this problem by restating it in terms of adistance.

is the distance between a and b. |a − b|

τ = the distance between x and|x +13 | = |x − (−

13 )|

-ƒ.τ

|x +13 | <

32

iff the distance between x and -1/3 is less than 3/2 iff x is between -1/3 - 3/2 and -1/3 + 3/2.

iff −13 −

32 < x < −

13 +

32

iff − 116 < x <

76

Answer: iff x ι .|x +13 | <

32 (−

116 ,

76 )

If the problem had been , the answer would|x +13 | >

32

have been x ι (−∞, −116 ) 4 (

76 ,∞)

N. |x − 5| m 2.We’ll solve this one using inequalities. You can solveinequalities using the same rules you use for equalitieswith the following exceptions.When you multiply or divide by a negative number, youmust change the direction of the inequality.You may not multipy or divide by a number if you donot know its sign (positive or negative).

In general,(1) iff iff .|w| [ a −a [ w and w [ a −a [ w [ a

(2) iff . |w| m a w [ −a or a [ w

In case (2) we have an “or” rather than the “and” ofcase (1).τ

|x − 5| m 2 iff or (add 5 to both side)(x − 5) [ −2 2 [ (x − 5)

iff or x [ 3 7 [ x

Answer: x ι (-5, 3]υ[7,5)

If the problem had been , the solution would|x − 5| < 2have been

iff iff |x − 5| < 2 −2 < x − 5 < 2 3 < x < 7And the answer would have been: x ι (3, 7).

Math 140 Hw 0 Worked examples of selected recommended problems.

Math 140 Lecture 1 Intervals, absolute value, domains, polynomials INTERVAL NOTATION.

xi[1, 3] iff 1 < x < 3 solid dot includes

xi(1, 3) iff 1 < x < 3 hollow dot excludes

xi iff 1 < x 1,xi iff x <-1 or 1 < x,1 1,

ABSOLUTE VALUE.|3| = 3, |-3| = 3, |0| = 0.

Negating a negative makes it positive, -(-3)=3. In general|x| = x if x > 0. |x| = -x if x < 0. Alternatively, | x| = its distance to 0.

`Rewrite without | |’s: = since | 2 | 2 2 2 0 = | 2 | 2 2

`Rewrite without | |’s: for xi(3, 5). |x 2| 2|x 6| ˆ ˆ .3 x 5 3 2 x 2 5 2 1 x 2 |x 2| x 2 ˆ ˆ3 x 5 3 6 x 6 5 6 x 6 1

.|x 6| x 6 x 6t .|x 2| 2|x 6| x 2 2x 6 x 10DISTANCE. = the distance between a and b. |a b|

NOTE. .|x| 3 iff3 x 3 iff x 3, 3 .|x| 3 iff x 3 or 3 x iff x ,3 3,

`Write as an interval: x : |x 7| 3 |x 7| 3 iff 3 x 7 3 iff 10 x 4.

Answer: = (-10, -4). x : |x 7| 3`Write as a union of two intervals: x : |x 5| 2

|x 5| 2 iff x 5 2 or 2 x 5 . iff x 3 or 7 x

Answer: x : |x 5| 2 , 3 7,.DEFINITION. The domain of a function (real values, real

variables) is the set of numbers for which it is defined.

`Where is not defined? Its domain is x = 2, 3.x1x2x3

`Where is not defined? Its domain is x > 2.x 2`Solve . Get radical on one side. Square.3x 4 x 0

Check validity: put solutions into the original equation., , , 3x 4 x 3x 4 x2 x2 3x 4 0

, .x 1x 4 0 x 1, 4 , is invalid.31 4 1 1 1 0 x 1

, is valid.34 4 4 16 4 0 x 4

Know what the following terms mean: polynomial,degree, coefficients, quadratic function, expanded form,factored form mean. Know how to factor a polynomialand use the quadratic formula.

x 12 x2 2x 1 is the factored form, x 12

is the expanded form. x2 2x 1

Solving inequalitiesAs with equalities, you may add or subtract anything

from both sides. You may multiply or divide both sides by a positive number. If you multiply or divide by a negative number, you must change the direction of the inequality (multiplying 5>3 by -1 gives -5 < -3). If you don’t know if a number is positive or negative, don’t multiply or divide by it. } (divide by 2)2x 6 x 3} (divide by -2; change sign direction)2x 6 x 3} doesn’t imply (don’t know if x > 0 or x < 0)x 1

x x2 1

To solve problems like , use the key-numberx 1x

method:ƒ Rewrite with 0 on the right. Factor the f(x) on the left.

fx 0, fx 0, fx 0, or fx 0.ƒ Find key numbers x where f (x) = 0 or f (x) is undefined. ƒ On the key intervals before, between, and after key

numbers, f (x) is either > 0 or < 0. To find out which,evaluate f (x) at some point inside the interval. Youdon’t have to find the value, just the sign.

} Use ( )’s with <, with >, around +5, and where f (x)is undefined.

} Use [ ]’s if f (x) is defined and the inequality is< or > .

`Solve for x: . Write as a union of two intervals. x 1x

x 1x

iff x 1x 0

iff x21x 0

iff t x1x1

x 0 fx x1x1x

Key numbers: x = -1, 0, 1. Key intervals: (-5, -1), (-1, 0), (0, 1), (1, 5).

f (-2)= -, f (-½)= +, f (½)= -, f (2)= +. We picked -2i(-5,-1), -½i(-1,0), ½i(0,1), 2i(1,5).

tx < 1/x iff f(x) < 0 iff Answer: xi(-5, -1)u(0, 1)

`Solve for x: . Put answer in interval notation. x2 x 6 0First, factor the function. f (x) = . x2 x 6 x 3x 2f (x) = 0 iff x = -3, 2. These are the key numbers. The 3 key intervals: (-5, -3], [-3, 2], [2,5). Picking -4, 0, 3: f (-4) > 0, f (0) < 0, f (3) > 0. t iff x2 x 6 0Answer: xi[-3, 2]

`Write as a union of two intervals: . x1x2x2 0

iff . x1x2x2 0 x1

x1x2 0Key numbers: -1, 1, 2. Undefined at -1, 2 Key intervals: (-5,-1), (-1,1], [1,2), (2,5) f (-2) < 0, f (0) > 0, f (3/2) < 0, f (3) > 0t Answer: xi(-1, 1]u(2, 5)

oo o

Try to do practice quizzes in the time indicated at the top right corner. These problems are liked similarlynumbered problems on exams and practice exams.Classwork 1 is hard, work on it in advance. Your TA will give you an extra five minutes time for group work.

Practice 3(4). Find all real solutions x. Write Q if there are none. 3 2x x 0 3 2x x 0

Get radical by itself on the right side, everything else on the left. x 3 2x Since this is rather than , we must check for validity at the end. x2 3 2x

E.g. but doesn’t imply since we could have x 3 x2 3 x2 3 x 3 x 3 . x2 2x 3 0 x 1x 3 0 x 1, x 3Now check for validity.Stick the proposed solutions in the original equation. Check that the square roots are defined and that equalityholds.For , x 3

3 2x x 0 3 23 3 0 9 3 0

false 3 3 0 t is invalidx 3For x 1

3 2x x 0 3 21 1 0

true 1 1 0 Answer: x 1

Practice 4(7). Solve for x. Write the answer as a union of two intervals. x 2 3x

x2 3x

Get everything on the left side, 0 on the right. x2 3x 0

x22x3x 0

x1x3x 0

Key numbers: x = -1, 0, 3 Plot them of the real line. (two solid, one hollow dot).

-1 0 3

Key intervals: ,1, 1,0, 0,3, 3,Key values: Alternate shortcut method:

fx x1x3x 0 f4

, The factors for the key numbers all have odd degree 1.f2

Hence, working backwards from which is +, we getf 12

3,

, .f1 on ,1, + on 1, 0, - on 0, 3, + on 3,

f4

tfx 0 Answer: x ,1 0, 3

Math 140 Discussion Lecture 1 Exam problems points/time 11pts 9 minutes

Classwork 2 is hard, work on it before class.Problem 3. Find all real solutions x. Write Q if there are none. 5 4x x 0Get radical by itself on the right side, everything else on the left. Square and solve. __/2Then check for validity; show which is valid/invalid. Stick each solution into the original equation. If equality holds, the solution is valid; if not, it is invalid. The one correct answer is a single positive or single negative digit.Show why one works and the other does not. Show your work in the space below, not on a separate sheet of paper.

Problem 4. Solve for x. __/2x 2x 1

Get everything on the left side, 0 on the right. Combine fractions, factor. E.g. like or x 4x 5x 2 0

.x 3x 4

x 0List the 3 key numbers (where left side is 0 or undefined) and plot them on the line. Should be two solid circles, one hollow circle.

In at least one of the four intervals, pick a point and calculate the sign of its key value - show your work. On the number line, mark each key interval with “+” or “-”. The answer is a union of two intervals. 13 symbols after “xi”. Should be 2 brackets, 2 parentheses. chk=4

Math 140 Classwork 1 Score ____/4No credit unless you work with a group of three or more people. You get the lowest score in your group.

Simplify fully to a formula with fewer symbols. has 6 symbols.13 x

yCircle the 4 formulas (excluding the example) which can not be simplified. Do not write in the gray area. Go to the top menu in

www.math.hawaii.edu/140, read “Common Errors”.Example: 6x2

3x3 2x

Example: x2 1

1. a b

fully Don’t write insimplified gray areas.

2. x2 25 fully simplified

3. 25b

5 b

4. 25b4

5b2

5. 25a2 b22

5a2 b2

6. -- 5 symbols --9x16

3 x4

7. 3x53y7

fully simplified

8. 3x3y5

fully simplified

9. 3x

3x6

x

x2

10. -- 3 symbols --3x2y3y2x

xy

11. -- single exponent, e.g., --67

37 59

27

12. -- single digit --3837

37

213. -- 7 symbols, is wrong. -- x2

x3 2x3x

2xx3

14. -- 5 symbols -- x3x

xx3

Evaluate the following functions, simplify to the given # of symbols.Example. gx x2 1, gx h x h2 1 ggx x2 12 1 x4 2x2

Example. hx 1 2x, hy2 1 2y2

15. -- 2 symbols --f x x2, f x2

x4

16. -- 6 symbols --f x x2, f x h x h2

17. -- 3 symbols --gx x 3, ggx x 6

18. -- 1 symbol --gx x 3, gx h gx h

19. -- 5 symbols, not --f x 1x , f x h 1

x h

1xh

Write as a single reduced fraction. No fractions in numerator or denominator.

20. -- 5 symbols --a2b2c

a4bc

21. -- 11 symbols --4b 22 4

a

2aabab2b

Math 140- ___ Take-home Quiz 1 Name ____________________________ Score ____/21 Write your section number. Write your answers in the white space provided, don’t write in gray areas; never hand in scratch paper.

1=1+1 has 5 symbols, (-5,0] has 6 symbols, has 8. 12 x 3

chk = sum of digits. For chk=3+1+8=12. ,3 1,8Do not turn in scratch paper; put your work in the space provided.For superscript E problems, see notes in the section titled “Errors”.Recommended (don’t turn in) textbook problems: §1.5 page

55 75-78, 81-83, 94. Turn in only the problems below.Find all real solutions x. Write Q if there are none. One problem has no solutions, one has one, one has two.

1. x(x-1) = -4

2E. l3x+1¯¯ = 4One solution, a positive integer, chk=5

3. 5x 2 2x 1

5 0Two solutions: one integral, the other a rational, chk=14

§1.7 page 84 problems 47-60. Odds have answers in the back.Solve for x. Write answer in interval notation. Two pointseach. A number is integral if it is an integer (positive or negative).

4E(2). x 42x 5 0

Single interval, 8 symbols, chk=11

5(2). x2 1x2 8x 15 0

Union of three intervals, 21 symbols, integral endpoints, chk=10

In 6, A, rewrite with 0 on the right. List the key numbers.Solve for x. Write the answer in interval notation.6(2). 2x

x 2 3Union of two intervals with 12 symbols, integral endpoints chk=8

7(2). x 10x 1 4

Union of two intervals with 12 symbols, integral endpoints chk=8

Math 140 Hw 1 Name ____________________________________ Score ____/ 11

Find all real solutions x. Write Q if there are no solutions.A. 3x 22 3x 2

B. 3x2 4x 3 0

C. x2 24

D. xx 1 1

E. 12 x2 x 1

3 0

F. (a) x 8 4

(b) x 8 2x 1

G. 1 3x 2

H. 3x5

4x 2

I. 1 x 26x1 0

J. 3x26x3x1x2x3

52xx25x6 0

K. 2xx21

1x3 0

Solve for x. Write your answer in interval notation. L. x 42x2 6x 1 0

M. x3 2x2 x 2 0

N. x1x1 0

O. 2x32x 0

P. x28x9x 0

Q. 2x35x27x3x27x4 0

R. xx1 1

S 32x32x

1x

T. 1 1x 1

x1

AnswersA. B. x 2

3 , 1 x 2 13 /3C. x 2 6D. x 1 5 /2E. x 3 15 /3F(a). (b) Qx 24G. x 1H. I. x 5

2 ,4 x 13 , 1

2

J. K. x 1 x 3 2 2

L. ,3 11

2 3 11

2 , 4M. 2,1 1,N. 1, 1O. , 3

2 2,P. ,1 0, 9Q. 7

2 , 43 1, 0 1,

R. ,1S. 3

2 , 0T. ,1 0,

Math 140 Hw 1 Recommended problems, don't turn in.

Find all real solutions x. Write Q if no solution.A. 3x 22 3x 2

Don’t divide both sides by , you could be3x 2dividing by 0. First get everything on the left, 0 on theright. Then factor or use the quadratic formula.3x 22 3x 2 03x 23x 2 1 03x 23x 3 0

. 33x 2x 1 0Answer: x 2

3 , 1

D. xx 1 1xx 1 1 0

Doesn’t factor, so use quadratic formula.x2 x 1 0

Answer: x b b24ac

2a 1 1411

2 1 5

2

F. (a) x 8 4Square both sides (when you do this, you mayintroduce invalid answers. Delete any invalid answersat the end.)

x 8 16x 24Sticking x back into gives x 8 4 24 8 4which is true.Answer: 24.(b) x 8 2x 1x 8 2x 12

x 8 4x2 4x 1x 8 4x2 4x 1 04x2 3x 9 0

Doesn’t factor, so use quadratic formula.4x2 3x 9 0

.x 3 9449

24 3 135

8

Answer: no solutions: Q.

I. 1 x 26x1 0

6x11x6x1 2

6x1 06x25x12

6x1 06x25x1

6x1 06x25x1

6x1 0

A rational is zero iff the top is zero.3x12x1

6x1 0Anwer: x = ƒ, ½

J. 3x26x3x1x2x3

52xx25x6 0

First factor as much as possible and simplify.3x26x3

x1x2x3 52x

x3x2 03x26x3

x1x2x3 x152x

x1x2x3 03x26x32x23x5

x1x2x3 0x23x2

x1x2x3 0

x1x2x1x2x3 0

x1x1x3 0

Answer: x = 1.

Solve for x. Write your answer in interval notation. L. x 42x2 6x 1 0

is zero at 4.x 4 is zero at, use the quadratic formula,2x2 6x 1

x 6 36421

22 6 444 62 11

4 3 112

Key numbers: x = 3 11

2 ,3 11

2 , 4Key intervals:

,3 11

2 , 3 112 ,

3 112 , 3 11

2 , 4, 4,Pick a point from each interval and evaluate if

is >0 or <0.fx x 42x2 6x 1f10 0, f0 4 0, f7/2 0, f10 0

Answer: ,3 11

2 3 112 , 4

O. 2x32x 0

iff 2 x 0 x 2 iff 3 2x 0 x 3

2

Key numbers: x 32 , 2

Key intervals: , 32 ,

32 , 2, 2,

Exclude 3/2 because is undefined there. Include 2fxbecause is defined and the inequality is >.fxPick a point from each interval and evaluate if

is >0 or <0.fx 2x32x

f0 0, f 74 0, f10 0

Answer: , 32 2,

S. 32x32x

1x

32x32x

1x 0

32xx32xx

32x32xx 0

3x2x232x32xx 0

2x2x332xx 0

The top doesn’t factor, use the quadratic formula.

iff = Q2x2 x 3 0 x 1 1423

24 1 23

8

Bottom = 0 iff x 32 , 0

Key numbers: 32 , 0

Key intervals: , 32 ,

32 , 0, 0,

, fx 2x2x332xx f10 0, f1 0, f10 0

Answer: 32 , 0


Math 140 Lecture 2 Rectangular coordinates(a, b) is the point on the plane whose x-coordinate is a and

y-coordinate is b.

a

b

a,b

x

y

2

1 5

fig. 2( )

WARNING. (a, b) can mean a point or an interval. DEFINITION. The graph of an equation is the set of all

points which satisfy the equation. The graph of y = 0 is the x-axis. The graph of x = 0 is the y-axis. `Does (5, 6) lie on y = x2

+1? (x, y) = (5, 6) lies on y = x2

+1 iff 6 = (5)2 +1 iff 6 = 26 iff

no. DISTANCE FORMULA. The distance d between points (x, y)

and (a, b) is d = x a2 y b2

`The distance between (1, 3) and (2, 5) is 1 22 3 52 12 22 5

Circles and completing the squareCIRCLE EQUATION. The circle with radius r and center (a, b)

has equation: . x a2 y b2 r2

Proof. (x, y) is on this circle iff the distance between (x, y) and (a, b) is r iff x a2 y b2 riff . x a2 y b2 r2

`Find the equation of the circle of radius 3 around (1, -2): x 12 y 22 32

FACTS. For positive a, . For all a, .a a 2 a2 |a|`Find the center and radius of the circle with equation:

x 12 y2 8 x 12 y 02 8 2 4 2 2 2 2 2

tcenter = (-1,0), radius = 2 2COMPLETING THE SQUARE. To make a perfect square,x2 ax

add . . The latter is a “perfect” a2 2 x2 ax a

2 2 x a2 2

square.`Find the center and radius of the circle with equation:

4x2 4y2 4y 79 0x2 y2 y 79

4 0x2 y2 y 79

4x 02 y2 y 1

4 794 1

4 804 20

x 02 y 12 2 20 2 4 5 2

x 02 y 12 2 2 5 2

tcenter = (0, -½), radius = 2 5

DEFINITION. An x-intercept is the x-coordinate of a pointwhere the graph crosses the x-axis. A y-intercept is they-coordinate of a point where the graph crosses they-axis. In fig. 2, 2 is the y-intercept; 1, 5 are the x-intercepts.

The graph crosses the x-axis when the y-coordinate is 0. } To find the x-intercepts, set y to 0 and solve for x. } To find the y-intercepts, set x to 0 and solve for y. `Find the x and y-intercepts.

y x2 4 0x-intercepts (set y = 0):

0 x2 4 0, x2 4, x2 4, x 2, 2The x-intercepts are -2, 2.

y-intercepts (set x = 0): y 02 4 0, y 4

The y-intercept is -4. `Find the x and y-intercepts.

y x2 4 0x-intercepts:

x-intercepts: none0 x2 4 0, x2 4, x2 4,y-intercept: 4. Straight lines: slopes and equationsDEFINITION. The slope of a line is the ratio of the vertical

(height) change in y over a horizontal change in x.THEOREM. } The slope of the line through (x, y) and (x1, y1):

m yy1xx1

} The equation of the line through (x1, y1) with slope m: y y1 mx x1

} The equation of the line with slope m and y-intercept b: y mx b

} The slope of a horizontal line is 0. } The slope of a vertical line is undefined. } Equation for the horizontal line through (a, b): y = b. } Equation of the vertical line through (a, b): x = a.

Estimatetheslopes

x

y

`Find the slope of the line through (2, 3) and (4, 3). 0LINE EQUATION FORMAT. In homework and tests, line

equations must be in one of these four forms: y = mx+b, y = mx, y = b, x = a.

`Find the line through (2, 5) and (4, 1). Check your answer. m = (5-1)/(2-4) = -2, y-5 = -2(x-2), y = -2x + 9 `Eq. of line with slope -8, y-intercept -6. y = -8x - 6

Practice problem numbers correspond to numbers of similar problems on the practice exams.Try to do practice quizzes in the time indicated at the top right corner.Classwork 2 is hard, work on it in advance. Your TA will give you an extra five minutes time for group work.

Theorem. equals the perfect square x2 ax minus the constant ,x a

2 2 a2 2

x2 ax x a2 2 a

2 2

Practice 5(10). For the circle : x2 6x y2 4y 4 0(a) Find the center and radius. (b) Find the x and y-intercepts.

(a) Given equation: x2 6x y2 4y 4 0Write the equation in the form: x a2 y b2 r2

Get the constant on the right, group the variables together. x2 6x y2 4y 4

Complete the square: x 32 9 y 22 4 4Add the 9, 4 to both sides.

x 32 y 22 4 9 4 x 32 y 22 32

In class work, make sure you write it like this, not like . x 32 y 22 9Answer: (a) center = (3, -2), radius = . 3

(b) x-intercept (set ): Use the original equation instead of the circle equation. That way you can get credit for the intercepts even ify 0your circle equation is wrong.

x2 6x 02 40 4 0 x2 6x 4 0

Try factoring first. Then try the quadratic formula.

x b b2 4ac2a

6 62 442

6 202

6 2 52 3 5

Answer: x-intercepts: x 3 5

y-intercept (set x 0 : 02 60 y2 4y 4 0

y2 4y 4 0 y 2y 2 0

Answer: y-intercept: y 2

Math 140 Discussion Lecture 2 Exam problems points/time 14pts 10mins

5. Write in circle-equation form: x a2 y b2 r2

x2 y 32 16

5b. For the circle x2 8x y2 6y 9 0(a) Write the equation in circle-equation form:

Complete the squares.x a2 y b2 r2

The form must be not .x a2 y b2 r2 x a2 y b2 c19 symbols , chk=17. (b) Find the center. Remember “( )”, e.g., “center = (1,2)” not“center = 1,2”. 6 symbols, chk=7 Center = ___ , ___ (c) Find the radius. 1 symbol.

Radius = ___ (d) Find the x-intercept (use original equation)

. Set and solve for x using thex2 8x y2 6y 9 0 y 0quadratic formula. . x

b b2 4ac2a

4,5 symbols after “x=”, chk=11, no fractions.

(e) Find the y-intercept (use original equation) . Set and solve for y by factoring.x2 8x y2 6y 9 0 x 0

2 symbols after “y=”, chk=3.

6. Find the equation for the line through (3, -1) and (1, 2).Write the equation in the form y = mx+b. 4 symbols.

(a) Find the slope: 4 symbols. chk=5m y2 y1x2 x1 ?

(b) Find the equation. Use one of the given points, say and the slope m above. x1, y1 3,1

In the point-slope formula ,y y1 mx x1fill in the values for . y1, x1, mRewrite in the form . 9 symbols after “y =”, chk=14.y mx b

Math 140 Classwork 2 Score ____/4

No credit unless you work with a group of three or more people. For examples, see discussion lecture 2. Hard. chk = sum of digits.

For superscript E problems, see notes in the section titled “Errors”.Do not turn in scratch paper; put your work in the space provided.§1.8 page 99 problems 89-94. Odds have answers in the back.In 1, 2, 3, find the center and radius.1(2). x2 y 12 20

Integral center, radius has a radical, chk=8

2E(2). x2 y2 10x 2y 17 0Integral center and radius, chk=9

3(2). 4x2 4x 4y2 63 0Integral radius, center has one fraction, chk=7

4. Does (4, -2) lie on ? yes? no?3x2 y2 52

§1.8 98:44, 45, 51-60. Odds have answers in back.Find the x-intercept(s) and y = intercept(s). Write “none”if there are none. You don’t need to draw the graphs. Two problems have 2 x-intercepts, one has none. The first threeproblems have integral intercepts. The forth has radicals.

5. 3x 4y 12chk=7

6. y x2 4x 12chk=11

7. y x2 4x 12chk=11

8. x 22 y 12 9chk=3

§1.10 120:1-8, 15-22. (recommended text problems)9E. Find the slope of the line through (-3, 0) and (4, 9).

The slope is a fraction, chk=16

Write the equations for the lines below in the one of thefollowing forms: y=mx+b, y=mx, y=b or x=a. In the remaining problems, the coefficients are all integers.10. The line with slope 22 through (0, 0). chk4=

11. The line through (7, 9) and (-11, 9).chk=9

12. The line through (12,13) and (13,12).chk=7

13. The line with slope 0 and y-intercept 14.chk=5

Math 140 Hw 2 Name __________________________________ Score ____/ 16

Find the center and radius.

A. x 32 y 12 25

B. x2 y2 2

C. x2 y2 8x 6y 24

D. 9x2 54x 9y2 6y 64 0

Find the x-intercept(s) and y-intercept(s). Write “none” ifthere are none. You don’t need to draw the graphs.

E. 3x 4y 12

F. y 2x 4

G. x y 1

H. y x2 3x 2

I. y x2 x 1

Find the slope of the line through the given points.

J. (a) 3, 2, 1,6

(b) 2,5, 4, 1

(c) 2,7, 1, 0

(d) 4, 5, 5, 8

K. (a) 1, 1, 1,1(b) 0, 5, 8, 5(c) 1, 1, 1,1(d) a, b, b, a

Write the equations for the lines below in the one of thefollowing forms: y=mx+b, y=mx, y=b or x=a.

L. (a) Slope m = -5, (-2,1) is on the line.

(b) Slope m = 4, (4,-4) is on the line.

(c) Slope m = ƒ, (-6,-2/3) is on the line.

(d) Slope m = -1, (0,1) is on the line.

M. (a) (4,8), (-3,-6) are on the line.

(b) (-2,0), (3,-10) are on the line.

(c) (-3,-2), (4,-1) are on the line.

N. (a) Slope m = -4, y-intercept = 7

(b) Slope m = 2, y-intercept = 3/2

(c) Slope m = -4/3, y-intercept = 14

AnswersA. center 3,1, radius 5B. center , radius 0, 0 4 2C. center , radius 14, 3D. center , radius 3, 1

3 2

E. x-intercept = 4; y-intercept = 3F. x-intercept = 2; y-intercept = -4G. x-intercept = 1; y-intercept = 1H. x-intercept = -1,-2; y-intercept = 2I. x-intercept = ; y-intercept = -11 5 /2

J. (a) -2, (b) 3, (c) -7/3 , (d) 3K. (a) 1, (b) 0, (c) -1, (d) -1L. (a) (b) y 5x 9 y 4x 20

(c) (d) y 13 x 4

3 y x 1M. (a) (b) y 2x y 2x 4

(c) y 17 x 11

7

N. (a) (b) y 4x 7 y 2x 32

(c) y 43 x 14


Find the center and radius.

C. x2 y2 8x 6y 24 x2 8x y2 6y 24

82 2 42 16, 6

2 2 32 9x 42 16 y 32 9 24

x 42 y 32 24 16 9 24 25 1 x 42 y 32 12

Answer: center , radius 14, 3

D. 9x2 54x 9y2 6y 64 0 9x2 54x 9y2 6y 64

Divide by 9 to make to coefficients of and are 1.x2 y2

x2 6x y2 23 y 64/9

62 2 32 9, 2/3

2 2 13 2 1

9x 32 9 y 1

3 2 19 64

9x 32 y 1

3 2 649 9 1

9 x 32 y 1

3 2 639 9 7 9 2

x 32 y 13 2 2 2

Answer: center , radius 3, 13 2

Find the x-intercept(s) and y-intercept(s). Write “none” ifthere are none. You don’t need to draw the graphs.

H. y x2 3x 2x-intercepts (set y = 0):

0 x2 3x 2 x 2x 1 0

Answer: x intercepts are: x 2,1y-intercepts (set x=0):

y 2Answer: y-intercept is: y 2

I. y x2 x 1x-intercepts (set y = 0): 0 x2 x 1

Doesn't factor, use the quadratic formula.x2 x 1 0

x b b24ac2a

1 141121

1 52

Answer: x intercepts are: x 1 52

y-intercepts (set x=0): y 1Answer: y-intercept is: y 1

Find the slope of the line through the given points.J. (a) 3, 2, 1,6

Answer: m y1y2x1x2

2631

84 2

Write the equations for the lines below in the one of thefollowing forms: y=mx+b, y=mx, y=b or x=a.

L. (d) Slope m = -1, (0,1) is on the line.

y mx b y 1x 1

Answer: y x 1

M. (c) (-3,-2), (4,-1) are on the line.

m y1y2x1x2

2134 1

7 17

y y1 mx x1 y 2 1

7 x 3 y 2 1

7 x 37 y 1

7 x 37 2

Answer: y 17 x 11

7

N. (c) Slope m = -4/3, y-intercept = 14.

y mx bAnswer: y 4

3 x 14


Math 140 Lecture 3 Reminder: Gateway Exam next week.Factoring and rootsTHEOREM. If , . a 0 x 2 a x a x a But has no roots and can’t be factored any more.x2 aDIVISION LAW. If p(x)/d(x) has quotient q(x) andremainder r(x) then pxdx qx rx

dx . Multiply by dx to get px dxqx rx.

d(x) divides into p(x) evenly iff the remainder is 0 iff iff d(x) is a factor of p(x).px dxqx

If d(x) is a factor of p(x), the other factor of p(x) is q(x),the quotient of p(x)/d(x).

`Given p(x)/d(x), divide to get the quotient q(x) andremainder r(x). Write the answer in division law form:

.px dxqx rxx31x1 , x31

x1 x2 x 1 2x1 , x3 1 x 1x2 x 1 2

Check that the answer is correct for x = 0. For x = 0, we get 0+1 = (-1)(0+0+1)+2, 1 = 1. ,

X-INTERCEPT. a is a root or zero of p(x) iff p(a) = 0.

THEOREM. a is a root of p(x) iff is a factor of p(x). x aNote, rewrite .x 3 as x 3To find all roots of p(x), completely factor p(x). Factor the polynomial and find all roots.` Root: -2x 2

` Fully factored as is, no roots.x2 2

` Rts: x2 2 x 2 x 2 2 , 2

` One repeated factor. Root: 2x2 4x 4 x 22

` given that -1 is a root. x3 5x2 8x 4

twe divide by (x+1).x a x 1 x 1

x3 5x2 8x 4x 1 x2 4x 4 x 2x 2 x 22

x3 +5x2 +8x +4 (=x +1())x +2)2 Roots: -2, -1.

`x3 −x2 −2x +2 given that 1 is a root.

.x3 x2 2x 2x 1 x2 2 x 2 x 2

x3 x2 2x 2 x 1x 2 x 2 Roots: . 2 , 1, 2

` . Find the roots with the quadratic formula.2x2 2x 1There will be three factors: one for each root and one forthe coefficient 2 of .x2

2x2 2x 1

Roots: x b b24ac2a

2 2242122

2 124

1 32

Factorization: 2x 1 32 x 1 3

2

Functions DEFINITION. For sets A and B, a function from A to Bassigns a value in B to each x in A. The domain of f f xis A; the range of f is the set of all possible values fx

`f (x) = x2 is a function from real numbers to realnumbers.

domain = since x2 is defined for all numbers.,range = since x2 can never be negative. 0,

NOTATION. Sometimes, instead of writing f (x) = x2, we define a function by writing y = x2.

Thus y is the value of the function. Since it depends onx, y is the dependent variable. Since x ranges freely overthe domain, it is the independent variable.

A function may assign only one value to each x. Thus y = is not a function. x`Of f and g, which are functions? (f isn’t, g is)

f g

`Write each domain in interval notation.} y 1 x ,

} y 11x , 1 1,

} y 1 x , 1

` . First add ( )’s around each x: . Simplify tofx x2 x2

an expanded polynomial.

} fx fax a x2a2

xa x a

} fx h fxh xh2x2

h 2x h

To get , replace x in by to get fx h fx x2 x h. Note, .fx h (x h)2 fx h x2 h2

` . Rewrite as . Simplifygx 1x x 1

x x

ggx 1gx gx 1

( 1x x) ( 1

x x) . x

1x2 1x x x2

x1x2 1x2

x1x2 x21x2x1x2 x43x21

x1x2

` . Simplifyhx 1xx

.hx h 1xhxh 1xh

xh

. hhx 1hxhx 1( 1x

x )( 1x

x ) ... 2x11x

Try to do practice quizzes in the time indicated at the top right corner.Practice problem numbers correspond to numbers of similar problems on the practice exams.

7(6). Find the domain of f (x) = . Interval notation.11 x 30

v What is under the radical, , must be > 0. Note x 30 x 30 0 not x 30 0x 30 0iff x 30

v The denominator must not be 0.1 x 301 x 30 0iff 1 x 30iff 1 x 30iff x 29Check validity validx 29 1 x 30 1 29 30 1 1 0,

Hence since -29 makes the denominator 0. x 29, v The domain consists of points satisfying both conditions.Draw the number line. Cross off all points which don’t satisfy both conditions. What is left is your answer.

the domain x iff . Deleting 6 and numbers < -30 leavesx 29 and x 30Answer: x 30,29 29,

Problems like this are often missed. Practice on homework 3’s recommended problems.

8(6). Factor and find all roots given that -1 is one root.x3 x2 3x 3

v Use the give root to get one factor. a root ˆ is a factor. 1 x 1 x 1

v Divide by this factor to get the quotient which is the other factor.x3 x2 3x 3

x 1 x2 3

v Factor this quotient if possible.x2 3 x 3 x 3

v Complete factorization: x3 x2 3x 3

x 1x 3 x 3

v List the roots: x 3 ,1, 3

Math 140 Discussion Lecture 3 Exam problems points/time 12pts 8 mins

17. Find the domain of . f x 2 xx2 9

Radical arguments must be > 0. Denominators must not be 0. w When is the radical argument > 0? The radicalargument is In order for to be defined, we2 x 2 xmust have ..* Solve this for x. If you get ,2 x 0 x 2you have an error. *Note 2 x 0 not 2 x 0.

w When is the denominator = 0? The denominator is . Set it equal to 0 and solve. Should get twox2 9

solutions.

w Hence the denominator 0

when and .x ____ x ____

w Find the domain. On the line below there should be two hollow circles, one solid.Cross off points not in the domain.

Now write the answer. It is a union of two intervals. 14 symbols. 3 ( )'s, and 1 [ ]. If your answer includes numbers > 2, it is wrong. chk=8

18. Factor and find all roots given that is one root.x 3x3 3x2 3x 9

w What is the factor determined by the root ?x 3

w Divide the given polynomial by thisx3 3x2 3x 9factor to get the quotient which is the other factor. 4 symbols

w Factor this new factor.

w Find the factorization and the roots.

Factorization: Should be three factors. =x3 3x2 3x 9

List the 3 roots:3 roots, 2 with radicals. chk=9

x = ___________, ___________, ___________

Math 140 Classwork 3 Score____/4 No credit unless you work with a group of three or more people. For examples, see Discussion Lecture 3.

For superscript E problems, see notes in the section titled "Errors",§3.2. page 270 problems 1-6. Odds have answers in back.Divide to get the quotient q(x) and remainder r(x). Write your answer in the form .px dxqx rx

2(1). x3 4x2 x 2x 5

2 digit positive remainder

10(1). x6 64x 2

3 digit positive remainder

§3.2. 271:55,56. §3.2. 279:11-20.Factor the two polynomials and find all the roots (zeros).28(2). x3 7x2 11x 5 0. 1 is one root.

2 roots (both integral),chk=6. 2 factors (one repeated), chk=8

4E(3). x3 8x2 3x 24 0. 8 is one root.3 roots (one integral),chk=14. 3 factors, chk=14

§3.1. 85: 87-90. §2.1 156:41-52, 54-57.5(1). Find the domain. y 3x 12

1 big interval

6E(1). Find the domain. y 13x 12

union of 2 intervals, chk=

7(1). Find the domain. y 3x 121 interval, chk=

§2.2. 268:55-56.8(1). Of f, g, F, G, circle the three which are functions.

f g F G

9(½). f(x) = . f(x 2) =4 3x

2 term polynomial, 5 symbols, chk=9

10(½). f(x) = . f(1/x) =4 3x2 terms, one fractional, 5 symbols, chk=7

11(1). f(x) = . f(f(x)) =4 3x2 term polynomial, 4 symbols, chk=17

12(½). f(x) = . x2 f(x) =4 3x

2 term polynomial, 7 symbols, chk=12

13(½). f(x) = . 1/f(x) =4 3xSingle fraction, 6 symbols, chk=8

14(1). . fx 2x 9fx h fx

h Assume h=0. 2-symbol integer, chk=2

15(1). . gx 2x2 9gx ga

x a Assume x=a. 6-symbol polynomial, chk=4

Math 140 Hw 3 Name ____________________________Score____/ 16

Divide to get the quotient q(x), remainder r(x). Write inthe form: .px dxqx rx

A. x28x4

x3

B. x26x2

x5

C. 6x32x3

2x1

D. x52x3

E. x664x2

Factor the polynomials and find all the roots (zeros).F. x3 4x2 9x 36 0 given 3 is a root.

G. x3 x2 7x 5 0 given 1 is a root.

H. 3x3 5x2 16x 12 0 given -2 is a root.

I. 2x3 x2 5x 3 0 given -3/2 is a root.

Write the domain in interval notation.J. (a) .y 5x 1

(b) y 1/5x 1.

(c) y 5x 1 .

(d) y 3 5x 1 .

K. (a) ft t2 8t 15.

(b) gt 1/t2 8t 15.

(c) ht t2 8t 15 .

(d) kt 3 t2 8t 15 .

L. Let Find the following.fx 3x2.(a) f2x.

(b) 2fx.

(c) fx2.

(d) fx2.

(e) fx/2.

(f) fx/2.

M. Let . Find the following.Hx 1 2x2

(a) H0.(b) H2.(c) H 2 .(d) H5/6.(e) Hx 1.(f) Hx h.(g) Hx h Hx.

(h) HxhHx

h .

N. Find the quotient for fxhfx

h fx 8x 3.

O. Find the quotient for .gxga

xa gx 4x2

P. Find the quotient for .gxga

xa gx x2 2x 4

AnswersA. x2 8x 4 x 3x 5 11

B. x2 6x 2 x 5x 11 53

C. 6x3 2x 3 2x 13x2 32 x 1

4 134

D. x5 2 x 3x4 3x3 9x2 27x 81 241

E. x6 64 x 2x5 2x4 4x3 8x2 16x 32F. G. H. 3, 4. 1,1 6 . 2, 2

3 , 3.

I. 32 , 1 5 /2.

J. (a) (b) . ,. , 15

15 ,

(c) (d) , 15 . ,

K. (a) (b) ,. , 3 3, 5 5,.(c) (d) , 3 5,. ,.L. (a) (b) (c) (d) (e) (f) 12x2. 6x2. 3x4. 9x4. 3

4 x2. 32 x 2

M. (a) 1. (b) -7. (c) -3. (d) 718 .

(e) (f) .2x2 4x 1. 1 2x2 4xh 2h2

(g) (h) 4xh 2h2. 4x 2h.

N. . O. P. 8 4x 4a x a 2.


Divide to get the quotient q(x), remainder r(x). Write in

the form: .p(x) = d(x)q(x) + r(x)

B. x2−6x−2x+5

x − 11 +53x+5

x + 5 ) x2 − 6x − 2 x2 + 5x

−11x − 2 −11x − 55

53

τ Multiplying by x+5 givesx2−6x−2

x+5 = (x − 11) +53x+5

Answer: x2 − 6x − 2 = (x + 5)(x − 11) + 53

Factor the polynomials and find all the roots (zeros).

G. x3 + x2 − 7x + 5 = 0 given 1 is a root.Since 1 is a root, divide by .x3 + x2 − 7x + 5 x − 1

The quotient is .x3+x2−7x+5

x−1 = x2 + 2x − 5Hence x3 + x2 − 7x + 5 = (x − 1)(x2 + 2x − 5)

Now we must factor . x2 + 2x − 5The first factors to try are and . Neither work,x ! 1 x ! 5thus we must use the quadratic formula.

x =−b! b2−4ac

2a =−2! 22−4(1)(−5)

2(1) =−2! 4+20

2

=−2! 24

2 =−2! 4$6

2 =−2!2 6

2 = −1 ! 6Hence the factorization of isx2 + 2x − 5

x2 + 2x − 5 = [x − (−1 − 6 )][x − (−1 + 6 )]

Hence factors intox3 + x2 − 7x + 5

Answer: (x − 1)[x − (−1 − 6 )][x − (−1 + 6 )]

The three roots are

Answer: 1, −1 ! 6 .

H. 3x3 − 5x2 − 16x + 12 = 0 given -2 is a root.Since -2 is a root, we divide by .x − (−2) = x + 2

The quotient is 3x3−5x2−16x+12

x+2 = 3x2 − 11x + 6Now we must factor 3x2 − 11x + 63x2 − 11x + 6 = (3x − 2)(x − 3)

Hence factors into3x3 − 5x2 − 16x + 12

Answer: (x + 2)(3x − 2)(x − 3)= (x − (−2))(3x − 2)(x − 3)

The three roots are

Answer: −2, 23 , 3.

Write domain in interval notation.

K. (a) f(t) = t2 − 8t + 15.There are no divisions by 0, no square roots of negativenumbers. Hence is defined everywhere.f(t)

Answer: Domain (−∞,∞).

(b) g(t) = 1/(t2 − 8t + 15).Factoring gives

. Thus is undefined at t=3, 5.g(t) =1

(t−3)(t−5) g(t)

Answer: Domain = (−∞, 3) 4 (3, 5)4 (5,∞).

(c) h(t) = t2 − 8t + 15 .This is defined provided > 0.t2 − 8t + 15By part (b), the key numbers are 3, 5.The key intervals are (-5, 3], [3, 5], [5, 5).We use [ ] since the inequality is > rather than >.Now evaluate at a point in each interval.t2 − 8t + 150ι(-5,3], 02 − 8 $ 0 + 15 = 15 = +

4ι[3,5], 42 − 8 $ 4 + 15 = −1 = −

10ι[5,5), 102 − 8 $ 10 + 15 = 35 = +

Hence h(t) is undefined between 3 and 5.

Answer: (c) Domain = (−∞, 3] 4 [5,∞).

(d) k(t) =3 t2 − 8t + 15 .

You can't take the square root of a negative number butyou can take the cube root of any number. Hence isk(t)

defined everywhere.

Answer: Domain = (−∞,∞).

L. Let Find the following.f(x) = 3x2.(c) f(x2).

Answer: f(x2) = 3(x2)2 = 3x4 3x4.

(d) [f(x)]2.Answer: [f(x)]2 = [3x2]2 = 9x4 9x4.

(e) f(x/2).

Answer: f(x/2) = 3(x/2)2 = 3 x2

434 x2.

M. Let . Find the following.H(x) = 1 − 2x2

(f) H(x + h).H(x + h) = 1 − 2(x + h)2 = 1 − 2(x2 + 2xh + h2)

Answer: .1 − 2x2 − 4xh − 2h2

(h) H(x+h)−H(x)

h .H(x+h)−H(x)

h =[1−2x2−4xh−2h2]−[1−2x2]

h

=1−2x2−4xh−2h2−1+2x2

h

=−4xh−2h2

h = −4x − 2hAnswer: −4x − 2h.

O. Find the quotient .g(x)−g(a)

x−a

g(x) = 4x2.g(x)−g(a)

x−a =[4x2]−[4a2]

x−a =4(x2−a2)

x−a =4(x+a)(x−a)

x−a

Answer: 4x + 4a


Math 140 Lecture 4 Gateway exam. Practice exams included with lecture notes.DEFINITION. The graph of a function f is the set of all

points (x, y) such that y = f (x). The height y is thefunction’s value f (x).

VERTICAL LINE TEST. A curve is the graph of a function iffno vertical line intersects it more than once.

`Which curve is the graph of a function?

`Find the domain and range of each function.

Domain: (-1,3]Range: {2}Find the values. f(1) = 2

f(5) = undefined For what x is f(x) = 2?

(-1,3]

Domain: [-1,3]Range: [1,4]Find the values. g(1) = 4

g (2) = 1For what x is g (x) = 1?

x = 2Compute [g (x)-g (1)]/[x-1] when x = 3.

-½

Domain: [0, 2)u(2, 4]Range: [1, 3)u(4, 5]Find the values. h(2) = undefined

h(4) =1

DEFINITION. v f increases on an interval if the value f (x) increases as

x moves from left to right in the interval. v f decreases if f (x) decreases as x goes left to right. v c is a turning point of f if f increases on one side of c

and decreases on the other.v f (a) = c is a maximum value iff c is > all other values.v f (a) = c is a minimum value iff c is < all other values. `For g and h graphed above, fill in the table. Write “none” if there

are none. Separate successive intervals of increase or decreasewith “,” not “u”, e.g., “[0,1], [3,4)” not “[0,1]u[3,4)”.

[0, 2), (2, 4][1, 2]interval(s) of decrease

none[-1, 1], [2, 3]interval(s) of increase

h(4) = 1g(2) = 1minimum value

h(0) = 5g (1) = 4maximum value

nonex = 1, 2turning points

hg

`Do the same for the functions f and g below.

f g

[2, 4][0, 2]interval(s) of decrease

[0, 2][-2, 0], [2, 5]interval(s) of increase

g(0) = g(4) = 0f (2) = -1minimum value

g(2) = 3f (0) = 4maximum value

x = 2x = 0, 2turning points

[0, 3][-1, 4]range

[0, 4][-2, 5]domain

gf

`Fill in the table. fx x2 1

(-5, -1]interval(s) of decrease

[1,5)interval(s) of increase

f (-1) = f (1) = 0minimum value

nonemaximum value

noneturning points

[0,5)range

(-5,-1]u[1,5)domain

`Graph.

fx

1x if x 0x if x 0

`Graph.

fx

1 if x 0x if 0 x 11 if 1 x

1

1

o

`Gateway problem. Complete the square of the formula. = ... = 2x2 3x 5 2x 3

4 2 318

oo

o

h

g

f

Practice problem numbers correspond to numbers of similar problems on the practice exams.Try to do practice quizzes in the time indicated at the top right corner.Classwork 4 is hard, work on it in advance.

9(4). Simplify to a single reduced polynomial:

g(x) = , rewrite g(x) as . 11 x

11 x

gx h gxh

11xh

11x

h 1

1xh 1

1x h

1 x h1 x1 x h1 x

1 x 1 x hh1 x h1 x h

h1 x h1 x h

1 x h1 x

11(16). Fill in the table. For domains and ranges, use interval notation, e.g., [1,5) or set notation, e.g., {1,3,5}. Sometimes the correctanswer is “none”. h(x) is graphed on the right. f(x) is listed on the left, it is not graphed.

-2-1012

-2 -1 0 1 2 3 4 5 6h(x)

[-2, 0], [4, 6][0, 5)interval(s) of decrease[0, 4](-5, 0]interval(s) of increase

h(0) = -2nonemin value(s) f(a)=bh(4) = 2f(0) = 1max value(s) f(a)=bx= 0, 4x= 0turning point(s), x=[-2, 6](-5, 5)domain

x= -2, 2, 6x= -1, 1roots, x=h(x), see graphf(x) = 1- x2function

v In intervals of increase and decrease, include the endpoints if the function is defined, exclude them if not.v For domains we take the union of the intervals .2, 0 4, 6When listing the intervals of increase and decrease, list the intervals separately .2, 0, 4, 6


9. Simplify the fraction over h to a reduced ratio of two polynomials, e.g. , not .2x 13x h

x 1/x3 1/x h

, rewrite as . Thengx 13 x gx 1

3 x

Note: and are wrong. gx h 13 x h

13 x h

13 x h 1

3 x hFind the following:

The answer isgx h gx

h 1 (3 −x()3 −x −h)

To get credit, you must show your work.

11. Fill in the table. For domains and ranges, use interval notation [1,5) or set notation {1,3,5}. Sometimes the correct answer is none.Some answers are given. Note the format.

-2-101

-2 -1 0 1 2 3 4 5 6h(x)

2

7 8f(x)

0, 2, 6, 8 note , not interval(s) of decrease__________, _________ 2 intervals(-5, 0]interval(s) of increase

h2 h2 2nonemin value(s) f(a)=bh______ ______f (0) = 0max value(s) f(a)=b

x= 3 turning points x= 0turning point(s) [-2, 2](-5, 0]range

domain x= 3 rootsx=roots

h(x), see graph abovef(x) = - x2 see graph abovefunction

The graph of h has no hollow circles, hence all endpoints are included and should be [, ], not (, ) .In intervals of increase and decrease, include the endpoints if the function is defined, exclude them if not (when graph has a hollow circle). When listing the intervals of increase and decrease, list the intervals separately . In contrast, for domains we take the union 2, 0, 4, 6

. For roots and turning points, we list only the x coordinate, e.g. not both coordinates . 2, 0 4, 6 x 3 3, 8

Math 140 Classwork 4 Score _______/5No credit unless you work with a group of three or more people. For examples, see Practice Quiz 4. Hard.

Number of points is in ( )’s. The answer in 4 has a radical, all other numeric answers are digits or negative digits.

1(1). In the graph of , find the y-coordinate ofy 1 x2

the point with x-coordinate . 4 symbols, exact answer, no 12

decimals.

§2.2 167:24,23. 169:57-60.In 2-4, list the domain and range of each graph (see text).Each domain is a single interval. For the ranges: 3, 4 useset notation e.g., {1,3,5}, 2 uses a combination e.g., (-1,1]u{2}.

2(1).

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

O

domain = chk=8

range = chk=7

3(1).

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3O

OO

domain = chk=6

range = chk=6

4(1).

1 2 3 4 5-1-2-3-4

1

2

3

4

-1

-2

-3

6

OO

domain = chk=9

range = chk=1

5(2). For the function h pictured,

1

2

3

4

-1

-2

a b c d

(a)(½) h(a) =

h(b) =

h(c) =

h(d) =

(c)(½) For which values of x does h(x) = 0?

(e)(½) As x increases from c to d, does h(x) increase or decrease?

(f)(½) As x increases from a to b, does h(x) increase or decrease?

6(1). For the f pictured, compute

1 2 3 4-1-2-3-4-5

1

2

3

4

(c)(½) when x = 3.fx f2/x 2

(d)(½) when x = -3.fx f2/x 2

Math 140 Hw 4 Name ___________________________________ Score _____/ 21

Continued on next page

7(7). Fill in the table for f (x) = 1/x and lx. Seven of the answers are “none”. ½ point for each entry.

function 1/x x



min value

max value

turning point

range

domain

§2.2 168:38-44.8(2). Graph U(x) where

Ux

x3 if 1 x 0x if 0 x 1

1/x if 1 x 3

1 2 3 4 5-1-2-3-4

1

2

3

-1

-2

-3

§2.3 179:1-4. §2.5 201:47-50.In 9, 10, find the range, max value, min value, intervals of

increase, and intervals of decrease. ½ point each part.

Separate two intervals of increase with “,” not “u”[ ] with lines and solid circles, ( ) only with hollow circles

9(5/2). For the function f pictured, find the

1 2 3 4 5-1-2

1

2

3

4

-1

-2

-3

range (single interval)

max value

min value

interval of increase

interval of decrease

10(5/2). For the function f pictured, find the

1 2 3 4 5-1-2

1

2

3

-1

-2

-3

range (single interval)

max value

min value

two intervals of increase

two intervals of decrease

Continued from previous page

In A-D, list the domain and range of each graph.

A. For the function pictured,

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

domain=

range=

B. For the function pictured,

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

O

domain=

range=

C. For the function pictured,

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

domain=

range=

D. For the function f pictured,

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

(a) Is f(0) positive or negative?

(b) Find f(-2) f(1)

f(2) f(3)

(c) Which is larger, f(2) or f(4)?

(d) Find f(4)-f(1).

(e) Find |f(4)-f(1)|.

(f) Find (in interval notation) the domain and range of f.

domain

range

E. Fill in the table for . |x|, x2, x3

function |x| x2 x3

interval(s) ofdecrease

interval(s) ofincrease

min value

max value

turning point

range

domain


Continued on next page.

In F,G, find the range, max value, min value, intervalsof increase, and intervals of decrease. ½ point each part.

F. For the function pictured,

243

10

1

-1

range

max value

min value



G. For the function pictured,

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

range

max value

min value



In H, I, A, graph the given function.

H. Ax

x3 if 2 x 1x2 if x 1

I. Cx

x3 if x 1x if x 1

J. fx

1/x x 1x 1 x 1

1/x 1 x


Continued from previous page.

In A-D, list the domain and range of each graph.

A. (see the graph in Hw 4 recommended problem 9). domain = the set of x-coordinates of points on thegraph. The smallest x-coordinate is -3 from the point(-3,-2). The largest x-coordinate is 4 from the point(4,1).Answer: domain = [-3, 4]

range = the set of y-coordinates of points on the graph.The lowest y-coordinate is -2 from the point (-3,-2).The highest y-coordinate is 2 from the point (1,2).Answer: range = [-2, 2]

B. (see the graph in Hw 4 recommended problem11).domain = the set of x-coordinates of points on thegraph. The smallest x-coordinate is -4 from the point(-4,2). The largest x-coordinate is 4 from the point(4,-2). There is hollow circle above x = -1, thus thefunction is undefined here and x = -1 is not in thedomain. Answer: domain = [-4, -1)υ(-1,4]

range = the set of y-coordinates of points on the graph.The smallest y-coordinate is -2 from the point (4,-2).The largest y-coordinate would be 3 from the point(-1,3). However there is a hollow circle here, so thispoint is excluded.Answer: range = [-2,2)

D. (see the graph of f in Hw 4 recommended problem11).Find f(-2)To find , find the point on the graph withf(−2)x-coordinate -2. The point is (-2,4). Then = thef(−2)y-coordinate of this point which is y = 4. Answer f(−2) = 4

In F, find the range, max value, min value, intervals ofincrease, and intervals of decrease. ½ point each part.

F. For the f function pictured,

243

10

1

-1

range = [-1, 1]max value f(1) = 1min value f(3) = -1interval(s) of increase [0,1], [3,4]Note, we don't write [0,1]υ[3,4] as we would if we were finding adomain. The function increases on each of the two intervalsseparately but not on their union.

interval(s) of decrease [1,3]. We include the endpoints in an interval of increase or decreasewhen the function is defined at those points.If there had been a hollow circle at (3,-1), then 3 would not beincluded in the interval of decrease (or increase) and the interval ofdecrease would then have been [1,3).

Graph the given function.

J. f(x) =⎧

⎩ ⎨ ⎪ ⎪

1/x x < −1x −1 [ x [ 1

1/x 1 < xOn the plane to the left of x = -1, and to the right of x =1, the graph is the same as the graph of 1/x. On thevertical strip of points between the vertical lines x = -1and x = 1, the graph is the same as the graph of x whichis the major diagonal.

1

-1


Math 140 Lecture 5

Exam 1, next week, Lectures 1-6Basic graphsKnow these graphs. Note: tangent to x3 at the origin is horizontal.

1, x, x2, x3, 1/x.

hor.tangent

1/x2, , |x|, .x 1 x2

Translations and reflections THEOREM. Changing the value f(x) changes the vertical

position; changing the argument x changes thehorizontal position in *opposite the expected direction.

f (-x)-f (x)f (x-1)f (x+1)f (x)-1f (x)+1

reflect iny-axis

reflect inx-axis*right 1*left 1down 1 up 1 unit

`Given f(x), find the functions for the other graphs.

f(x)

1 2-1

-1

1

2right 2, up 1

right 2

down 1

reflect in x-axis

reflect in y-axis

reflect in x and y-axis

f(-x)

-f(-x) -f(x)

f(x)-1

f(x-2)

f(x-2)+1

y-axis

x-axis3

VERTICAL MOVES WITH VALUE CHANGES `Given f (x) = |x|, graph f (x)+2, f (x)-2, -f (x). f (x) = |x|, f (x)+2= |x|+2, f (x)-2= |x|-2, -f (x) = -|x|.

The value f (x) = the height = the vertical position of apoint on the graph. Changing f (x) changes the verticalposition of the graph. Adding 2 raises the graph 2 units. Negating f (x) reflects the graph vertically across the x-axis.

`Given f (x) = , graph f (x+2), f (x-2), f (-x).xf (x) = , f (x+2) = , f (x-2) = , f (-x).x x 2 x 2

HORIZONTAL MOVES WITH ARGUMENT CHANGES x = the x-axis position. Changing x, changes the

horizontal position of the coordinate system. Replacing x

by x+2 shifts the coordinate system 2 units to the left.

For a formula with several shifts and reflections of afunction f, rewrite it in the graph-translation form

a f bx c dThen the shifts and reflections occur in the left-to-rightorder:

A negative a gives a vertical reflection.A negative b gives a horizontal reflection. The horizontal shift is determined by the c: right if c is

positive, left if c is negative. The vertical shift is determined by d: up if d is positive,

down if d is negative.

` . Describe the following shifts. fx 1/x

f (x) +1 = (1/x) +1 Shift up 1

-f (x) = -1/x Reflect in x-axis-f (x) +1 = -1/x +1 Up 1, reflect in x-axisf (x-1) = 1/(x-1) Shift right 1

f (x +1) = 1/(x +1) Shift left 1

f (-x) = 1/-x Reflect in y-axisf (1-x) = 1/(1-x) Rewrite: f (1-x) = f (-x+1) = f (-(x-1))

Reflect in y-axis, shift right 1

f (-x+1) = 1/(-x+1) Rewrite: f (-x-1) = f (-(x-1)) Reflect in y-axis, shift right 1

Rewrite as1 fx 1 1 fx 1

= fx 1 1= fx 1 1 Reflect in x-axis, reflect in y-axis, left 1, up 1.

` Graph Parabola with roots 0, 1gx x2 x.Graph Shift left 1g1 xGraph Reflect in y-axis, shift right 1.g1 x

Practice 14(6). The graph of f (x) is given. Graph (a) . y f x and (b) y f 2 x(a) is already in the right form: reflect vertically, then horizontally. f x(b) First rewrite: factor out any coefficient of x. f 2 x f x 2 f x 2 isf x 2

for a=1, b=-1, c=2, d=0. af bx c d no vert. refl. hor. reflect. right shift 2, no upward shift.

(a) -f(-x) (b) f(2-x)

1 2

f(x)

13(6). Graph . 1 1 x 22

The known graph is (upper unit semicircle). 1 x2

Rewrite the function in the form . a 1 bx c2 d1 1 x 22 1 1x 22 1

Hence a = -1, b = 1, c = -2, d = 1Start with the known function . 1 x2

Then apply, in order, vert. reflect. (a = -1) , no hor. reflect. (b = 1), left shift 2 (c = -2), shift up 1, (d = 1).

1 2 3 4Number the graphs.

1

23

4


13. Graph .1 1 x

Factor out the “-” under the radical (you can’t pull it outside of the radical) Rewrite it in the form .a bx c d

__ __ x __ __

Start with the graph (1) of and apply the shifts and reflections needed to get the graph of . x 1 1 xGraph 1 is given. Number the other graphs, 2, 3, 4, 5. 5 being the final answer. The graph has an x-intercept. Be careful to get the x-intercept right (it is not a fraction).

1

Math 140 Classwork 5 Score _______/4

There is an example on the next page

13. Graph .1 |2 x|

Factor out the “-” under the radical. Rewrite it in the form .a|bx c| d

1 |2 x| | x 2| 1 | x 2| 1

a =-1 b = -1 c = 2 d = 1ƒ ƒ ƒ ƒ |x| |x| | x| | x 2| |2 x| 1

vert. reflection hor. reflection right 2 up 1Graph number: 1 2 3 4 5

2=3

1 3

4

5

Math 140 Classwork 5 Example for the problem on the previous page.

Simplify fully to a formula with fewer symbols. has 6 symbols.13 x

y

Circle the three formulas which cannot be simplified. Go to

www.math.hawaii.edu/140 and read “Common Errors”.

Example: 3 symbols Example: 6x2

3x3 =2x x2 − 1 =

The following are variants of often missed problems

from Take-home Quiz 1.

2. 9 − x2 =

fully Don’t write in

simplified shaded areas.

7. 2x+32x+2 =

fully

simplified

8. 4x

2x+6 =

2xx+3

9. 4x

2x+5 =

fully

simplified

12. -- single digit --210−28

27 =

6

13. -- 7 symbols, is wrong. --−3−x2−x = x−3

x−2

x−32−x

14. -- 5 symbols --−x

x−2 =

x

2−x

Evaluate the following functionsExample. g(x) = x2 − 1, g(x + h) = (x + h)2 − 1 g(g(x)) = (x2 − 1)2 − 1 = x4 − 2x2

Example. h(x) = 1 − 2x, h(y2) = 1 − 2y2

17. -- 4 symbols --g(x) = h − 2x, g(g(x)) =

4x − h

19. -- 5 symbols --f (x) =1x , f (h − x) =

1h−x

Write as a single reduced fraction. No fractions in numerator or denominator.

20. -- 4 symbols --

c2b

2c =

14b

21. -- 11 symbols --

2a +4

4+2b

=

b+2ab2ab+a

Math 140- ___ Take-home Quiz 2 Name ______________________________ Score _____/11 Write your section number. Write your answers in the white space provided; do not hand in scratch paper.

§2.4 190:1-20. Odds have answers in back.

1(5). List the combination of translations and/orreflections needed to get the given function from f (x).Careful with k.

A. Left 2, down 3. G. Reflect in x-axis, right 2.

B. Left 3, up 2. H. Reflect in x-axis, left 2.

C. Right 3, up 2. I. Reflect in y-axis, right 2.

D. Left 3, down 2. J. Right 2, up 3.

E. Right 3, down 2. K. Left 2, up 3.

F. Reflect in y-axis, up 2. L. Right 2, down 3.

(a) f (x+2)+3

(c) f (x-2)+3

(d) f (x-2)-3

(j) -f (x-2)

(k) f (2-x)

(l)* f(-x)+2 F: Reflect in y-axis, up 2. *Example

§2.4 191:33-48.Graph the functions.

2(2). y x 42 1

3(2). y x2 3

4(2). y x 32 3

The graph of f (x) is a line segment from (-4,1) to (-1,2).

Graph the following.

5(1). y fx 2

6(1). .y fx 4 2

Math 140 Hw 5 Name ________________________________ Score ____/ 13

A. Describe the combination of translations and/orreflections needed to get the given function from f (x). (a) right 1y fx 1(b) down 1y fx 1(c) up 1y fx 1(d) left 1y fx 1(e) y-axis reflection, up 1y fx 1(f) y-axis reflection, down 1y fx 1(g) x-axis reflection, up 1y fx 1(h) left1, x-axis reflectiony fx 1(i) x-axis reflection, down 1y fx 1(j) y-axis reflection, right 1, up 1y f1 x 1(k) x-axis reflection, y-axis refl., up1 y fx 1

Graph the functions.B. y x3 3

C. D. y x 42 y x 42

E. F. y x2 y x 32



G. (a) (b) y fx y fx

(c) (d) y fx y fx 1

(e) (f) y f1 x y 1 f1 x

H. Given g(x), graph (a), (b), (c).

g(x)

(a) (b) (c) y gx y gx y gx


A. Describe the combination of translations and/or

reflections needed to get the given function from f (x). (a) right 1y = f(x − 1)

(b) down 1y = f(x) − 1(c) up 1y = f(x) + 1(d) left 1y = f(x + 1)

(e) y-axis reflection, up 1y = f(−x) + 1(f) y-axis reflection, down 1y = f(−x) − 1(g) x-axis reflection, up 1y = −f(x) + 1(h) left1, x-axis reflectiony = −f(x + 1)

(i) x-axis reflection, down 1y = −f(x) − 1(j) y-axis reflection, right 1, up 1y = f(1 − x) + 1Note: the order is important: y-axis reflection, left1, up 1 is wrong.

(k) x-axis reflection, y-axis refl., up1 y = −f(−x) + 1

Graph the functions.

B. y = x3 − 3Start with the graph of , then move it down 3 units.x3

-3

C. y = (x + 4)2

Start with the graph of then shift if left 4 units.x2,

-4

F. y = −(x − 3)2

Start with reflect it vertically around the x-axis tox2,get . Shift it right 3 units to get . −x2 −(x − 3)2

3



G. (a) (b) y = −f(x) y = f(−x)

(c) (d) y = −f(−x) y = −f(x − 1)

(e) (f) y = −f(1 − x) y = 1 − f(1 − x)


Math 140 Lecture 6 Study Practice Exam 1 and the recommended exercises.

Functions can be added and multiplied just like numbers.DEFINITION. For functions f, g, define f +g, f - g, f.g, f /g by

(f + g)(x) = f (x) +g(x), Note: (f +g)(x) is not (f +g).(x)

(f - g)(x)= f (x)- g(x), The first is function application.

(fg)(x) = f (x)g(x), The second is multiplication.

(f /g)(x) = f (x)/g(x).

`If f (x) = x- 2, g(x) = 6, then (f + g)(x) = f (x) + g(x) = (x- 2) + (6) = x + 4, (f-g)(x) = f (x) - g(x) = (x - 2) - (6) = x - 8, (fg)(x) = f (x)g(x) = (x-2)(6) = 6x-12, (f /g)(x) = f (x)/g(x) = (x-2)/6 = .1

6 x 13

DEFINITION. For functions f and g, define fog, the composition of f and g, by

(fog)(x) = f (g(x)) Apply g to x. Get g(x). Apply f to g(x). Get f (g(x)). f is the outer function; g is the inner function.

`Suppose f (x) = x-2 and g(x) = x2. (a) Find fog and gof.

(fog)(x) = f(g(x)) = f (x2) = x2-2.(gof)(x) = g(f(x)) = g(x-2) = (x-2)2 = x2- 4x + 4.

Note that f og = gof. For composition, order matters.(b) Find (fog)(2).

Since (fog)(x) = x2-2, (fog)(2) = 22-2 = 4-2 = 2. (b’) Find (fog)(2) and (gof )(2) directly without (fog)(x),

(gof )(x). (fog)(2) = f(g(2)) = f(22) = f(4) = 4-2 = 2. (gof)(2) = g(f(2)) = g(2-2) = g(0) = 02 = 0.

`If h(x) = c, then h(8) = c, h(y) = c, h(x2-1) = c, h(g(x)) = c.

`For f (x) = 3x+4, g(x) = 5, find fog and gof. f(g(x)) = f(5) = 3.5+4 = 19.g(f(x)) = g(3x+4) = 5.

f(x)

g(x)

For f and g above, note that f(-3) = 1, f(-1) = 2, f(2) = 3, f(4) = 2, and g(-2) = -1, g(-1) = -2, g(1) = -1, g(2) = 2.

Find (fog)(2) = f(g(2)) = f(2) = 3. (gof)(2) = g(f(2)) = g(3) = undef. (fof)(-1) = f(f(-1)) = f(2) = 3.

`fx x 1x , ffx x 1

x 1/x 1x x 1

x xx21

Write each function below as a composition f (g(x)) of two simpler functions,

an outer function f and an inner function g.

Find the inner function first.

`Write as a composition . x2 26 fgxinner function x2 26 gx x2 2

„ outer function does what remains fx to be done. tx6

.fx x6

check: . fgx f x2 2 x2 26

`Write as a composition . 4 1x 3 f gx

inner function 4 1x 3 gx 1

x

„ outer function does what remains f x to be done. t4x 3

. f x 4x 3check: . f gx f 1

x 4 1x 3

`Write as a composition. x 1x 1

„ gx x 1 x f x x

check: .f gx f x 1 x 1

`Write as a composition.x4 x2 1. x4 x2 1 x22 x2 1

gx x2

f x x2 x 1check: .f gx fx2 x4 x2 1

`Write as a composition of 2 functions.x /1 x `Write as a composition of 3 functions.1/1 |x|

Ans hfgx, gx |x|, fx 1 x, hx 1/x

DEFINITION. id(x) = x is called the identity function.

id(x)=x

Hence id(5) = 5, id(y) = y, id(x2-1) = x2-1, ... .

THEOREM. For any function f (x), foid = f and idof = f.PROOF. ( foid)(x) = f (id(x)) = f (x).

(idof )(x) = id(f (x)) = f (x).

0 is the identity for addition, since f + 0 = 0 + f = f .

1 is the identity for muliplication, f .1 = 1. f = f .

id(x) is the identity for composition, since foid = idof = f .

15(4). f (x) = , find ( fof )(x). Simplify to a single polynomial or fraction.x 1x

f f x f x 1f x x

1x

1x 1

x x 1

x xx21

x2x2 1 x2 1 x2

xx2 1 x4 3x2 1xx2 1

16(2). .f 1 2, f 2 4, f 3 1, f 4 3

. g1 4, g2 3, g3 1, g4 2f g f4 ?

Work from inside out. .f g f 4 f gf 4 f g3 f 1 2

17. (a)(3) Write as a composition f (g(x)) of two polynomials f (x) and g(x) of lower degree. g(x) is theGx (x4 3)x2

inner function.Write this as x22 3x2

The inner function g(x) = x2

The outer function applies to the inner function; it does what remains to be done.To get the outer function, replace each occurrence of the inner function by x.

f (x) = (x2 3)x

(b)(3) Write as a composition f(h(g(x))) of three simpler functions f(x), h(x) and g(x). Fx 1 x1 x

10

Inner function: g(x) = x

Replace each occurrence of by x.xIntermediate function h(x): h(x) = 1 x

1 x

Outer function f(x) f x x10

1. To get back x from x+2, you have to subtract 2. Hence the inverse of x+2 is x-2.To get back x from 3x, you have to divide by 3. Hence the inverse of 3x is x/3.To get back x from , you have to take the cube root. Hence the inverse of , is .x3 x3 3 x

Math 140 Discussion Lecture 6 Exam 1 next Exam problems points/time 12pts 8mins

17. (a) Write as a composition f (g(x)) of two polynomials f (x) and g(x) of lower degree. g(x) is thex12 3x4 6inner function.

g(x) = chk=4

f (x) = chk=12

17. (b) Write as a composition f (h(g(x))) of three nontrivial simpler functions f (x), h(x), and g(x).2 x3

The functions must not be the trivial function x. Hence is always a wrong answer.gx xg(x) is the innermost function, find it first. Circle it.h applies next. To find h, look at the circled inner function, and ask what happens next to this inner function? f is the outermost. It does what remains to be done after g and h have been applied.

is wrong. f does only the remaining step, not all the accumulated steps.fx 2x3

Of the functions g, h, f , two have 3 symbols, one has 2 symbols.

g(x) = 3 sym

h (x) = 2 sym

f (x) = 3 sym

17. (c) Write as a composition f (h(g(x))) of three nontrivial simpler functions f (x), h(x), and g(x). 3 1 x2 21 x2

g(x) = 2 sym

h (x) = 3 sym

f (x) = 7 sym

1. To get back x from , you have to subtract 2. x 2Hence the inverse of is .x 2 x 2

To get back x from , you have to divide by 3. 3xHence the inverse of is .3x x/3

To get back x from , you have to ? 5 xHence the inverse of is 2 sym5 x

Math 140 Classwork 6 Score _______/4For worked examples of similar problems, see practice quiz 6. Easy.

Ignore * problems; they are just examples. Exam 1.

§2.7 219:1-6.f(x)=2x-1, g(x)=x2-3x-6, h(x)=x3, k(x)=2, m(x)=x2-9.

* (g.h)(x) = g(x)h(x) = x2 3x 6x3 x5 3x4 6x3

example

1. (f.h)(x) = 2 terms, 6 symbols, chk=9

2. (h/f )(x) = fraction, 7 symbols, chk=6

3. (f /h)(1) = digit

§2.7 220:17-22.f(x) = 1-2x2, g(x) = x+1.

* (gog)(x)=g(g(x)) = . --gx 1 x 1 1 x 2

example

(gog)(3)=g(g(3)) = g3 1 3 1 1 5

4. (fog)(x) = 3 terms, 9 symbols, chk=9

5. (fog)(-1) = digit

6. (gof)(x) = 2 terms, 5 or 6 symbols, chk=6

7. (gof)(-1) = digit

8. (fof)(x) = 3 terms, 10 symbols, chk=23

Given:G(-4) = -3, G(-3) = 1, G(-1) = 2, G(1) = 5.F(-3) = -4, F(1) = -3, F(2) = -1, F(5) = 1.

* (FoG)(-3) = F(G(-3)) = F(1) = -3 -- example

Three of the next 4 answers are digits, the other a negative digit.

9. (GoF)(1) =

10. (FoG)(1) =

11. (FoF)(1) =

12. (GoG)(-3) =

Given: a(x)=x2, b(x)=|x|, c(x)=3x-1. Express 13, 14, and 15 as one of the compositions:

aoa, aob, aoc, boa, bob, boc, coa, cob, coc.

* -- example3 x 1 3bx 1 cbx.

13. (3x-1)2 =

14. |3x-1|=

15. 3x2-1=

§2.7 220:45-54.

* Write F(x) = as a composition of a1x4 x2 1

3-symbol function f(x), a 6-symbol g(x) and a 2-symbolh(x). F(x) = f(g(h(x))) where f(x)=1/x, g(x)=x2+x+1, h(x)=x2.

16(2). Write G(x) = 1/(1+x4) as a composition of a3-symbol function f(x) and a 4-symbol function g(x).G(x) = f(g(x)) where

f(x) =

g(x) =

Math 140 Hw 6 Name _______________________________ Score ___/17 Exam 1

f(x)=2x-1, g(x)=x2-3x-6, h(x)=x3, k(x)=2, m(x)=x2-9.

A. (a) (f gx)

(b) f gx

(c) f g0

B. (a) m f x

(b) f mx

C. (a) (f kx)

(b) k fx

(c) f k1 k f 2

D. f x 3x 1, gx 2x 5

(a) f gx

(b) f g10

(c) g f x

(d) g f 10

E. Compute f gx and g f x

(a) fx 1 x, gx 1 x

(b) fx x2 3x 4, gx 2 3x

(c) fx x/3, gx 1 x4

(d) fx 2x, gx x2 1

Write each function as a composition of simpler functions f and g.f g

F. (a) 3 3x 4

(b) |2x 3|

(c) ax b5

(d) 1/ x

G. (a) 3 2x 1

(b) 1/x2

(c) 2x2 1

(d) 2 3 x 1

H. Write each function as a composition f g h of three simpler functions f , g and h.3x 44

Answers A. (a) x2 x 7

(b) x2 5x 5(c) 5

B. (a) x2 2x 8(b) x2 2x 8

C. (a) (b) (c) 4x 2 4x 2 4D. (a) (b) 6x 14 74

(c) (d) 6x 7 67E. (a) x, 2 x

(b) 9x2 3x 6, 3x2 9x 14(c) 1

3 1 x4, 1 181 x4

(d) 2x21, 22x 1F. (a) fx 3 x , gx 3x 4

(b) fx |x|, gx 2x 3(c) fx x5, gx ax b(d) fx 1/x, gx x

G. (a) fx 3 x , gx 2x 1(b) fx 1/x, gx x2

(c) fx 2x 1, gx x2

(d) fx 2x 1, gx 3 xH. fx x4, gx x 4, hx 3x


f(x)=2x-1, g(x)=x2-3x-6, h(x)=x3, k(x)=2, m(x)=x2-9.

D. f x 3x 1, gx 2x 5

(c) g f x

g fx gfx g3x 1 23x 1 5= 6x 2 5 6x 7

Answer: 6x 7

(d) g f 10

g f10 gf10 g3 10 1 g31= 231 5 62 5 67

Answer: 67

E. Compute f gx and g f x

(a) fx 1 x, gx 1 x

f gx fgx f1 x 1 1 x xg fx gfx g1 x 1 1 x 2 xAnswer: x, 2 x

(c) fx x/3, gx 1 x4

f gx fgx f1 x4 1 x4/3g fx gfx gx/3 1 x/34

Answer: 13 1 x4, 1 1

81 x4

(d) fx 2x, gx x2 1

f gx fgx fx2 1 2x21

g fx gfx g2x 2x2 1 22x 1Answer: 2x21, 22x 1

Write each function as a composition of simpler functions f and g.f g

F. (a) 3 3x 4The largest inner function if gx 3x 4After , what remains to be done is taking the cube3x 4root. Thus the outer function is .fx 3 xCheck: fgx f3x 4 3 3x 4Answer: fx 3 x , gx 3x 4

(b) |2x 3|Inner function: gx 2x 3

is also an inner function but not the largest.gx 2xOuter function: fx |x|Check: fgx f2x 3 |2x 3|Answer: fx |x|, gx 2x 3

(c) ax b5

Inner: gx ax bOuter: fx x5

Answer: fx x5, gx ax b

(d) 1/ xInner: gx xOuter: fx 1

x

Answer: fx 1/x, gx x

G. (a) 3 2x 1Inner: gx 2x 1Outer: fx 3 xAnswer: fx 3 x , gx 2x 1

(b) 1/x2

Inner: gx x2

Outer: fx 1/xAnswer: fx 1/x, gx x2

(c) 2x2 1Inner: gx x 2

Outer: fx 2x 1Answer: fx 2x 1, gx x2

(d) 2 3 x 1Inner: gx 3 xOuter: fx 2x 1Answer: fx 2x 1, gx 3 x


Math 140 Lecture 7 Inverse functionsDEFINITION. f

-1, the inverse of f , is the function, if any,such that, f ( f

-1(x)) = x when f -1(x) is defined and

f -1( f (x)) = x when f (x) is defined.

This says that f and f -1 undo each other: f

-1 undoes what f (x) does and gives you back x.

`f (x) = 2x, f -1(x) = ½x. Verify: f (f -1(x)) = x & f

-1(f (x)) = x

`g(x) = x +3, g-1(x) = x-3. `h(x) = 2x +3, h

-1(x) = (x-3)/2 = . 12 x 3

2

Verify that h -1(h(x)) = x.

To undo a sequence of operations, you must undo them inthe reverse order: the inverse of .gf x is f 1g1x

y xf

f -1

THEOREM. y = f -1(x) iff f (y) = x.

To find f -1(x) for complicated functions:

Start with f (y) = x. Solve for y to get y = f

-1(x).

` , find . f x x3 f 1xf y xty3 xt .y 3 x f 1x 3 x

WARNING. f -1(x) and (f (x)) -1 are not the same.

f -1(x) = the inverse

(f (x)) -1 = the reciprocal = 1/f (x) .If f (x) = x3

f -1(x) = f

-1(0) = 03 x(f (x)) -1

= 1/x3 (f (0)) -1 = undefined.

` = . Find .f x x1x1 f 1x

f y x,y1y1 x, y 1 xy 1

y 1 xy x, y xy x 1y1 x x 1, y x1

1x x1x1

tf 1x x1x1

` for . Find .gx x2 7x x 72 g1x

for y2 7y x y 72

, for .y2 7y x 0 ay2 by c 0 a 1, b 7, c x

y 7 7241x

21 7 494x

2

. Choose “-” not “+” since .y 7 494x

2 y 72

tg1x 7 494x

2

`If f (x) = x+3 then f -1(x) = x-3.

If g(x) = x/2 then g -1(x) = 2x.If h(x) = then h -1(x) = x2 for x > 0. x

Note how the graph of f is related to the graph of f -1.

f f-1

-1

-1g

g h

h

By the Theorem, y = f -1(x) iff f (y) = x. Thus the graph of

y = f -1(x) is the graph of f (y) = x which is just the graph

of f (x) = y with x and y interchanged. Interchanging xand y reflects the plane around the major diagonal y = x.Hence

THEOREM. The graph of y = f -1(x) is the reflection of the

graph of y = f (x) across the major diagonal y = x.

For each function, draw the three graphs y = f (x), y = x, y = f

-1(x) on the same coordinate system.

`f (x) = x3 `f (x) = -x

3

DEFINITION. f is 1-1 (“one-to-one”) iff x = y implies f (x) = f (y).

`f (x) = 3x is 1-1 x = y ˆ 3x = 3y f (x) = x

2 is not 1 = -1 but (-1) 2

= 12.

THEOREM. The following are equivalent:v f has an inverse

v f is 1-1 v no horizontal line intersects its graph more than once.

`Which of the following functions has an inverse?

THEOREM. The domain of f -1 is the range of f. The range of

f -1 is the domain of f .

Proof. The reflection around the major diagonal whichcarries the graph of f to the graph of f

-1 also carries thedomain of f to the range of f

-1 and the range of f to thedomain of f

-1.

Stated in full, the inverse is the compositional inverse. For addition, the inverse in negation. For multiplication, the inverse is the reciprocal.

1(6). , find . fx 2x12x f 1x

2y12y x2 y x1 2y2 y x 2xyy 2xy x 2y x2

2x1

f 1x x22x1

1(6). , find . Hint, use the quadratic formula.fx x2 3x for x 3/2 f 1x

y2 3y x for y 3/2

y2 3y x 0

y 3 941x

21

f 1x 3 94x

2

2(4). In (a), (b), the dotted line is the main diagonal, the solid line is the graph of f (x). If f -1(x) exists, draw its graph.If not, draw a horizontal line which proves it is not 1-1. (a) (b)


2. In (a), (b), the dotted line is the main diagonal, thesolid line is the graph of f (x). If f -1(x) exists, draw itsgraph. If not, draw a horizontal line which proves it is not1-1.

(a) (b)

Here are two examples from the lecture.

` = . Find .f x x 1x 1 f 1x

Interchange variablesf y x

Now solve for y.y 1y 1 x

y 1 xy 1y 1 xy x

y xy x 1y1 x x 1

y x 11 x x 1

x 1

tf 1x x 1x 1

` for . Find .gx x2 7x x 72 g1x

is for . Hencegx y x2 7x y x 72

is gy x

for . The equation is quadratic in y.y2 7y x y 72

As usual for quadratics, move things to the left.y2 7y x 0

This is for .ay2 by c 0 a 1, b 7, c x

y 7 72 41x

21 7 49 4x

2

. Choose “-” not “+” since .y 7 49 4x2 y 7

2

tg1x 7 49 4x2

1a. , find . chk=6, 8 symbols.f x x 2x 3 f 1x

How to reduce 9 symbols to 8 symbols: .2x 51 x 2x 5

x 1

1b. , find . chk=6, 6 symbols.gx x2 4x x 2 g 1xAnswer has one “x”, no “+”.

If doesn’t involve x, it is wrong. If it has it isg1x wrong.

Math 140 Classwork 7 Score ___/6For worked examples of similar problems, see practice quiz 7.

§2.8 230:1-6, 31-50.

1(2). .gt 1t 1

(a) Find g 1tchk=2

(b) Find the inverse g 12chk=1

and the reciprocal 1/g 2chk=5

2(3). . f x 13 x 2

(a) Find .f 1xchk=9

(b) Verify that f f 1x x

and that f 1fx x

(c) Graph, on same coordinatey f x, y x, y f 1xsystem.

y=x

3(3). . f x 1x

(a) Find f 1xchk=1

(b) Verify that f f 1x x

and that f 1f x x

(c) Graph, on same coordinatey f x, y f 1xsystem.

y=x

4(1). . Find . Omit the range & domain.f x 2x 3x 4 f 1x

chk=9

5(1). Given the graph of , graph .hx y h1x

y=xh

§2.8 230:7-12, 15,16. In 6, 7, 8, graph the function. Circle “1-1” if 1-to-1;circle “not 1-1” if not.6(1). 1-1? not 1-1?y x

7(1). 1-1? not 1-1?y x

8(1). 1-1? not 1-1?y 1 x3

Math 140 Hw 7 Name ___________________________________ Score____/ 13

A. fx 3x 1(a) Find f 1x

(c) Graph, on same coordinatey f x, y x, f f 1xsystem.

B. . f x x 1

(a) Find .f 1x


y=x

C. . f x x2x3

(a) Find the domain and range of f.

(b) Find f 1x

(c) Find the domain and range of f -1.

D. . Find . f x 2x3 1 f 1x

E. Given the graph of , graphgx

y=xg(x)

(a) (b) y g1x y g1x 1(c) (d) y g1x 1 y g1x(e) (f) y g1x y g1x

In E, F, G, graph the function. Circle “1-1” if 1-to-1;circle “not 1-1” if not.

F. 1-1? not 1-1?y x2 1

G. 1-1? not 1-1?y 1/x

H. 1-1? not 1-1?y x3

AnswersA. f 1x 1

3 x 1B. for x>0f 1x x2 1C. (a) dom: , 3 3,, rng: (, 1 1,

(b) f 1x 3x 2/x 1(c) dom: , 1 1,, rng: (, 3 3,

D. f 1x 3 12 x 1

E. Not shown.F. Not 1-1. G. Is 1-1 H. Is 1-1


y=x

A. f x 3x 1(a) Find f 1xSwitch x and y in to getf x y

ifff y x3y 1 x3y x 1y 1

3 x 1Answer: f 1x 1

3 x 1

(c) Graph, on samey f x, y x, y f 1xcoordinate system.Answer:

y=xf(x)

f (x)-1

B. . f x x 1(a) Find .f 1x

ifffy xNote that x is >0 since square roots >0.y 1 x

y 1 x2

for x > 0y x2 1Answer: for x>0f 1x x2 1


f(x)f (x)

-1

C. . f x x2x3

(b) Find f 1x ifffy x

y2y3 xy 2 y 3xy 2 xy 3xy xy 3x 2y1 x 3x 2y 3x 2/1 x 3x 2/x 1Answer: (b) f 1x 3x 2/x 1

D. . Find . f x 2x3 1 f 1x ifffy x

2y3 1 x2y3 x 1y3 1

2 x 1y 3 1

2 x 1

Answer: f 1x 3 12 x 1

E. Given the graph of , (a) graph gx g1x

y=xg(x)

g (x)-1

In F and G , graph the function. Circle “1-1” if 1-to-1;circle “not 1-1” if not.

F. y x2 1A horizontal crosses it twice, t it is not 1-1

G. y 1/xNo horizontal line crosses the graph twice, t it is 1-1.


Math 140 Lecture 8 DEFINITION. A quadratic function is a degree-2 polynomial

, . y ax2 bx c a 0The graph is a parabola.v If a > 0, the horns point up. v If a < 0, the horns point down. v If |a| > 1, the parabola is narrower than y = x2. v If |a| < 1, the parabola is wider than y = x2.

Find the roots by factoring or using the quadratic formula:

. No roots if .x b b24ac

2a b2 4ac 0COMPLETING THE SQUARE THEOREM. Every quadratic

function may be written in the form: y ax x02 y0

where (x0, y0) is the vertex (nose) of the parabola. Proof. Factor the a out of the ax2+bx part of ax2+bx+c .

Complete the square. Anything which is added mustalso be subtracted to preserve equality.

EXAMPLES

v Find the roots (they are the x-intercepts). v Write in completed square form: . y ax x02 y0

v Graph. On the graph list both coordinates of thevertex.

` y 12 x 12

Roots: x = -1 vertex: (-1, 0)y 1

2 x 12 0

(-1, 0)

` y 2x2 2x y 2x2 2x 2xx 1,

Roots: x 0, 1 Factor out coefficient of x

2.y 2x2 x -1,;by2,ƒ(-1/2),square,ƒ (1/4)y 2x2 x 1

4 2 14

The added 1/4 is multiplied by 2, thus so is the subtracted 1/4.. vertex: (1/2, -1/2)y 2x 1

2 2 12

(1/2,-1/2)

0 1

`y x2 2x 1

Roots: x b b24ac

2a 2 44

2 22 11

2 1 2 y x2 2x 1 1 1

y x 12 2

. vertex: (-1, -2)y x 12 2

(-1,-2)

Word problemsv Draw the picture. Indicate the variables in the picture.v Write the given equations which relate the variables. v Solve for the wanted quantities. List Given and Answer.

`The perimeter of a rectangle is 10 feet. Express the area A in terms

of the width x. Picture:

xy

10 = perimeterA = area

Given: 10 = 2x + 2y

A = xy

Want A in x, need y in x, need an equation in x and y:

2x 2y 10 t x y 5

ty 5 x

tA xy x5 x

Answer: A = square feet.x5 x

`The corner of a triangle lies on the line y = . Express the area4 x

A and perimeter P of the triangle in terms of the base x.

Given: y 4 x

A xy2

P x y z

z2 x2 y2

Want A in x, need y in x, need an equation in x and y:

Since , y 4 x A xy/2 x4 x/2

Answer: A x4x2

Want P in x, have y in x, need z in x, need equation in x, y and z:

z x2 y2 x2 4 x2 x2 16 8x x2

tP x y z x 4 x 2x2 8x 16

Answer: .P 4 2x2 8x 16`The area of an isosceles triangle is 16. Express the height of the

triangle in terms of its width x.

h

x

16 = area

Given: 12 xh 16

Want h in x. Answer: h 32x

xy

y = 4-x

A = area

P = perimeter

z

4&5(8). Write the distance between the point (1, 0) and apoint (x, y) on the curve in terms of x.y x3

You get credit for drawing the picture indicating the variables, forlisting the given facts (explicit and implicit) and for the answer.Picture: Draw the picture. On the picture indicate your variables.

(1,0)

(x,y)

d

Given: Write the equations which relate the variables.

d x 12 y 02

y x3

Answer: Want d in y. Have d in x and y. Need y in x (see given).y x3

d x 12 y2

x2 2x 1 x32

x6 x2 2x 1

Recall. Theorem. x2 ax x a2 2 a

2 2

3(8). . Write in completed-square form. y 2x2 4x 1Find the vertex, intercepts, graph.The vertex must be an order pair with ()'s, e.g., (3, 4) not just 3, 4.The completed-square form must be . Instead of y ax b2 c

write 4x 32 3 4x 32 3

Completed square: 2x2 4x 1 2x2 2x 1 2x 12 1 1 2x 12 2 1 2x 12 1

vertex:1,1

x-intercepts:y 2x2 4x 1

x 4 16 421

22 4 8

4 2 2

2y-intercept:

y 202 40 1 1Draw the graph. Label the vertex with its coordinates.


1

(-1,-1)

Recall. Theorem. x2 ax x a2 2 a

2 2

3. . Rewrite in the completed-square form . Hint, can be negative y 2x 42 3 ax x02 y0 x0, y0

3. . Find the vertex, intercepts, graph.y 2x2 8x 3

Do the “horns” of the parabola point up u or down n ?

Leave the constant 3 alone. Factor the 2 out of then complete the square using the theorem above.2x2 8x

If your equation looks like , rewrite it in the completed-square form . ax x02 y0 ax x02 y0vertex = You must use “( )”. E.g., vertex= (3,4), not vertex = 3,4. 7 symbols.

x-intercept(s)? Set y = 0.

Either factor or, if not possible, use the quadratic formula or set the completed square2x2 8x 3 x b b24ac2a

form equal to 0. No roots if the radical is undefined.7 or 8 symbols counting as 1 symbol. chk=5 or 7 or 9.

y-intercept? Equation has 3 symbols

Draw the graph. Label the vertex with its coordinates.

Math 140 Classwork 8 Score____/5

§2.5 201:5-18.v Write the function in the form: y ax h2 k.

E.g., rewrite as y x 12 y x 12 0.v List the roots. One has no roots, one has 1, two have 2.

v Graph. On the graph mark the vertex and give both coordinates.

1(3). y = y x 22

chk=4,2 roots:

2(3). y = y 2x 22 4chk=10,0 roots:

3(3). y = y x2 6x 1Radical roots , chk=6, 4 roots:

4(3). y = y 3x2 12xchk=10, 4 roots:

§2.6 210: 1-12.v In 5 and 6, draw the picture and indicate your variables

on the picture. v Write the given equations which relate the variables. v Solve for the wanted quantities. 5(4). A rectangle is inscribed in a circle of diameter 12.

(a) Express the perimeter P as a function of the width x.(b) Express the area A as a function of the width x.P has 11 or 12 symbols, A has 8.

Given:

(a) P =

(b) A =

6(4). The top of a right triangle lies on the curve asy xshown. (a) Express the area A of the triangle in terms of x. (b) Express the perimeter P of the triangle in terms of x.A has 5 symbols, P has 10.

Given:

(a) A =

(b) P =

7(2). The product of two numbers is 16. Express the sum Sof the squares of the two numbers as a function of oneof the numbers. S has 9 symbols.

S =

Math 140 Hw 8 Name ___________________________________ Score____/ 22

P=perimeter

A=area

y

x

v Write the function in the form: y ax h2 k.E.g., rewrite as y x 12 y x 12 0.

v List the roots. v Graph. On the graph mark the vertex and give both coordinates. A. y x2 4x

B. y 1 x2

C. y x2 2x 3

D. y x2 6x 2

v Draw the picture and indicate your variables on the picture. v Write the given equations which relate the variables. v Solve for the wanted quantities. E. (a) The perimeter of a rectangle is 16. Express the area of the

rectanble in terms of the width x.(b) The area of a rectangle is 85. Express the perimeter of therectangle in terms of the width x.

F. P = (x, y) is a point on the curve .y x2 1(a) Express the distance of P from (0,0) as a function d(x) of x.(b) Express the slope of the line segment from (0,0) to P as afunction m(x) of x.

G. A piece of wire is py inches long. It is first bent into a circle. (a) The wire is bent into a circle. Express the area A(y) of the circlein terms of y. (b) The wire is bent into a square. Express the area A(y) of thesquare in terms of y.

H. The sum of two numbers is 16.(a) Express the product of the two numbers in terms of a singlevariable.(b) Express the sum of the squares of the two numbers in terms of asingle variable.

AnswersA. , roots: 0,4, vertex: (2,-4).x 22 4B. , roots: +1, vertex: (0,1).x 02 1C. , roots: -1,3, vertex: (1,-4).x 12 4D. , roots: , vertex: (3,11). x 32 11 3 11E. (a) (b) 8x x2 2x 170/xF. (a) (b) x4 3x2 1 x2 1/xG. (a) sq. inches (b) sq. inchesy2/4 2y2/16H. (a) (b) 16x x2 2x2 32x 256


v Write the function in the form: y ax h2 k.E.g., rewrite as y x 12 y x 12 0.

v List the roots. v Graph. On the graph mark the vertex and give both

coordinates.

C. y x2 2x 3roots: -1, 3y x 3x 1-2 ƒ½(-2)=-1ƒ(-1)2=1y x2 2x 3

y x2 2x 1 3 1 x 12 4vertex: (1,-4)y x 12 4

(1,-4)

3-1

Answer: , roots: -1,3, vertex: (1,-4).x 12 4

D. y x2 6x 2

. roots: y 6 62412

21 6 442 3 11

3 11. -6ƒ½(-6)=-3ƒ(-3)2=9 y x2 6x 2

y x2 6x 9 2 9vertex: (3,11)y x 32 11

(3,11)

Answer: , roots: , vertex: (3,11). x 32 11 3 11

v In 2 and 4, draw the picture and indicate your variables on thepicture.

v Write the given equations which relate the variables. v Solve for the wanted quantities.

F. P = (x, y) is a point on the curve .y x2 1(a) Express the distance of P from (0,0) as a function d(x) of x.(b) Express the slope of the line segment from (0,0) to P as afunction m(x) of x.

Given

y x2 1d x 02 y 02

m y0x0

(a) Want d in x, need y in x but we have y = x2+1d x2 x2 12 x2 x4 2x2 1

Answer: d x4 3x2 1 d x2 3x2 1

(b) Want m in x Answer: m y0

x0 x21x m x2 1/x

G. A piece of wire is py inches long. It is first bent into a circle.(a) The wire is bent into a circle. Express the area A(y) of the circlein terms of y. (b) The wire is bent into a square. Express the area S(y) of thesquare in terms of y.

Given Let C= circumference (a) C=py, C=2pr, A=pr2

(b) C =py, C=4x, A=x2

(a) Want A in y, need r in y.2r y, r y/2

Answer: sq. inchesA r2 y2

2 y2

4 A y2

4

(b) Want A in y, need x in y.4x y, x y/4

. Answer: sq. in.A x2 y/42 2y2/16 A 2y2/16

H. The sum of two numbers is 16.(a) Express the product of the two numbers in terms of a singlevariable.(b) Express the sum of the squares of the two numbers in terms ofa single variable.

Let x and y be the two numbers. Let P be the product and S thesum of the squares.Given: x y 16

P xyS x2 y2

(a) Want P in x, need y in x.x y 16y 16 x

Answer: P xy x16 x P 16x x2

(b) Want S in x, need y in x.S x2 y2 x2 16 x2 x2 256 32x x2

Answer: S 2x2 32x 256


(x,y)

(0,0)

(a) (b)

A P

Math 140 Lecture 9 See inside text’s front cover for area and volume formulas.

`A 6'x6' tarp forms the top of a pup tent. Write theheight h of a pup tent as a function of the floor area A.

h 6b/2 b/2

3

Given: A 6b, h2 b2

2 32

Want A in h. Need b in h.

b2

2 32 h2, b2 9 h2 , b 2 9 h2

. A 6b 62 9 h2 12 9 h2

`The height of a can (right circular cylinder) is three timesthe radius.

r

hS = curved surface area

hp2 r

Given: h = 3r, S = 2prh (a) Express the curved surface area as a function of theradius. 4 symbols

S 6r2

(b) Express the radius as a function of the curvedsurface area. 5 symbols

, . r2 S6 r S

6

`Three sides of a 500 square foot rectangle are fenced.Express the length of the fence as a function of side x.

x

y

Area = 500f = length of fencex

Given: xy 500 f 2x yWant f in x, t need y in x, tneed an equation in x and y.

, feet. 8 symbols + unitsy 500x f 2x 500

x

The iClicker quiz and the classwork for Lecture 10 ishard. Study both in advance.

Graphing polynomials A polynomial graph is smooth: no breaks, no sharp corners.For an expanded polynomial with axn the termaxn ... c

of highest degree: axn is the leading term, a is theleading coefficient, and n is the degree. The y-interceptis the constant term c.

` , y 3x4 x2 5

leading term = -3x4, leading coeff.= -3, degree= 4, constant term = -5.

` , y x x3

leading term = -x3, leading coeff. = -1, degree = 3, constant term = 0.

y= x y= -xy= x 2 2

y= x4

y= x3 3y= -x

v For large x (near +5), graph looks like the leading term axn.

v As x goes to 5, y goes to +5 if a > 0, to -5 if a < 0. v Graphs of odd degree go to +5 in one direction, -5 in

the other, like y = x3, y = -x3. v Graphs of even degree either go to +5 in both

directions or to -5 in both directions, like y = x2, y = -x2.

Use the factored form to get the roots and their degrees(the degree of a root is the exponent of its factor).

v At roots of degree 1, the graph crosses x-axis like y = x or y = -x.v At roots of odd degrees 3, 5, 7, ..., the graph crosses the x-axis

like y = x3 or y = -x3.v At roots of even degrees, 2, 4, 6, ..., the graph touches but doesn’t

cross the x-axis, like y = x2 or y = -x2.At a root, the graph looks like the graph of the root’s factor.The sign of the other factors determines the + or .To get the leading term of a factored polynomial, replace each factor

by its leading term. To get the constant term (y-intercept), set x = 0.When graphing, find the x and y-intercepts and also calculate a key

value in each of the key intervals before, after, and between roots.

`Graph . y x 234 x2x 12

-2 is a root of degree 3, 4 is a root of degree 1, ½ has degree 2.

lead term = (x)3(-x)(2x)2 = -4x6, degree = 6, lead coeff .= -4.

constant (y-intercept): set x = 0, y = (2)3(4)(-1)2 = 32.

Key values in the key intervals (-5,-2], [-2,½], [½,4], [4,5):f(-3) = -343, f(-1) = 45, f(2) = 1152, f(5) = -27783.

41/2-2

32

deg=1deg=2deg=3

100

-200

-300

1000

The dotted lines indicate the portion of the graph around f(2)= 1152which exceeds our coordinate limits.

4&5(10). A square is inscribed inside a circle. Write the area of the inscribed square as a function of the circle’s area.Picture: Indicate variables in the picture.

x

xr

r A=circle area

S = square area

Given: Write the equations which relate the variables. A r2

S x2

x2 x2 2r2

Want S in A. Have S in x. Need x in r and then r in A.2x2 4r2

x2 2r2

r2 A/S x2 2r2 2A

Answer: S 2A

4&5(12). Four pieces of some length x are cut from a 4 ft length of wire. The four pieces make the four sides of asquare. The remaining piece of wire is bent into a circle. Write the area of the square as a function of the circle’s area.Picture: Indicate variables in the picture.

S=square area A=circle area

x

x

x

r

x

x x x x 4-4x

Given: Write the equations which relate the variables. S x2

A r2

4x 2r 4

Want S in A. Have S in x. Need x in r and r in A.4x 4 2rx 1 1

2rr2 A/r A/S x2 1 1

2r2

1 12 A/ 2

1 12 2A/ 2

Answer: include units sq. ft.S 1 1

2 A 2

A common error involves solving for the wrong variable. If and you are to write x in terms of y, the answer is2x 3y not . x 3/2 y y 2/3 x

The most common error is forgetting the units (one point loss on an exam).


Classwork 10 is hard, work on it before class.4&5. The volume of a cone is . The curved surface (or side) area of the cone is . Suppose the volume1

3r2h r r2 h2

is 2 cubic inches. Write the curved surface area as a function of the radius r.

Picture: Draw the picture. On the picture indicate your variables. (Is the volume a variable? Should have three variables.)

Given: Write the equations which relate the variables. (Should have two given equations.)

Answer: Is your answer a formula in h or r? It should involve p, 36, a radical and a fraction. Remember the units. 11 to 14 symbols after “=”.


§2.6 210: 13-18,23b,24a,25a,26a,29a,30a,31a,33a,34a.Draw the picture; indicate your variables on the picture. Write the equations (the “given”) which relate the variables. Solve for the wanted quantities.

1(3). Let 2s be the length of the side of an equilateraltriangle.(a)(2) Express the height of the triangle as a function ofs. 3 symbols, chk=3

(b)(1) Express the area of the triangle as a function of s.4 symbols

2(3). The height of a right circular cylinder is twice theradius. Express the volume as a function of the radius

4 symbols, chk=5

3(3). For the rectangle pictured, express the area as afunction of x.

y=10-x2

(x,y)

A=area

xx

Given:

Want A in x. 7 symbols, chk=5 or 7

4(4). For the shaded triangle below, find the area A andthe perimeter P.

x

y1

A=area

P=perimeter

Given:

(a) Want A in x, need y in x. 8 symbols, chk=5

(b) Want P in x, have y in x. 9 symbols, chk=4

5(3). Three sides of a rectangle are fenced with 500 feet offencing. Express the area of the rectangle as a functionof x.

x x

y

A = area

Given:

Want A in x, need y in x. 8 symbols, chk=7 or 9

§3.1 262:11-22. Graph and on the graph mark the x andy-intercepts. Label the degrees of the roots. No credit if thegraph isn’t smooth. E.g., the graph of |x| isn’t smooth.

6(2). y x 23

7(2). y x 3x 2x 1

Check your graphs at https://www.desmos.com/calculator

Math 140 Hw 9 Name _________________________________ Score ____/ 20

Draw the picture; indicate your variables on the picture. Write the equations which relate the variables. Solve for the wanted quantities.

A(3). Express the area of an equilateral triangle as afunction of a side x.

B(3). The height of a right circular cylinder is twice theradius. Express the radius as a function of the volume.

C. The volume of a right circular cylinder is 12p. (a) Express the height as a function of the radius.

(b) Express the total surface area as a function of theradius.

D. The volume V and the surface area S of a sphere ofradius r are given by the formulas and V 4

3 r3

. Express V as a function of S.S 4r2

E. The hypotenuse of a right triangle is 20 cm. Expressthe area of the triangle as a function of the length x ofone of the legs.

F. In the figure. Express AB as a function of x.

x 45

A

B

Graph and on the graph mark the x and y-intercepts.Label the degrees of the roots. No credit if the graphisn't smooth. E.g., the graph of |x| isn't smooth.

G. y x 43 2

H. y 2x 54

I. y 12 x 15

J. y x 13 1

K. y x 2x 1x 1

L. y 2xx 2x 1

AnswersA. A 3 x2/4B. r 3 V/2C. (a) (b) hr 12/r2 Sr 2r2 24/rD. VS S S /6 E. Ax 1

2 x 400 x2

F. dx x 4 x2 25 /xSee Hw 9 worked examples for remaining answers.


Draw the picture; indicate your variables on the picture. Write the equations which relate the variables. Solve for the wanted quantities.

D. The volume V and the surface area S of a sphere ofradius r are given by the formulas and V 4

3r3

. Express V as a function of S.S 4r2

Given: , V 43r3 S 4r2

Want V in S, need r in S.4r2 Sr2 S

4

r S4

V 43r3 4

3S

4

3

Answer: VS S S /6

E. The hypotenuse of a right triangle is 20 cm. Expressthe area of the triangle as a function of the length x ofone of the legs.

GivenA 1

2 xyy2 x2 202

Want A in x, need y in x y2 400 x2

y 400 x2

A 12 xy 1

2 x 400 x2

Answer: Ax 12 x 400 x2

F. In the figure. Express AB as a function of x.

x 45

A

B

h

Let h be the height pictured.Given: Pythagorean TheoremAB2 x 42 h2

Similar triangleshx4

5x

Want AB in x, need h in x.

h 5x4x

AB x 42 h2 x 42 25x42

x2

= = x 4 1 25x2 x 4 x225

x2

Answer: dx x 4 x2 25 /x

Graph and on the graph mark the x and y-intercepts.Label the degrees of the roots. No credit if the graphisn't smooth. E.g., the graph of |x| isn't smooth.

G. y x 43 2Start with the graph of , shift right 4 units, then downx3

2 units. The y-intercept is -66. There is one root x = of degree 1.4 3 2

H. . Start with (looks like ), shift left 5y 2x 54 x4 x2

units, reflect vertically around the x-axis, stretchvertically away from the x-axis by a factor of 2.

I. . Start with (looks like ), shift left 1y 12 x 15 x5 x3

units, compress vertically toward the x-axis.

J. . Start with , shift right 1 unit, reflecty x 13 1 x3

vertically around the x-axis, shift down 1 uint.

K. y x 2x 1x 1The y-intercept is 2, the roots are x = 2, 1, -1, all ofdegree 1.

-1 deg 1 1 deg 1

2 deg 1

L. y 2xx 2x 1The y-intercept is 0, the roots are x = 0, 2, 1, all ofdegree 1.

0 deg 1 1 deg 1

2 deg 1


20

x

y

Math 140 Lecture 10 Rational functions and their graphs

DEFINTION. A rational function is a ratio of twopolynomials. It is reduced if the top and bottom have nocommon factors.

Like polynomials, rational functions have smooth graphs.But they may have asymptotes.

In the graphs below, x = 0 (the y-axis) is the verticalasymptote, y = 0 (the x-axis) is the horizontal asymptote.

Let v.a. and h.a. abbreviate vertical and horizontal asymptotes.

y=1/x

y= -1/x y= -1/x

y=1/x

y= -1/x

y=1/x 2

2 3

3

y=0

y=0

x=0

x=0

x=0

y=0

x=0

y=0

x=0

y=0

y=0

x=0v.a.

v.a.

h.a.

h.a.

v For odd degree vertical asymptotes, one side goes to

+5, the other to -5. See 1/x and -1/x.

v For even degree vertical asymptotes, both sides go to

+5 or both go to -5. See 1/x2 and -1/x2.

DEFINITION. For rational functions, the leading term is thereduced ratio of the leading terms of the top andbottom.

Recall, to get the leading term of a factored polynomial, replace

each factor to its leading term and then simplify.

Rational functions:

Leading terms:

Hor. asymptote:

For a reduced rational function: v x-intercepts (roots) occur where the top is 0.

If the root has degree n, the x-intercept looks like that of y = xn or y= -xn.

v If the bottom is 0 at a, then x=a is a vertical asymptote.If the factor has degree n, the vertical asymptote looks like that of y=1/xn or y = −1/xn.

v As xƒ +5, the graph resembles the graph of the

leading term which is either a constant b, a fractiona/xn or a term axn of positive degree.

(1) If a constant b, then y = b is a horizontal asymptote.

(2) If it is a/xn, then y = 0 is a horizontal asymptote.

(3) If it is axn, there is no horizontal asymptote.

Graph. On the graph mark the x and y-intercepts. Mark

the vertical and horizontal asymptotes with their

equations ( y = a or x = a ) .

a is a key number iff . f(a) = 0 or f(a) = undefinedKey intervals lie before, between and after key numbers. In each key interval, calculate a key value.

. y =2x−6

2x3−8x2

reduce and factor: = = = 2(x−3)

2(x3−4x2)

(x−3)

x3−4x2

x−3

x2(x−4)

y-intercept: none x-intercept: 3 (deg 1)

vertical asymptotes: x = 0 (deg 2) x = 4 (deg 1)

lead term: 2x2x3 =

1x2

horizontal asymptote: y = 0

key values: f (-1) = 4/5, f (1) = 2/3, f (7/2) = -4/49, f (5) = 2/25

0

x=4

4y=0h.a.

v.a.

x=0v.a.

3

. y =2x3−8x2

2x−6

reduce and factor: x2(x−4)

x−3

y-intercept: 0 x-intercepts: 0 (deg 2)

4 (deg 1)

vert. asymp.: x = 3 (deg 1)

lead term: 2x3

2x = x2

hor. asymp.: none

key values: f (-1) = 5/4, f (1) = 3/2, f (7/2) = -49/4, f (5) = 25/2

0

x=3

4

v.a.

3

x1−x

x2+13x2

(x−1)(5x+2)

(x−3)2(1−2x)3

(1−2x)3

2x−6

−1 13

−5

8x3 −4x2

y = −1 y =13 y = 0 none

6(a)(11) . Graph, intercepts, asymptotes. Note: asymptotes are lines not numbers, write VA x=0, HA y=0, not VA 0, HA 0y x 1x1 x

h.a.: y = 0lead. term: -1/x

v.a.: x = 0, x = 1 degree 1, 1

y-intercept = nonex-intercept = -1 degree=1

Draw the graph. Indicate the asymptotes with dotted lines.

y=0 hax=1 va

x=0

-1

va

Graph pieces have been moved and exaggerated. The center piece has been moved down to be visible.The vertical scale of the left piece has been exaggerated.On tests and homework, you must show all pieces. If you use a graphing calculator for homework, you may have to shrink the coordinate system to see all pieces.

(b)(12) . Graph. List intercepts, asymptotes. Again, asymptotes are lines, indicate them with equations, not numbers. y x1 x

x 1

h.a.: y = nonelead. term.: -x

v.a.: x = -1 degree 1

y-intercept = 0x-intercept = 0, 1 degrees 1, 1

Draw the graph. Indicate any horizontal./vertical asymptote.

x=-1 va1

0


6(a) . y x 22

x2 2x x 22

xx 2

Key values: f (x) = (x-2)2 / [x(x+2)] Plot a value (not just +, -), for each of the four key intervals. f ( -3 ) = 25/3

f ( -1 ) = (-1-2)2/(-1)(-1+2) = (-3)2/-1 = -9

f ( 1 ) = (1-2)2/1(1+2) = (-1)2/3 = 1/3

f ( 3 ) = 1/15

Horizontal asymptote: y =

Leading Term:

Vertical asymptotes: x = degree = x = degree =

y-intercept (set x = 0) : y =

x-intercepts: x = degree =

Draw the graph. Indicate the horizontal asymptote and

two vertical asymptotes with dotted lines.Label the asymptotes with their equations.Label the root with its x-coordinate. The graph has three pieces. One piece crosses the

horizontal asymptote.Graphs never cross vertical asymptotes but they can cross horizontal asymptotes as does your graph. There is only one root, the graph does not cross the x-axis

anywhere else. The root has an even degree and, like , the graph touches but does not cross the x-axis.x2

Math 140 Classwork 10 Score ___/8 Hard.

§3.1 262:5-10, 23-30. Graph and on the graph mark the x and y-intercepts. Label thedegrees of the roots. No credit if the graph isn't smooth.

1(2). There are two turning points.y x3 9x roots:

key values:

3-3

-10

10

1 2-1-2

2(2). There are two turning points.y x 1x 42

roots: key values:

48

1216

-4-8

-12-16

1 42 3 5-2

§3.6 313:11-14, 33-46. Graph. On the graph mark the x and y-intercepts. Mark the verticaland horizontal asymptotes with their equations (y=a or x=a). Justwriting "a" won't do. Graphs must be smooth.

3(4). One y-intercept; one x-intercept.y x1x1

roots: vert. asym: leading term: hor. asym:

1-1

1

2

-1-2 2

-2

4(4). There is just one intercept.y 1x22

roots:

vert. asym:

leading term:

hor. asym:

key values:

2

-1

-2

1 2 3 4-1-2-3

1

5(4). y xx1x3

Two vertical, one horizontal asymptote. One intercept. roots:

vert. asym:

leading term:

hor. asym:

key values:

1 2 3 4-1-2-3

1

2

-1

-2



Graph and on the graph mark the x and y-intercepts. Label thedegrees of the roots. No credit if the graph isn't smooth.

A. y x3 4x2 5x

B. y x3 3x2 4x 12

C. y x3x 2

D. y 2x 1x 43

E. y x 12x 1x 3

Graph. On the graph mark the x and y-intercepts. Mark the verticaland horizontal asymptotes with their equations (y=a or x=a). Thegraphs must be smooth.

F. y 2/x 3 roots:

vert. asym:

leading term:

hor. asym:

G. y x 3/x 1 roots:

vert. asym:

leading term:

hor. asym:

values:

H. y 4x 2/2x 1 roots:

vert. asym:

leading term:

hor. asym:

values:

I. y 1/x 22

roots:

vert. asym:

leading term:

hor. asym:

values:

J. y 3/x 22

roots:

vert. asym:

leading term:

hor. asym:

values:

K. y 1/x 23

roots:

vert. asym:

leading term:

hor. asym:

values:

L. y 4/x 53

roots:

vert. asym:

leading term:

hor. asym:

values:

M. y x/x 2x 2 roots:

vert. asym:

leading term:

hor. asym:

values:

AnswersFor answers to B33, E39, G15, J21, M27, see Hw 10

worked examples. For the others, see the next 2 pages. Continued on next two pages.



A. y x3 4x2 5x

-1 deg 1

0 deg 1 5 deg 1

y intercept=0

C. y x3x 2

-2 deg 1 0 deg 3

y intercept=0

D. y 2x 1x 43

1 deg 1

4 deg 3

y intercept=128


F. y 2/x 3 roots: none

vert. asym: x 3

leading term: 2/x

hor. asym: y 0

x=3 v.a.

y=0 h.a.

H. y 4x 2/2x 1 roots: x 1/2 vert. asym: x 1/2leading term: 2 hor. asym: y 2

x=-1/2 v.a.

y= 2 h.a.

1/2

I. y 1/x 22

roots: none vert. asym: x 2leading term: 1/x2

hor. asym: y 0

x=2 v.a.

y=0 h.a.

Continued on next page.

Math 140 Hw 10 Recommended problems, answers

K. y 1/x 23

roots: none

vert. asym: x 2

leading term: 1/x3

hor. asym: y 0

x= -2 v.a.

y=0 h.a.

L. y 4/x 53

roots: none

vert. asym: x 5

leading term: 4/x3

hor. asym: y 0

x= -5 v.a.

y=0 h.a.



B. y x3 3x2 4x 12For large x, graph looks like x 3.y-intercept = constant term = -12Factored form: y x 2x 2x 3Roots: x =-3, -2, 2 each with degree 1.

-3 deg 1

-2 deg 1 2 deg 1

y intercept=-12

E. y x 12x 1x 3lead term = x2xx x4

constant term = 1213 3roots: x = -1 deg 2, 1 deg 1, 3 deg 1.


G. y x 3/x 1 roots: x = 3 vert. asym: x = 1leading term: 1 hor. asym: y = 1 values: f(0) =3, f(2) =-1, f(4) =1/3

x=1 v.a.

y=1 h.a.

3

J. y 3/x 22

roots: none vert. asym: x = -2leading term: 3/x2

hor. asym: y = 0 values: f(0)=3/4, f(-3)=3/25

x=-2 v.a.

y=0 h.a.

M. y x/x 2x 2 roots: x = 0 vert. asym: x = -2, x = 2leading term: -1/x hor. asym: y = 0 values: f(-3)=3/5, f(-1)=-1/3, f(1)=1/3, f(3)=-3/5

x=-2 v.a.

y=0 h.a.

x=2 v.a.


Math 140 Lecture 11 Exponential functions DEFINITION. An exponential function is of the form y = bx

with the base b > 0. b0 1, b1 b, b2 b b, ...bn 1/bn

, the nth root of b b1/n n bbp/q bp1/q bp1/q b1/qp

EXPONENT RULES Examples bnm bnm 523 525252 56

bnbm bnm 5253 5 55 5 5 55

, bn

bm bnm 53

57 54 154

57

53 54

abn anbn 2535 65

ab n an

bn 23 5 25

35

Simplify to an integer or a single exponent bn.

` 21/2

23/2 2

1/23/2 21/24/2 2

12 2 21 2

` 2 3 3 2 3 3 23 8

` (32 5 )(32 5 ) 32 5 2 5 34 81

` 81

81 811 811 82

PROPERTY. If b = 1, . Hence is 1-1. bx by x y bx

Gateway problems with the variable in the exponent: putboth sides above a common base then equateexponents.

` 27x 9 33x 32

t t3x = 2, tx = 2/3 33x 32

` 8z1 32/ 2 8z1 2521/2

23z1 25 12

23z3 292

3z 3 92 6z 6 9

6z 3 z 1/2

` , t z8 32/ 2 z8 25/212 25 1

2 292 z 2

916

Note, this is different, the variable is in the base rather than the exponent. Raise both sides to the 1/8 power.

means a is approximately equal to b. a bFact: , i.e., 210 is approximately equal to 103. 210 103

`Estimate 250 and 2-20. 250 = (210)5 (103)5 = 1015 2-20 = (210)-2 (103)-2 = 10-6

The graph of y = 1x is the horizontal line y = 1. Otherwise, the graph of y = bx v has y-intercept 1 but no x-intercept, v it goes to 5 in one direction, v it has the horizontal asymptote y = 0 in the other.

For b > 1, the graph of bx is like the graph of 2x as below. For 0 < b < 1, the graph is like (½)x.

y=2x

xy=(1/2)

1

y=0 hor asymp

`Graph Reflect 2x across y-axis. y 2x

`Graph Shift right 1 unit, down 1 unit. y 2x1 1

`Graph Reflect in y-axis, shift right 1.y 21x

21x 2x1 2x1, 2x 2x 2x1

The function ex DEFINITION. ex is the unique exponential function bx

whose the tangent at (0, 1) has slope 1.

Fact: Again, means approximately equal. e 2.7

Thus 2 < e < 3. Hence the graph of ex lies between the graphs of 2x and 3x.

Similarly, e-x lies between 2-x and 3-x.

y = e x

-x y= e

1

y=0 h.a.

slope = 1

x-intercept none y-intercept 1 hor. asym. y = 0 domain (-5, 5) range (0, 5)

`True or false? RECALL: e 2.7false since e 1 e 2.7 1 1true since e2 9 e2 2.72 32 9

9(3). Simplify (a)(b) to at most 3 symbols.(a) (c) 5 2 2

273

33

31/2 331/2 36/21/2 35/2

(b) (d) 5 1 3

5 1 3

3 24 21/3

22 21/32 21/36/3 25/3

5 3 3 5 2 3

( 5 2) 3 5 3

7(2). Estimate as a power of 10. 280

280 (210 )8

(103 )8

1024

10(5). Graph . Find, if any, the horizontal and vertical asymptotes. y 5 10x2

Asymptotes must be equations, not numbers. E.g. x =3, y =3, not just 3.horizontal asymptote y = 5

vertical asymptote none

Graph 5 10x2 10x2 5

Gateway problems. Solve for x

A. 5 26x 5

25

B. x 3 5

25First simplify the right hand side to a single power.

.525 51/2

52 51/22 51/24/2 53/2

A. 51/226x 53/2 51/226x 53/2 513x 53/2

1 3x 3/2 2 6x 3 6x 5 x 5/6

B. x1/23 53/2 x3/2 53/2 x3/22/3 53/22/3 x 51 1/5


7. Estimate as a power of 10: 250 First write as a power of , e.g. .250 210 210n

Then use the approximation .210 103

4 symbols, chk=7.

9. Simplify to at most 3 symbols.

(a) 3 6 6

Initially, leave the radical alone.3Recall that nested exponents multiply.1st get rid of the radical in the exponent.2nd get rid of the radical in the base.3The answer is a two-digit integer. 2 symbols, chk=9.

(b) 3 1 2

3 1 2

Bring the denominator upstairs by negating its exponent.Simplify the combined exponent.Get rid of the radical . 3The answer has an integer in the base, a radical in the exponent. 3 symbols, chk=5.

Gateway problemsSolve for x First write as a power of 11. Hint: .121

113 11 111/3, 112 121, 11 111/2

A. 3 11 2x2 12111

4 symbol fraction, chk=8.

B. 3 x 4 12111

5 symbols, chk=19.

Math 140 Classwork 11 Score ___/5

§3.6 313:47-56. Graph. On the graph mark the x andy-intercepts. Mark the vertical and horizontal asymptotes with theirequations (y=a or x=a). The graphs must be smooth.

1(4). 1 point graph, 1 root, 1 v.a., 1 h.a.y 2xx 12

roots: vert. asym: lead term: hor. asym:

2(1). Estimate as a power of 10. 290 4 symbols

§1.2 21:31-70In 3, 4, 5: simplify to an integer or single exponent bn.*9. 5 2 2 51/2 2 2 5 2

3(1). 2 2 2

single digit

4(1). 102

102

3 symbols, chk=5

5(1). 3 4

2 symbols, chk=9

6(1). Solve for t: 2t 14

2 symbols, chk=2

7(1). Solve for y: 23y1 24 symbol negative fraction, chk=7

§4.1 337:19-32. 8(3). Graph and on the graph mark the x-intercept,

y-intercept, and horizontal asymptote.y 3x 3

9(3). Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote.y 3x 3

10(3). Graph and on the graph mark, if any, thex-intercept, y-intercept, and horizontal asymptote.

y ex



A. Estimate as a power of 10:(a) (b) 230 250

In B-E: simplify to a number of the form bn.

B. 5 3 3

C. 41 2 41 2

D. 24

21

E. 52 2

F. Solve for y: 312y 3

G. Solve for z: 3z 9 3

Answers

A(a). (b) 230 109 250 1015

B. 125 C. 16D. 8 E. 5 2

F. G. y 1/4 z 5/2

H. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote.y 2x 1

I. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote.

y 3x 1

J. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote.y 2x1

K. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote. y ex2

Answers

See Hw 11 worked examples.


A. Estimate as a power of 10:(a) 230

230 2103 1033 109

Answer: 230 109

In B-E: simplify to a number of the form bn.B. 5 3 3

= 1255 3 3 5 3 3 53

Answer: 125

C. 41 2 41 2 41 2 41 2 41 2 1 2

42 16Answer: 16

D. 24

21

24

21 241 241 23 8Answer: 8

E. 52 2

52 2 5 2 2 5 2 2 5 2

Answer: 5 2

F. Solve for y: 312y 3 iff 312y 3 312y 31/2

iff iff 1 2y 1/2 2 4y 1 4y 1iff y 1/4Answer: y 1/4

G. Solve for z: 3z 9 3 iff 3z 9 3 3z 3231/2

iff iff 3z 35/2 z 5/2Answer: z 5/2

H. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote. y 2x 1Start with , reflect around x-axis, shift up 1.2x

y=1 h.a.

0

I. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote. y 3x 1Start with , reflect across y-axis, shift up 1.3x

y=1 h.a.

J. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote.

y 2x1

Start with and shift right 1.2x

y=0 h.a.0

K. Graph and on the graph mark the x-intercept,y-intercept, and horizontal asymptote. y ex2

Start with , shift right 2 units, reflect across x-axis.ex

y=0 h.a.


Math 140 Lecture 12 Exam 2 covers Lectures 7 -12. Study the recommended exercises.Review area, circumference, volume formulas - inside front cover.RECALL. The graphs of .ex and ex

y = e x

-x y= e

1

y=0 h.a.

slope=1

x-intercept none y-intercept 1vert. asym. none hor. asym. y = 0

domain (-5,5) range (0,5),

`Graph . y ex1 1Give the domain, range, intercepts and asymptotes.

1

-1 y=-1h.a.

x-intercept x = 1y-intercept y =

1e 1 1

2.7 1 .63hor. asym. y = -1domain (-5,5) range (-1,5).

LogarithmsAssume b > 0, b = 1. Thus bx is 1-1 and it has an inverse.DEFINITION. , the log of x to the base b, is thelogbx

inverse of the exponential function bx. ln(x), the natural logarithm, = = the inverse of ex.logexNote, “ln” is “el-n” not “one-n”, not “eye-n”.

Inverses act in opposite directions and inverses cancel. t iff . y = ln(x) iff . y logbx by x ey x

ln(ex) = x. logbbx x

eln(x) = x.blogbx x

If we have ln(bx) instead of ln(ex), then ln and bx don’tcompletely cancel and ln(ex) = x becomes:

. The exponent comes down to the outside.lnbx x lnb

FACT. e0 = 1 ˆ 0 = ln(1). e1 = e ˆ 1 = ln(e)

`Simplify to a rational..log55 5 log55 5 log55151/2 log553/2 3

2

.log218 log2

18 log2

123 log223 3

1st solve . , , log82 8x 2 23x 21 3x 1 x 13

.log82 log881/3 1/3

Write in log form. Take the appropriate log of both sides and cancelinverses. Or use the fact that exponentiation and logarithms act inopposite directions. .

` 2x sends 3 to 823 8 Since 2x and log2(x) act in opposite directions,

log2(x) sends 8 to 3.3 log28

` 52x1 6Take the log of both sides.

log552x1 log56 so 2x 1 log56

Solve for x, write the answer using logarithms. `5x2 4

x2 log54Leave answers in exact form, no decimals.x log54

` e3t1 8 3t 1 ln8 3t ln8 1

t 13 ln8 1

` Use natural logarithms and method 2.32x 5x1

ln32x ln5x1 2x ln 3 x 1 ln 5 x2 ln3 ln 5 ln5

= See Practice Exam 2, prob. 11b.x ln5/2 ln3 ln 5ln 5

ln 9/5

Since they are inverses, the graph of is thelogbxreflection of bx across the major diagonal and ln(x) isthe reflection of ex.

y=ex

1 y=ln(x)

1

x=0 v.a.

y=0 h.a.

For the graph of ln(x)x-intercept 1 y-intercept nonevert. asym. x = 0. hor. asym. nonedomain (0,5) range (-5,5).

`Graph . y lnx 1 1

x=-1vert asym

1

-..63

y-intercept: ln(0+1)+1 = ln1 +1 = 0+1 = 1x-intercept: ln(x+1)+1 = 0 iff ln(x+1) = -1 iff x +1 = e-1 iff

x = 1/e - 1 1/(2.7)-1 -.63vert. asym: x = -1 hor. asym: nonedomain: (-1, 5) range: (-5,5).

`Graph . refl. across y, right 1, refl. across x, up 1.y 1 ln1 xThe 5 log properties of Lecture 13 are needed for Exam 2.

9(2). Simplify to at most 3 symbols. log8 2

log8 81/3

log881/31/2

log881/6 1/6

11(6). . Solve for x using natural logarithms. 5x1 2x1

ln(5x1) ln(2x1)(x 1) ln5 (x 1) ln 2x ln5 ln5 x ln 2 ln2x ln5 x ln 2 ln 2 ln5xln5 ln 2 ln 2 ln5

x ln 2 ln 5ln 5 ln 2

ln 10ln 5/2

10(8). . For an asymptote, you must give an equation, not just a number.y 3 ln2 x

Horizontal asymptote: none Vertical asymptote: x = 2

domain = (-5, 2)y-intercept = 3-ln(2)

Rewritten value: 3 ln2 x lnx 2 3 lnx 2 3 reflect vertically in x, reflect horizontally in y right 2 up 3

Draw the graph. Indicate the asymptote with a dotted line.

Draw the graph. Indicate the asymptote with a dotted line.

x=2

Math 140 Discussion Lecture 12 Exam 2 next week Exam points/time 16pts 12mins

9. Simplify to at most 3 or 4 symbols. (a) 4 symb, chk=3log9

13

(b) 3 symb, chk=5log9 3

Hint for (a): write the argument as a power of the base: solve .9x 13

Replace the argument of the logarithm with this power of its base: log9

13 log99x

Use the fact that . Do (b) similarly.logbbx x

11. . Solve for x using natural logarithms. 3x1 2x4

Take the natural logarithm of both sides.

Bring the exponents inside the logarithm outside.They become coefficients on the outside.Use .logxn n logx

Get terms involving x on the left, everything else (the constants) on the right.

Factor out x and then divide to solve for x. 16 symbol fraction.Use the following properties to simplify the answer to a ratio of two logarithms, e.g, . chk=17ln 40

ln5/3

logb xn = n logb x logb x/y = logb x - logb y logb xy = logb x + logb y Log properties


Exam 2 next week.§4.1 337:33-38. Graph. List the intercepts, the asymptotesand the domain and range. Write “none” if they don’t exit.1(2). Two “none” ’s, three “5” symbols.y ex1

x-intercept y-intercept

hor. asym. vert. asym.

domain range

In 2, 3, and 4, circle true, or false.

2(1). true? false?e2 4



§4.2 349:3-14. Rewrite each equation (do not solve). One answer has a fraction.

5(1). Rewrite using logarithms.1125 53

6(1). Rewrite using exponents.log525 x

8(1). Rewrite using logarithms.e3t 8

7(1). Rewrite using exponents.ln3t 8

§4.2 349:15-24. Simplify to a rational number. Two fractions, one digit.

9(1). log16164

chk=5

10(1). log10(10)

11(1). log28 2 chk=9

§4.2 350:41-46, 49-56.Graph. List the intercepts, asymptotes, domain and range.12(2). Two “none” ’s, three “5” symbols.y lnx



domain range

13(2). y lnx



domain range

Simplify to a single-digit integer.

14(1). lne

15(1). lne2

16(1). lne2

§4.4 366:1-22. Solve for x. Write the exact answer (nodecimal answers) using a logarithm with the appropriate base.

Two answers have fractions, all have a logarithm

17(1). x =102x1 145

18(1). x =10x2 40

19(1). t =et1 16

20(2). Solve for x using ln. . x = 74x 213x

Fractional answer with 3 ln's.

Math 140 Hw 12 Name ____________________________ Score ____/ 24 Exam 2

Rewrite each equation using logarithms rather thanexponentials (no decimal answers).

A. 9 32

B. 1000 103

C. 73 343

D. 2 21/2

Simplify to a rational number.

E. log927

F. log41/32

G. log55 5

Graph. List the intercepts, asymptotes, domain and range.H. For answer, see Hw 12 worked examples.y log2x



domain range

Answers

A. log39 2B. log101000 3C. log7343 3D. log2 2 1/2E. 3/2F. -5/2G. 3/2

I. See Hw 12 worked examples.y log3x 2 1



domain range

J. y lnx e



domain range

Simplify.

K(a). lne4

(b). ln1/e

(c). ln e

Solve for x. Write the exact answer (no decimal answers)using a logarithm with the appropriate base.

L. x =10x 25

M. x =10x2 40

N. t =e2t3 10

Answers

J. Shift the graph of e units to the left. lnxK(a). 4 (b). -1 (c). 1/2L. M. or log1025 1 log104 log1040

N. 3 ln10/2


Rewrite each equation using logarithms rather thanexponentials (no decimal answers).

A. 9 32

If exponential to the base 3 sends 2 to 9, then itsinverse, , does the opposite: it sends 9 to 2. log3

Answer: log39 2

The basic principle here is: y bx iff logby x

B. 1000 103

Answer: log101000 3An alternate way to do these problems is to take the

logarithm of both sides and then use the fact that logbbx x

C. 73 343Take of both sides. log7

log773 log7343t 3 log7343Answer: log7343 3

D. 2 21/2

log2 2 log221/2

Answer: log2 2 1/2

Simplify to a rational number. For these, use the principle: .logbbx xE. log927

= 3/2log927 log933 log991/23 log993/2

Answer: 3/2

F. log41/32 = -5/2log41/32 log425 log44

1/25 log445/2

Answer: -5/2

G. log55 5log55 5 log55 51/2 log553/2 3/2Answer: 3/2

Graph. List the intercepts, asymptotes, domain and range.H. y log2x

x-intercept -1 y-intercept nonehor. asym. none vert. asym. x = 0domain (-5, 0) range (-5, 5)

I. y log3x 2 1

x-intercept x = 5 y-intercept nonehor. asym. none vert. asym. x = 2domain (2,5) range (-5, 5)

Simplify.Use the principle: ln(ex) xK(a). lne4

lne4 4Answer: 4

(b). ln1/eln1/e lne1 1Answer: -1

(c). ln e ln e lne1/2 1/2Answer: 1/2

Solve for x. Write the exact answer (no decimal answers)using a logarithm with the appropriate base.

L. 10x 25log10x log25Answer: x log1025

M. 10x2 40log10x2 log40x2 log40Answer: log1040

N. e2t3 10lne2t3 ln102t 3 ln102t 3 ln10Answer: t 3 ln10/2


Math 140 Lecture 13 Recall: Since and are inverses of each other:logbx bx

v iff logbx y x by

v logbbx xv blogbx x

logb1 = 0, ln(1) = 0 b0 1logbb = 1, ln(e) = 1 b1 bSimplify. `lne2 lne4 ln 1 lne ln2

2 4 0 1 ln2 3 ln2

èln2 eln 4 e1 e0 e2

= 2 4 e 1 e2 3 e e2

Most properties of logarithms are the same as forexponentiation but with

( _ ) n;xexchanged with

n( _ )-+the symbols

Assume b > 0, b = 1. On the log side, assume x, y > 0.

bnx bxnlogb xn = n logb x

bxy bx/bylogb x/y = logb x - logb ybxy bxbylogb xy = logb x + logb yEXPONENT PROPERTIESLOG PROPERTIES

Proofs. (In these proofs, the “iff”s are required, -1 if missing.) iff logbxy logbx logby

iff blogbxy blogbxlogby

iff xy blogbxblogby

iff true. xy xyLikewise for . logb

xy logbx logby

iff logbxn n logbx iff blogbxn bn logbx

iff xn blogbxn

iff xn blogbxn

iff true. xn xn

Combine into a single logarithm. ` 2 log10x log10y

log10x2 log10y log10x2y

`log2x 4 log2y log2x log2y4 log2

xy4

` lnx2 y2 lny3 a ln x2y2

y3a

Write as a sum / difference / multiple of the simplestpossible logarithms.

`logb 42y

y3a

logb2y

y3a

14

14 logb

2yy3a 1

4 logb2y logby3 a 1

4 logb2 logby logby3 a`ln 1

x2y2

lnx2 y21/2 12 lnx2 y2

Note: . The lastlogbx y logbx logby logbx yterm cannot be broken into simpler pieces.

SOLVING FOR x.v Get terms involving x on the left, the rest on the right.v Combine into a single logarithm. v Exponentiate both sides to the base of the logarithm. v Solve. v Delete solutions that give undefined logarithms

(logarithms with arguments <0). `log24x log23 log2x 2

log24x log2x 2 log23 log2

4xx2 log23

2log24x

x2 2log23

4xx2 3, 4x 3x 6, x 6Validity: for x = 6, both log24x log224 and log2x 2

are defined, thus the solution is valid. log28`lnx 1 2 lnx 1

lnx 1 lnx 1 2 ln[x 1x 1] 2

lnx2 1 2 elnx21 e2

x2 1 e2

x2 1 e2

x 1 e2

Validity: is not valid since x 1 e2

lnx 1 ln 1 e2 1 undefinedBut is defined since .ln 1 e2 1 1 e2 1 0ANSWER: The only valid solution is . x 1 e2

CHANGE OF BASE FORMULA. .logax logbxlogba

Proof: . logbxlogba

logbalogax

logba logax logba

logba logax

Èxpress in terms of log5: ln2

ln2 loge2 log52log5e

Èxpress in terms of natural logarithms ln: log5tlog5t

ln tln5

1(2). Combine into a single logarithm:lnxy x ln2 1

2 lnx y

lnxy ln2x ln x y ln xy

2x x y

2(2). Write as a sum and/or difference of multiples of :log5x, log5y 1, log5x 1

log5y 13

x x 14

log5y 13 log5x1/2 log5x 14

3 log5y 1 12 log5x 4 log5x 1

4(3). Express in terms logarithms to the base 2: ln13

log213log2e

3(6). Solve for x. Show your work. The solution(s) must be valid.Write "none" if neither solution is valid.lnx 4 ln6 ln1 x

lnx 4 ln1 x ln6lnx 41 x ln6x 41 x 6x x2 4 4x 6x2 3x 10 0x 5x 2 0x 5,2At the equationx 5

ln5 4 ln6 ln1 5holds with all terms definedt5 is validAt the equationx 2ln2 4 ln6 ln1 2has undefined termst-2 is invalidAns. x 5


Here is an example from the lecture. Your problem is in the next column.

`Solve for x: log24x log23 log2x 2

log24x log2x 2 log23

log24x

x 2 log23

2log24x

x 2 2log23

4xx 2 3

4x 3x 6x 6

Validity check for x = 6Substitute 6 into the original equation.

log24x log23 log2x 2This gives

log24 6 log23 log26 2All logarithms have positive arguments. Hence they are defined and the solution is valid.

3. Solve for x: . Show yourln1 x ln6 ln2 xwork. The solution(s) must be valid. Write "none" if neither solutionis valid.

Get the terms involving x on the left, everything else (the ) on the right.ln6

Combine the logarithms into a single logarithm.

Exponentiate both sides to eliminate the logarithms.

Multiply out. For this problem, the result should be aquadratic equation. To solve, get everything on the left, 0on the right. Then factor.

There should be two solutions: a positive digit and anegative digit.

Determine the validity of each solution. Substitute each solution into the original equation.

ln1 x ln6 ln2 xCheck if the logarithms all have positive arguments.If so,the solution is valid. If not, the solution is invalid.


§4.3 356:1-12. Simplify. 1. lne3 lneSingle digit

2. eln3 eln2 elne

3 symbols, chk=5

§4.3 357:39-46. Combine into a single logarithm.3. 2 log10x 3log10y10 symbols, chk=6

4. lnx3 1 lnx2 x 113 symbols, 7 if simplified, chk=1 or 7

§4.3 357:13-34. Write as a sum / difference / multiple ofthe simplest possible logarithms.

5. logb3 x 3

x20 symbols, chk=6 or 7

6. ln 1x2 x 1

12 or 14 symbols, chk=5 or 6

§4.3 357:49-52. 7. Express in terms of log10.log2b13 symbol fraction, chk=4

8. Express in terms of natural logarithms, ln.log2108 symbol fraction, chk=3

§4.4 367:35-50. Solve for x.9. lnx lnx 1 ln12Warning: only the positive 1-digit solution is in the domain of ln.

10. lnx 1 2 lnx 19 symbol fraction, chk=6


Simplify. A. log1070 log107

B. log7 7

C. log3108 log33/4

D. 12 ln e

E. 2log25 3 log53 5

Combine into a single logarithm.F. log1030 log102

G. log56 log51/3 log510

H. logb4 3logb1 x 12 logb1 x


I. log10x2

1x2

J. ln x2

1x2

K. log10 9 x2

L. ln4x2

x1x13/2

M. Express in terms of log10.logb2

Express in terms of natural logarithms ln.N. .log106

O. log10e

P. Solve for x using ln. .5x 32x1

Solve for x.Q. log102x 4 log10x 2 1

R. log10x 3 log10x 2 2

S. log10x 1 2 log10x 1

Answers

A. 1 B. 1/2C. 4 D. 0E. 4 F. log1060

G. H. log520 logb41x3

1x3/2

I. 2 log10x log101 x2J. 2 ln x 1

2 ln1 x2K. 1

2 log103 x 12 log103 x

L. 12 ln2 x 1

2 ln2 x lnx 1 32 lnx 1

M. N. O. log102

log10bln 6

ln 101

ln 10

P. x ln3/2ln3 ln5Q. R. x 3 x 203/99S. x 3


Simplify. A. log1070 log107

log1070 log107 log10707 log1010 1

Answer: 1

B. log7 7log7 7 log771/2 1/2Answer: 1/2

C. log3108 log33/4log3108 log33/4 log3108 3

4 log327 3 log334 4Answer: 4

D. 12 ln e

12 ln e 1

2 ln e1/2 12

12 0

Answer: 0

E. 2log25 3 log53 5

=42log25 3 log53 5 5 3 log551/3 5 31/3

Answer: 4

Combine into a single logarithm.F. log1030 log102

log1030 log102 log1030 2 log1060Answer: log1060

G. log56 log51/3 log510log56 log51/3 log510= =log56

13 10 log520

Answer: log520

H. logb4 3logb1 x 12 logb1 x

logb4 3 logb1 x 32 logb1 x

logb4 logb1 x3 logb1 x3/2

logb41x3

1x3/2

Answer: logb41x3

1x3/2


I. log10x2

1x2

=log10x2 log101 x2= 2 log10x log101 x2Answer: 2 log10x log101 x2

J. ln x2

1x2

lnx2 ln1 x21/2

= 2 ln x 12 ln1 x2

Answer: 2 ln x 12 ln1 x2

K. log10 9 x2

log109 x21/2 log103 x3 x1/2

12 log103 x3 x 1

2 log103 x 12 log103 x

Answer: 12 log103 x 1

2 log103 x

L. ln 4x2

x1x13/2

ln4 x21/2 lnx 1 lnx 13/2

12 ln2 x2 x lnx 1 3

2 lnx 1Answer: 1

2 ln2 x 12 ln2 x lnx 1 3

2 lnx 1M. Express in terms of log10.logb2

logb2 log102log10b

Express in terms of natural logarithms ln.N. . log106 log106 ln 6

ln 10

O. log10e log10e ln eln 10

1ln 10

Solve for x.P. Solve for x using ln. .5x 32x1

iff iff5x 32x1 ln5x ln32x1

iff iff x ln5 2x 1 ln3 x ln5 2x ln3 ln 3 iff iffx ln5 2x ln3 ln3 xln5 2 ln3 ln3

x ln3/ln5 2 ln 3Answer: x ln3/2 ln3 ln5

Q. log102x 4 log10x 2 1iff log102x 4x 2 1iff iff 10log102x4x2 101

iff iff 2x 4x 2 10 2x2 8 10 iff 2x2 18 0 x2 9 0

iff x 3, 3-3 is not valid since x=-3 implies x-2 = -5 and so undefinedlog10x 2 log105 Answer: x 3

S. log10x 1 2 log10x 1iff log10x 1 log10x 12 0iff log10x 1/x 12 0iff iff x 1/x 12 100 x1

x12 1iff iff x 1 x 12 x 1 x2 2x 1iff iff x2 3x 0 x 0, 3x = 0 is invalid since x= 0 implies x-1= -1 and so undefinedlog10x 1 log101 Answer: x 3


Math 140 Lecture 14 Exponential growth

A bacteria colony starts with 10 bugs. Each bug splits intotwo bugs every hour. How many bugs are there after t hours?

10.2tt hours

......

(10.22).2 = 10.23

3

(10.2).2 = 10.22

2 10.21

100

Number of bugsNumber of hours

Exponential functions measure the size of a growing population,the amount of money in a compound interest account, the number ofatoms left after radioactive decay, etc. N(t) = the amount at time t.

BASE-e FORM LEMMA. Every exponential function can bewritten in the form

. Nt N0ekt

N0 = N(0) is the initial amount. k, the coefficient of t, is the growth constant. If k > 0, N(t) measures exponential growth. If k < 0, N(t) measures exponential decay.

FACT: Every a > 0 is a power of e: . a eln a

Example: 2 = .eln2

`Write the bug population in base-e form.Nt 10 2t

Nt 10 2t 10 eln 2t 10 eln2 t

Hence, in base-e form, . Nt 10eln2 t

Thus N0 = 10, the initial amount and k = ln(2), the natural log of the initial base. Ìf you deposit P dollars at 5% interest per year. How

much is there after one year?Answer: P .05P P1 .05 P 1.05`You put $4,000 in an account at 5% interestcompounded annually. Write the amount N(t) of money inthe account after t years in base-e form.

4000.(1.05)tt years

......

4000.(1.05)33

4000..(1.05).(1.05) =4000.(1.05)2

2

4000.(1.05)1

4,0000

Amount N(t) in accountNumber of years

Hence dollarsNt 40001.05t

4000eln1.05t 4000eln1.05 t

t . Nt 4000eln 1.05 t

tN0 = 4000 and k = ln(1.05).

Let = the initial amount, the amount at a laterN0 N1 time t, and k = the growth constant. Then

Nt N1N0ekt N1

ekt N1/N0t kt lnN1/N0

GROWTH EQUATION. kt lnN1/N0

For exponential work problems, first find k and find N(t) inbase-e form. Then answer the questions.

À bacteria colony starts with 1000 bugs. Four hourslater it has 5000 bugs. (a) What is the population two hours after the start? (b) How long will it take for the population to triple?

Solution: First find the growth constant k and Nt., N0 1000 t 4, N1 5000

Growth equation: kt lnN1/N0

k lnN1/N0

t ln5000/1000

4 ln 54

bugsNt N0ekt 1000e ln 54 t

(a) After two hours the population is bugs.N2 103e ln5

4 2 103e ln52

(b) When the population has tripled . N1 3N0 3000Growth equation: kt lnN1/N0ln 54 t ln3000/1000

ln5t 4 ln3

hourst 4 ln 3ln 5

Ìnitially a sample has 8 lbs of a radioactive substancewith a half-life of 5 days, i.e., half decays after 5 days.How many pounds remain after 3 days?

Solution: First find the growth constant k and Nt., N0 8 t 5, N1 8/2 4


k lnN1/N0

t ln4/8

5 15 ln 1

2

Nt N0ekt 8e 15 ln 1

2 t

N3 8e 15 ln 1

2 3 8e35 ln 1

2

Answer: lbs remain after 3 days. 8e 35 ln 1

2

This is a decay since ln(1/2) = ln(2-1) = -ln(2) is negative.

Suppose the initial amount is . Suppose that at a later time t the amount is . Then N0 N1Nt N1N0ekt N1ekt N1/N0

growth equationkt lnN1/N0

5. A compound interest account starts with $4000. 6 years later it is $6000. Show work; only 3 points per part without work. Remember the units (-1 point if omitted).

Given: N0 4000, t 6, N1 N6 6000, Nt N0ekt

(a) How much will there be in the account after 12 years? Write your answer in base-e form. Remember the units.

(b) When will the account have $8000? Remember the units.

First ignore the questions, first find k and Ntkt lnN1 / N0

k6 ln6000/4000

k ln6/46

ln3/26

Nt N0ekt 4000et ln3/2

6

(a) Amount after 12 years?

dollars N12 4000e12 ln3/2

6 4000e2 ln3/2

(this is the required base-e form, it simplifies to which is also accepted)4000eln3/22 4000 32 2 4000 9

4 $9000

(b) When will it be $8000? Here Nt N1 8000 kt lnN1 /N0ln3/2

6 t ln8000/4000

years t ln8/4ln3/2/6 6 ln 2

ln3/2


Here is an example from the lecture. Your problem is in the next column.`A bacteria colony starts with 103 bugs. Four hours laterit has 5x103 bugs.

First find the growth equation and base-e form.Initial amount: N0 103

Later amount: N1 5 103

Time: . t 4

Growth equation: kt lnN1/N0 k4 ln5 103/103 4k ln5

Growth constant: k ln 5

4

Base-e form: bugsNt N0ekt 103e ln 5

4 t

What is the population two hours after the start? people.N2 103e ln 5

4 2 103e ln 52

How long will it take for the population to triple? Now the later amount is triple of the original amount, = and t is the time for this tripled amount.N1 3 103

Growth equation: kt lnN1/N0ln 54 t ln3 103/103

ln5t 4 ln33

hours.t 4 ln 3ln 5

5. A compound interest account starts with $2000. 3years later it has $4000.

Initial amount: N0

Later amount: N1

Time: t


Growth constant: k

Base-e form: Nt N0ekt

(a) How much is there after 6 years? (in base-e form)9 symbols + the units, e.g., inches, feet, lbs, dollars, seconds, ... . N6

(b) When will there be $6000 in the account?


8 symbols followed by the units, chk=8t


§4.5 379:5-22. Write N(t) in base e form. Then solve the problem.Except for 2, give exact answers, not approximate decimal answers.

1(8). Initially, 2x104 bacteria are present in a colony.Eight hours later there are 3x104. Hint, first find the growthequation, then the growth constant, then Nt.

(a)(3) What is the population two hours after the start?chk=11 or 12 or 16 or 17 fraction must be reduced.

(b)(5) How long will it take for the population to triple?chk=16 Remember the units

3(6). The half-life of strontium-90 is 28 years. Initially asample has 10 grams.

(a)(3) How many grams will remain after 1 year?chk=17 or 18 Remember the units

(b)(3) After 10 years?chk=14 fraction must be reduced.

4(9). The half-life of radium-226 is 1620 years. Initiallythe sample has 2 grams.

(a)(3) How many grams will remain after 10 years?chk=14

(b)(6) Find the time required for 80% of the 2-grams todecay? Hint, at this time only .4 grams are left. chk=18 No Nodecimals in the answer. Remember the units

Math 140 Hw 14 Name ________________________________Score ____/ 23

Write N(t) in base e form. Then solve the problem.A. A colony of bacteria starts with 2000 bugs. Two hours

later it has 3800.

(a) Determine the growth constant k and . Nt

k =

N(t) =

(b) What is the population 5 hours after the start?

(c) How long will it take to reach 10,000?

B. The half-life of iodine-131 is 8 days. Initially a samplehas 1 gram.

k =

N(t) =

How many grams will remain after 7 days?

C. The half-life of sodium-24 is 15 hours. Initially thesample has 40 grams.

k =

N(t) =

(a) How many grams will remain after 48 hours?

(b) How long will it be until only one gram remains?

D. The half-life of plutonium-241 in 13 years. Initially asample has 2 grams.

k =

N(t) =

(a) How many grams will remain after 5 years?

(b) How long will it take for 90% of the sample to decay?Hint, when 90% has decayed, 10% remains.

Answers

A. Nt 2000e 12 ln 19

10 t

(a) k = (b) 12 ln 19

10 2000e 52 ln 19

10

(c) t = hr2 ln5/ ln19/10B. k = Ans. g1

15 ln 1/2 Nt e 18 ln 1

2 t e78 ln 1

2

C. k = 115 ln 1/2 Nt 40e 1

15 ln 1/2 t

(a) g 40e165 ln1/2

(b) hr15 ln1/40/ ln1/2D. k = 1

13 ln 1/2 Nt 2e 113 ln1/2 t

(a) g (b) yr2e 513 ln1/2 13 ln.1/ ln1/2


Write N(t) in base e form. Then solve the problem.

A. A colony of bacteria starts with 2000 bugs. Two hourslater it has 3800.

(a) Determine the growth constant k and .Ntkt lnN1 / N0

k2 ln3800/2000

Answer: k ln38/20

2 ln19/10

2

Nt N0ekt 2000eln19/10t

2

(b) What is the population 5 hours after the start?N5 2000e 1

2 ln 1910 5

Answer: bugs2000e 52 ln 19

10

(c) How long will it take to reach 10,000?kt lnN1 / N0

tln19/10

2 ln10000/2000

Answer: shhourst ln5

ln19/102

2 ln5

ln19/20

B. The half-life of iodine-131 is 8 days. Initially a samplehas 1 gram. kt lnN1 / N0

k8 ln.5/1

k ln1/28

Nt N0ekt e 18 ln 1

2 t

How many grams will remain after 7 days?N7 e 1

8 ln 12 7

Answer: gramse78 ln 1

2

C. The half-life of sodium-24 is 15 hours. Initially thesample has 40 grams.kt lnN1 / N0

k15 ln20/40

k ln1/215

Nt N0ekt eln1/2t

15

(a) How many grams will remain after 48 hours? N48 40e 1

15 ln12 48

Answer: grams 40e 165 ln 1

2

(b) How long will it be until only one gram remains? kt lnN1 / N0

t ln1/215 ln1/40

hourst ln1/40ln1/2

15

15 ln1/40

ln1/2

D. The half-life of plutonium-241 in 13 years. Initially asample has 2 grams.

kt lnN1 / N0 k13 ln1/2

k ln1/213

Nt N0ekt 2eln1/2t

13

(a) How many grams will remain after 5 years?N5 2e 1

13 ln 12 5

Answer: grams 2e513 ln 1

2

(b) How long will it take for 90% of the sample to decay?Hint, when 90% has decayed, 10% remains.kt lnN1 / N0

t ln1/213 ln.1/1

yearst ln1/10ln1/2

13

13 ln1/10

ln1/2


Math 140 Lecture 15 Angles, arcs, and radiansRECALL. A circle of radius r has circumference 2pr.

p 3.14. Unit circle circumference = 2pr = 2p1 = 2p.q and w are the Greek letters “theta” and “omega”.

DEFINITION. Suppose the vertex of an angle is at the centerof a circle of radius r. Let s be the length of the arc theangle intercepts on the circle. Then

q = s/r is the radian measure of the angle. For unit circles, the radius r = 1 and radian measure equals arc length: q = s.

`Radian and degree measures on the unit circle.

p/2

p/4

0 = 0, 360 = 2p

= 45

= 90

-p/4= -45

oo

oo o

o

1

180 = p

Clockwise angles ?are negative.

CONVERSION FORMULAS. 180o = p radians. Thus 1o = p/180 radians; 1 radian = 180/p degrees.

`Convert 100o to radians. 1.74 radians 100o 100

180 radians 59 radians

`Convert p/6 radians to degrees. .

6 radians 6

180

o 30o

When q is in radians, we can solve q = s/r for arclength: s = qr

`Find the length of a 30o arc on a circle of radius 12inches. First convert to radians. By the above 30o

= p/6 radians.t . s r

6 12 2 inches

`Find the degree measure of an angle which intercepts a5 inch arc on a circle of radius 12 inches.

radians 512 5

12180o

75o

Speed DEFINITION. If an object travels a distance d in time t, its

linear speed is d/t. If an object rotates through an angle q in time t, itsrotational speed is w = q/t.

THEOREM. If a point rotates around a circle of radius rwith rotational speed w, then its linear speed is wr.

Proof. If a point rotates through an angle q on a circle ofradius r in time t, then w = q/t and the distance d ittravels = the length of the arc it traces = qr . tits linearspeed = d/t = qr/t = (q/t)r = wr . We assume q is in radians.

`A point revolves around a circle of radius 3 feet at 10revolutions per minute.

(a) What is its rotational speed (in radians)? 1 revolution = 2p radians, tw = 10 revs/min = 10.2p radians/min. = 20p radians/min.

(b) What is its linear speed? Its linear speed = w r = (20p rad/min). (3 feet) = 60p feet/min.

Radians are dimensionless; you may delete dimensionless units.

Trigonometric functions The unit circle has radius one and center (0, 0). There are

four quadrants I, II, III, IV as pictured. An angle is in standard position if its vertex is the origin

(0,0) and its initial side is the positive x-axis. The otherside of the angle is the terminal side. The terminal side

intersects the unit circle at a point Pq.

P

qI

II

III IV

x(0,0)

q

DEFINITION. Given a standard-position angle of radian

measure q, let (x, y) be the coordinates of Pq.

1q y = sinq

x=cosq

P = (x,y)q

The six trigonometric functions of q are:

cot q = cos q / sin q cotangenttan q = sin q / cos q tangent

sec q = 1/cos q secantcos q = x cosine

csc q = 1/ sin q cosecantsin q = y sine

`Draw an angle of p/2 radians in standard position. Findthe six trigonometric functions. sin(p/2) = 1 csc(p/2) = 1cos(p/2) = 0 sec(p/2) = undef tan(p/2) = undef cot(p/2) = 0

`A point (x, y) on the unit circle is in the secondquadrant and y = . Find the six trig functions for q.3

4

y = . First, find x.34

(x, y) on the unit circle ˆ x

2 + y

2 = 1.

t .x2 1 y2 1 34 2 1 9

16 716

x= . (x,y) in the second quadrant ˆ x is 7 /4negative.tx = . 7 /4sin q = 3/4 csc q = 4/3 cos q = sec q = 7 /4 4/ 7tan q = cot q = 3/ 7 7 /3

3/4(x,y)

x

1q

9. (a)(2) Find the length of the arc intercepted by an angle of 40o on a circle of radius 3 feet. Remember units

40o 40o rad180o

radians 29

length r ft. 2

9 3 2/3

(b)(2) Find the number of degrees in an angle which intercepts a 4 foot arc on a circle of radius 12 feet.

radians 412 1

3 1 rad3

180o

rad 60o

10. A point rotates around a circle of radius 10 inches at 3 revolutions per second. (a)(2) Find its angular speed in radians per second. Give the exact answer using p. Remember to include the units.

3 revs/sec 3 2 6p radians/sec

(b)(2) Find its linear speed. Exact answer using p, include units.

r 6 10 inches/sec60


9. (a) Find the length of the arc intercepted by an angle of 30o on a circle of radius 50 inches.

First convert the angle to radians. Recall , hence 180o radians 1o

180 radians.

30o

Now use the formula

length r Reduce the fraction, remember the units, 5 symbols, chk=10. .

(b) Find the number of degrees in an angle which intercepts a 30 inch arc on a circle of radius 50 inches.

First find the angle in radians using the formula

s/r

Now convert the radian measure to degrees. 5 symbols, chk=9Since , .180o radians 1 radian 180

degrees

10. A point rotates around a circle of radius 50 inches at 30 revolutions per second. (a) Find its angular speed in radians per second. Give the exact answer using p. Remember to the units, chk=6.

First convert 30 revolutions per second to radians per second. 1 revolution = radians. 2

Hence 30 revolutions / second = _____ x ______ radians / second = _______ radians / second

(b) Find its linear speed. Exact answer using p, include units.

Use the formula: linear speed r Remember the units, 5 symbols, chk=3.


Remember to include any needed units, e.g., mi/hr, rad/sec.Except for 1,2, give only exact answers. No decimals.

§6.1 474:1-24.1. Convert 30o to radian measure

exact answer using p: 3 symbols, chk=6

two-place decimal answer: 3 symbols, chk=7

2. Convert 150o to radian measure

exact answer using p: 4 symbols, chk=11

two-place decimal answer: 4 symbols, chk=10

3. Convert 3p radians to degrees 4 symbols, chk=9

4. Convert 3p/2 radians to degrees 4 symbols, chk=9

§6.1 475:49-56, 77-80.5. Find the radian measure of an angle which intercepts a

2 cm arc on a circle of radius 3 cm? Exact answer.

3 symbols+units, chk=9

6. Find the length of an arc which is intercepted by a p/5radian angle on a circle of radius 12 cm? Exact answer.

5 symbols + units, chk=8

7. A point rotates around a circle of radius 20cm at 15revolutions/sec. (a) Find its angular speed w. Give the exact answer using p.

w = 3 symbols + units, chk=3

(b) Find its linear speed. Give the exact answer using p.

4 symbols + units, chk=6

8. Sketch p/6, -p/6 and -5p/6 in standard position, allon the same graph.

§5.2 416:3-8, 23-26.9. Find the six trigonometric functions of -p/2.

4 integers, 2 undefined

cot(-p/2) = tan(-p/2) =

sec(-p/2) = cos(-p/2) =

csc(-p/2) =sin(-p/2) =

10. Find the six trigonometric functions of 4p. 4 integers, 2 undefined

cot(4p) = tan(4p) =

sec(4p) = cos(4p) =

csc(4p) = sin(4p) =

11. A point (x, y) on the unit circle and on the terminalside of an angle q is in the fourth quadrant. Find the six trigonometric functions if x = 1/3. 4 answers have a radical, give exact answers, not decimals.

cot q = tan q =

sec q = cos q =

csc q = sin q =

§5.1 406:13-18.12. A point (x, y) on the unit circle and on the terminal

side of an angle q is in the first quadrant. Find the six trigonometric functions if x = 3/5. Rational answers, give exact answers, not decimals.

Use improper fractions, not mixed fractions. E.g. 3/2, not 1½.

cot q = tan q =

sec q = cos q =

csc q = sin q =

13. Complete the table. Mark “+” where the functions arepositive, “-” where they are negative.

tan q, cot q

sin q, csc q

-+cos q, sec q

Quad IVQuad IIIQuad IIQuad I

Math 140 Hw 15 Name ________________________________ Score ____/14

Remember to include any needed units, e.g., mi/hr, rad/sec. Exceptfor 5ab, give only exact answers. No decimals.

A. Convert 45o to radian measure

exact answer using p:

two-place decimal answer:

B. Convert 90o to radian measure

exact answer using p:

two-place decimal answer:

C. Convert p/3 radians to degrees

D. Convert 5p/3 radians to degrees

E. Find the length of an arc intercepted by a 4p/3 radianangle on a circle of radius 3 ft.? Exact answer.

F(a). A point rotates around a circle of radius 12 cm at 6revolutions/sec. Find it's angular speed w. Give the exact answer using p.

w =

(b). A point rotates around a circle of radius 12 cm at 6revolutions/sec. Find it's linear speed. Give the exact answer using p.

G. Find the six trigonometric functions of -3p/2.

cot(-3p/2) = tan(-3p/2) =

sec(-3p/2) = cos(-3p/2) =

csc(-3p/2) =sin(-3p/2) =

H. Find the six trigonometric functions of 0.

cot(0) = tan(0)=

sec(0) = cos(0)=

csc(0) = sin(0)=

I. A point (x,y) on the unit circle and on the terminal sideof an angle q is in the first quadrant. Find the six trigonometric functions if x = 1/3.

cot q = tan q =

sec q = cos q =

csc q = sin q =

J. A point (x,y) on the unit circle and on the terminal sideof an angle q is in the third quadrant. Find the six trigonometric functions if x = -3/5.

cot q = tan q =

sec q = cos q =

csc q = sin q =

AnswersA. p/4 .79B. p/2 1.57C. 60o

D. 300o

E. 4p ft.F(a). w = 12p rad/sec (b). 144p cm/sec

G.

cot(-3p/2) = 0tan(-3p/2) = undef

sec(-3p/2) = undefcos(-3p/2) = 0

csc(-3p/2) =1sin(-3p/2) = 1

H.

cot(0) = undeftan(0)=0

sec(0) = 1cos(0)=1

csc(0) = undefsin(0)= 0

I.

cot q = 1/(2l2)tan q = 2l2

sec q = 3cos q = 1/3

csc q = 3/(2l2)sin q = 2l2/3

J.

cot q = 3/4tan q = 4/3

sec q = -5/3cos q = -3/5

csc q = -5/4sin q = -4/5


Remember to include any needed units, e.g., mi/hr, rad/sec. Exceptfor 5ab, give only exact answers. No decimals.

A. Convert 45o to radian measure.

q = 45o 180

4 .79

exact answer using p: p/4

two-place decimal answer: .79

C. Convert p/3 radians to degrees.

3

180 60o

D. Convert 5p/3 radians to degrees.

53

180 300o

E. Find the length of an arc intercepted by a 4p/3 radianangle on a circle of radius 3 ft.? Exact answer.

s r 3 43

Answer: 4p ft.

F(a). A point rotates around a circle of radius 12 cm at 6revolutions/sec.

Find it's angular speed w. Give the exact answer using p.

1 rev = 2 radians6revs/sec 62rad/sec=12 rad/secAnswer: 12p rad/sec

(b). A point rotates around a circle of radius 12 cm at 6revolutions/sec. Find its linear speed. Give the exact answer using p.

By part (a), 12linear speed =r 1212 144Answer: 144p cm/sec

G. Find the six trigonometric functions of -3p/2.

The unit-circle point with angle -3p/2 is (0, 1).

Answer:

cot(-3p/2) = 0tan(-3p/2) = undef

sec(-3p/2) = undefcos(-3p/2) = 0

csc(-3p/2) =1sin(-3p/2) = 1

H. Find the six trigonometric functions of 0.

The unit-circle point with angle 0 is (1, 0).

Answer:

cot(0) = undeftan(0)=0

sec(0) = 1cos(0)=1

csc(0) = undefsin(0)= 0

I. A point (x,y) on the unit circle and on the terminal sideof an angle q is in the first quadrant.

Find the six trigonometric functions if x = 1/3. First find y. x2 y2 1y2 1 x2

. It is the + since (x,y) is in quadrant 1.y 1 x2

y 1 x2 1 1/32 1 1/9 8/9, since 8 = 4.2, y 2 2 /3 8 4 2 2 2

Answer:

cot q = 1/(2l2)tan q = 2l2

sec q = 3cos q = 1/3

csc q = 3/(2l2)sin q = 2l2/3

J. A point (x,y) on the unit circle and on the terminal sideof an angle q is in the third quadrant.

Find the six trigonometric functions if x = -3/5. First find y. y2 1 x2

. It is - since (x, y) is in quadrant III.y 1 x2

.y 1 3/52 1 9/25 16/25 4/5

Answer

cot q = 3/4tan q = 4/3

sec q = -5/3cos q = -3/5

csc q = -5/4sin q = -4/5


Math 140 Lecture 16Fri. = last day to withdraw. Keller 419A secretary will sign for me.

FACTS. Know the sin, cos and tan of: 0, p/6, p/4, p/3, p/2.

0

p/6=30

p/4=45p/3=60

p/2=90oo

o

o

o

1

(0,1)

(1,0)

l

l l

( 3/2,1/2)

(1/ 2, 1/ 2)

(1/ 2, 3/ 2) l

sin(0) = 0 cos(0) = 1 tan(0) = 0sin(p/6) = 1/2 cos(p/6) = tan(p/6) = 1/l33 /2sin(p/4) = * cos(p/4) = * tan(p/4) = 11/ 2 1/ 2sin(p/3) = l3/2 cos(p/3) = 1/2 tan(p/3) = l3sin(p/2) = 1 cos(p/2) = 0 tan(p/2) = undef.

* is also ok. 2 /2

PROOFS. All points are on the unit circle.0 = 0o: The point for 0 = the rightmost point = (1, 0).

p/2 = 90o: The point for p/2 = the top point = (0, 1).

For the other angles, we have a right triangle whose hypotenuse is a radius of length 1.

p/6 = 30o: This is a 30o- 60o right triangle. Thus the smallleg is ½ since it is half the hypotenuse which is 1. Let xbe the third side.

11/2

x30

30o 60

60

o

oo

1x

Then x2 + (1/2)2 = 12

x2 + 1/4 = 1

x2 = 3/4

x = l3/2 t sin(p/6) = sin(30o) = 1/2 and cos(p/6) = 3 /2

p/3 = 60o: This is the 30o triangle again but with the 30o

and 60o angles swapped. Swapping sides 1/2 and 3 /2gives: sin(p/3) = sin(60o) = and cos(p/3) = 1/2. 3 /2

p/4 = 45o: If one angle of a right triangle is 45o, so is theother acute angle and the triangle is isosceles.

Let x be the side length.

1

xx

Then x2 + x2 = 12

2x2 = 1 x2 = 1/2 x = . 1/ 2tsin(p/4) = sin(45o) = = cos(p/4)1/ 2

DEFINITION. For any angle in standard position (vertex = the

origin, initial side = the positive x-axis), its reference angle isthe acute positive angle (in [0, p/2]) between the terminalside and the x-axis (positive or negative, whichever is nearest).

ref

ref ref

THEOREM. The sin and cos of q equals the sin and cos of its reference angle except for the sign which is determined by q’s quadrant.

`Rewrite in terms of reference angles, then evaluate. cos(120o) = -cos(60o) = -1/2 quadrant IIsin(120o) = sin(60o) = quadrant II3 /2cos(3p/4) = -cos(p/4) = quadrant II1/ 2sin(3p/4) = sin(p/4) = quadrant II1/ 2cos(-3p/4) = -cos(p/4) = quadrant III1/ 2sin(-3p/4) = -sin(p/4) = quadrant III1/ 2

`List three angles (in radian measure) whose cos is ½. p/3, -p/3, 5p/3

NOTATION. sin q means sin( q); sin2 q means (sin q)2.

THEOREM. Since sin q and cos q are the legs of a righttriangle of hypotenuse 1,

sin2 q + cos2

q = 1

`sin q = -2/3 and p < q < 3p/2. Find cos q and tan q. (-2/3)2

+ cos2 q = 1

cos2 q = 1 - 4/9 = 5/9

cos q = 5 /3p < q < 3p/2 is in quadrant III, thus cos q < 0 t

cos q = 5 /3tan q = sin q/cos q = 2/3/ 5 /3 2/ 5

`sec q = -3 and sin q < 0. Find tan q. First find sin, cos.

cos q = -1/3 sin

2 q + (-1/3)2 = 1

sin 2

q = 1-1/9 = 8/9 sin q = . Since sin q < 0, 8 /3sin q = = . ttan q = . 8 /3 2 2 /3 2 2

`Simplify (cos q + cotq)(tan q + sec q) = cos q tan q+cos q sec q+cotqtan q+cotqsec q = cos sin

cos cos 1cos

cos sin

sincos

cos sin

1cos

= sin q + 1 + 1 + 1/sin q = sin q + 2+csc q.

Note: y2 + 3y + 2 factors into (y +1)(y + 2).

`Factor: csc2 q + 3csc q + 2 = (csc q +1)(csc q + 2).

Note: x2- y2 = (x- y)(x+y).

`Factor: sin2 B - cos2

B = (sin B-cos B)(sin B+ cos B).

16(5). Simplify to 5 symbols: 4 tan 55 cot 4

4 tan 55 cot 4

4 tan 551/ tan 4

4 tan 551/ tan 4

tantan

4 tan 55 4 tan tan

4 tan 54 tan 5 tan

tan

11. Suppose sin q = 3/5.

cos2 sin2 1cos2 3/52 1cos2 1 9/25cos 16/25cos 4/5

(a)(2) Find cos q if 0 < q < p/2.

0 < q < p/2tcos 4

5

(b)(3) Find tan q if p/2 < q < p.

p/2 < q < ptcos 4/5tan sin/ cos

3/54/5 3

4


11. Suppose sin q = -1/4. Find tan q if 3p/2 < q < 2p.

To find tan, we need sin and cos. sin is given. We need cos q. Use the fact that . cos 1 sin2Use the inequality 3p/2 < q < 2p to determine if “+” is “+” or “-”. Then find tan q = sin q / cos q.

In the two figures write the sines and cosinesrespectively of the listed angles. Some havealready been done. See Lecture 16.


p/2p/3

p/4

p/6

-p/6

-p/4

-p/3

3p/4

p= -p

5p/6

2p/3

7p/6= -5p/6

5p/4= -3p/4

4p/3= -2p/33p/2= -p/2

0=2p 1

0

cosines

1/2

-1/2

p/2p/3

p/4

p/6

-p/6

-p/4

-p/3

3p/4

p= -p

5p/6

2p/3

7p/6= -5p/6

5p/4= -3p/4

4p/3= -2p/33p/2= -p/2

0=2p 0

1

sines 1/2

-1/2

III

III IV

All are positiveSin is positive

Tan is positive Cosine is positive

ASTC

All Students Take Calculus

AS

T C

Last day to withdraw. The Keller 419A secretary signs for me.

The parts of problems 1-4 are ½ point each.

For each angle, sketch the reference angle, draw anarrow to indicate the angle's direction. A reference anglemust be positive and < p/2 = 90o.1(1). (a) 300o (c) -15o

2(1). (a) 5p/6 (d) 5p/4

§5.1 407:33-50. Rewrite the functions in terms of theirreference angles, then evaluate. E.g.,

= -½= -cos(60o)cos(120o)

3(4).

== sin(-150o)

= =sin 150o

== cos(-150o)

= =cos 150o

4(4).

= =sin(-2p/3)

= =sin(2p/3)

==cos(-2p/3)

==cos(2p/3)

5(1). List three angles (radian measure) whose sin is -½

q1 = q2 = q3 =

§7.1 533:1-10. 6(1). Simplify .cos2 sin2 2sincos2

11 symbols

7(1). Simplify (5-2tan q)/(2tan q-5) to an integer.

8(1). Factor 3sec2 b + 2sec b - 8.

9(1). Factor 16sin3 B - 9sin2

B.

§5.2 417:63-70. §6.2 484:17,18,21,22. 10(4). cos q = 5/13 and 3p/2 < q < 2p.

Find sin q 6 symbols, chk=7

Find cot q 5 symbols, chk=8

11(2). sec q = and sin q > 0. Find tan q. 13 /24 symbols, chk=5

Math 140 Hw 16 Name _____________________________ Score ____/21

For each angle, sketch the reference angle, draw an

arrow to indicate the angle's direction. A reference

angle must be positive and < π/2 = 90o.

A. (a) 110o (b) 60o

B. (a) 3π/4 (b) 7π/6

Rewrite the functions in terms of their reference angles,

then evaluate. E.g.,

= -½= -cos(60o)cos(120o)

C.

== sin(-210o)

= =sin 210o

== cos(-210o)

= =cos 210o

D.

= =sin(-4π/3)

= =sin(4π/3)

==cos(-4π/3)

==cos(4π/3)

E. List three angles (in radian measure) whose cos is -½

θ1 = θ2 = θ3 =

F. Simplify . sin�−cos�cos�−sin�

G. Factor . tan2� + 8 tan � − 9

H. Factor . 4 cos2B − 1

I. Factor 9 sec2B tan3B + 6 sec B tan2B

J33. sin θ = -3/5 and π < θ < 3π/2.

Find cos θ

Find tan θ

K. sin t = and π/2 < t < π. Find tan t. 3 /4

AnswersA., B. See Hw 16 Worked examples

C. −3

2 , −3

2 , −12 ,

12

D. −12 , −

12 , −

3

2 ,3

2

E. −2�3 ,

2�3 ,

4�3

F. -1

G. (tan � − 1)(tan � + 9)

H. (2 cos B + 1)(2 cos B − 1)

I. 3 sec B tan2B(3 sec B tan B + 2)

J. cos � = −4/5, tan � = 3/4

K. tan t = − 3/13

Math 140 Hw 16 Recommended problems.

For each angle, sketch the reference angle, draw anarrow to indicate the angle's direction. A referenceangle must be positive and < p/2 = 90o.

A. (a) 110o (b) 60o

70o 60o

B. (a) 3p/4 (b) 7p/6

p/4

p/6

Rewrite the functions in terms of their reference angles,then evaluate. E.g.,

C.

= 1/2= sin 30osin(-210o)

= -1/2= -sin 30osin 210o

= -l3/2= -cos 30ocos(-210o)

= -l3/2= -cos 30ocos 210o

D.

= l3/2= sin(p/3)sin(-4p/3)

= -l3/2= -sin(p/3)sin(4p/3)

= -1/2= -cos(p/3)cos(-4p/3)

= -1/2= -cos(p/3)cos(4p/3)

E. List three angles (in radian measure) whose cos is -½

23 , 2

3 , 43

F. Simplify . sin cos cossin

=cossin

cossin 1

G. Factor . tan2 8 tan 9

tan 1tan 9

H. Factor . 4 cos2B 1

2cos B 12cos B 1

I. Factor 9 sec2B tan3B 6sec B tan2B

3sec B tan2B3sec B tanB 2

J. sin q = -3/5 and p < q < 3p/2. p < q < 3p/2 implies q is in quadrant III impliescos q is negative.

cos 1 sin2 1 sin2

= 1 3/52 1 9/25 4/5

Find cos q 4/5

Find tan q = sin/ cos 3/5/4/5 3/4

K. sin t = and p/2 < t < p. Find tan t. 3 /4p/2 < t < p implies t is in quadrant II implies

is negativecos t

cos t 1 sin2t 1 3 /42

1 3/16 13 /4

tan t sin t/ cos t 3 /4/ 13 /4

tan t 3/13


Math 140 Lecture 17RECALL. sin2

q + cos2 q = 1

tan q = sin q/cos q cot q = cos q/sin qsec q = 1/cos q csc q = 1/sin q

Simplify.` cot seccos cos

sin1

cos cos cot

`sin x1

sin x 1sin x1

sin x 1 sin x1sin x

sin x1sin x 2 sin x1

1 2 sin x 1

DEFINITION. An identity is an equation which is alwaystrue.

`x2 +y2

= 1, y = 2x aren’t identities: they aren’t always

true.sin2

q + cos2 q = 1, xy = yx are identities, i.e., always true.

You have two ways to prove an identity: (1) Start from one side (usually the more complicated side) and

work your way to the other side. (2) Show that the equality is equivalent to “true”. The text uses (1); we’ll give examples of (2).

`Prove cot2 csc2 1

iff cos2sin2

1sin2

1 iff cos2 1 sin2

iff iff true.sin2 cos2 1

`Prove cos csc

sin sec

(by cross multiplying)iff cossec cscsin

true. iff cos 1cos

1sin sin iff 1 1 iff

Right triangles

q q1

ya

c

bx

The trig functions were defined on the unit circle wherethe hypotenuse of the right triangle had length 1. Forthe angle pictured, sin q = y. Any other right trianglewith an angle q is similar to this unit-circle triangle.Hence the ratios of the corresponding sides are equal. Thus sin q = y = y/1 = a/c = side opposite/hypotenuse.

THEOREM. For any right triangle with an acute angle q:sin q = opposite/hypotenuse* cos q = adjacent/hypotenuse tan q = opposite/adjacent

* More precisely, “length of the side opposite/length of thehypotenuse”. We’ll abbreviate this to just: opp/hyp.

In a right triangle, the two complementary small anglesadd up to 90o. If one is q, the other is 90o

- q or p/2 - q.THEOREM.

sin 2 cos sin90o coscos 2 sin cos90o sin

Proof. In the picture below,

q

90-qc

a

b

sin q and are both a/c. cos90o cos q and are both b/c. sin90o

6

5q

x-1

9x2

B90- q

Fig. 1 Fig. 2

`In Fig. 1, find sin, cos and tan for q and . 90o First find the third side. The hypotenuse = 52 62 61sin q = 6/ 61cos q = 5/ 61tan q = 6/5 sin90o 5/ 61

cos90o 6/ 61 tan90o 5/6

`In Fig. 2, find sin B, cos B and tan B.

Let y be the third side. Then y2 x 12 9x22

y2 81x4 x 12 81x4 x2 2x 1 y2 81x4 x2 2x 1

y = 81x4 x2 2x 1sin B = y/9x2 81x4 x2 2x 1 /9x2

cos B = x 1/9x2tan B = y/x 1 81x4 x2 2x 1 /x 1

`tan q = 2/3, find sin q and cos q, where q is acute.

Since tan = opposite/adjacent = 2/3, draw a right trianglewith opposite = 2 and adjacent = 3 (the drawing need not be

exact). Then find the third side.

2

3

y

q

Then y = 32 22 13sin q = 2/ 13cos q = 3/ 13

10(4). In a right triangle, the side adjacent angle q is and the opposite side is . x 2 x 2Simplify to the specified number of symbols.

qx-2

x+2

y

y 2x2 8Find . Simplify to a 7 symbol answer.tan

x 2x 2

Locate the angle in the triangle which is complementary to .Find . Simplify to a 10 symbol answer.sin 2 x 22x2 8

11.(4) q is acute and . Locate . Find sin(q) and cos(q) .tan 5/3 90o

q

5

3

x

x2 32 52

x 9 25 34sin 5

34sin90o 3

34cos 3

34cos90o 5

34

12(3). Prove: tan sin sec cos-- rewrite everything in terms of and sin cos

iff sincos sin 1

cos cos -- multiply by too clear the denominatorcos

iff sin2 1 cos2

iff ,sin2 cos2 1

13(3). Prove: cos 1 sin

1 sincos 2sec

-- common denominatorcos 1 sin

1 sincos

cos2 1 sin2

1 sin cos

-- cos2 1 2 sin sin21 sin cos sin2 cos2 1

2 2 sin1 sin cos

21 sin

1 sin cos 2cos 2 sec

Note: Write “iff ” or between equations, not “=”. Write “=” between formulas, not “iff” or .


11. q is acute and . Find sin(q), cos(q), and .tan 3/5 sin 2 cos 2 First draw a large right triangle with angle q, and sides of lengths 3 and 5 in the correct positions.

In the triangle above, locate the angle which is complementary to q. Label it or 900 - q.2

Find the length of the third side of the triangle. 3 symb, chk=7

sin(q) = 5 symb, chk=10

cos(q) = 5 symb, chk=12

= 5 symb, chk=12sin 2

= 5 symb, chk=10cos 2

13. Simplify: . Simplify to a 5 symbol answer or 7 symbols if you include "( )".sin1 cos

1 cos sin

Note: 2/tanq = 2cotq, 3/cosq = 3secq, 5/sinq =5cscq.


§7.1 533:11-24.1(1). Simplify: sincsc tan

2(2). Simplify: sec A tanAsec A tanA

§7.1 533:25-50.Note: Write “iff ” or between equations, not “=”.

Write “=” between formulas, not “iff” or .3(2). Prove: tan2A 1 sec2A

4(3). Prove: 1sin sin cot cos

5(3). Prove: cot A 1cot A 1 1 tan A

1 tan A

§6.2 484:1-8. 6(3). Find sin A, cos A, tan A.

6

2

A

sin A = 5 symb, chk=2

cos A = 5 symb, chk=4

tan A = 3 symb, chk=4

7(3). Given x> 1, find sin f, cos f, tan f.

f

x +1

x -12

2

sin f = chk=6

cos f =

chk=5

tan f = chk=5

§6.2 484:19,20. The angles B and q below are acute.8(2). cos B = 3/8, find sin B and tan B.

sin B = 5 symb, chk=18

tan B = 5 symb, chk=13

9(2). sin q = 3x/5, find cos q and tan q.

9 symb,chk=23cos

10 symb,chk=21tan

Math 140 Hw 17 Name _________________________________ Score ____/ 21

Simplify.

A. Hint: factor the top.sin2Acos2Asin Acos A

B. sin2coscsc3sec

C. cot B sin2Bcot B

D. cos2Acos A12

cos A3

E. tan

sec1 tan

sec1

F. sec Acsc A tanA cot A

G. cot2csc2

tan2sec2

Prove the identities.H. sincosseccsc 1

I. cos Asec A cos A sin2A

J. 1 sinsec tan cos

K. sin A cos A sin A1cot A cos A

tan A1

L. sinAtanA 1cos2A

cos A

M. Find sin A, cos A, tan A.

A3

2

sin A =

cos A =

tan A =

q

b

3

2x

4x

3

N. Find sin q, cos q, tan q.sin q =

cos q =

tan q =

O. Find sin b, cos b, tan b.sin b =

cos b =

tan b =

The angles below are acute.P. cos B = 4/7, find sin B and tan B.

sin B =

tan B =

Q. sin q = , find cos q and tan q.2 3 /5cos tan

R. sin q = x/2, find cos q and tan q.cos q =

tan q = S. cos q = x

2, find sin q and tan q.sin q =

tan q = AnswersA. sinA cosAB. cscC. cos2BD. cos A 4E. 2cscF. 0 G. 1M. cos A 3

13, sin A 2

13, tan A 2

3

N. sin 2x4x29

, cos 34x29

, tan 2x3

O. sin 16x29

4x , cos 34x , tan

16x293

P. sin B 337 , tan B

334

Q. cos 135 , tan

2 3

13

R. cos 4x2

2 , tan x

4x2

S. sin 1 x4 , tan 1x4

x2


Simplify.

A. Hint: factor the top.sin2Acos2Asin Acos A

sinA cos AsinA cos A/sinA cos A sinA cos A

B. sin2coscsc3sec sin2 cos 1

sin3 1

cos

1sin csc

C. cot B sin2Bcot B

cosBsinB sin2B cosB

sinB cos2B

D. cos2Acos A12cos A3

cos A4cos A3cos A3 cos A 4

E. tansec1

tansec1

tansec1sec1sec1

tansec1sec1sec1

tan sectantan sectan sec21

2 tan sectan2 2 sec

tan 2/ cossin/ cos

2sin 2 csc

G. cot2csc2

tan2sec2

cos2/ sin21/ sin2

sin2/ cos21/ cos2

cos2 sin2 1

Prove the identities.H. sincosseccsc 1

sincosseccsc sincos1/cos1/sin sincos/cossin 1

I. cosAsecA cosA sin2AcosAsecA cosA sin2Aiff cos sec cos2A sin2Aiff cos/ cos sin2 cos2

iff iff true1 sin2 cos2

J. 1 sinsec tan cos1 sinsec tan 1 sin 1

cos sincos

1 sin 1sin cos

1sin 1sin cos 1sin2

cos cos2cos cos

K. sin A cos A sin A1cot A

cos Atan A1

iff 1 cot AtanA 1sinA cos A sin AtanA 1 cos A1 cot Aiff tanA 1 cot A tanA cot AsinA cos A sinA tanA sinA cos A cos Acot Aiff tanA 2 cot AsinA cos A

sinA tanA sinA cos A cos Acot Aiff

tan A sin A tan A cos A 2 sin A 2 cos Acot A sin A cos Acot A

= tanAsinA sinA cos A cos Acot AifftanAcos A 2sinA 2cos A cot AsinA sinA cos AiffsinA 2sinA 2cos A cos A sinA cos Aiff true

L. sin A tan A 1cos2Acos A

1cos2Acos A sin2A

cos A sin A sin Acos A sin A tan A

M. Find sin A, cos A, tan A.

A3

2

Let c be the hypotenuse. Thus c2 22 32 c 4 9 13

cos A 313

, sin A 213

, tanA 23

q

b

3

2x

4x

3

N. Find sin q, cos q, tan q.Let c be the hypotenuse. Thusc2 4x2 9, c 4x2 9sin 2x

4x29, cos 3

4x29, tan 2x

3

O. Find sin b, cos b, tan b.Let b be the side opposite b.

, 32 b2 16x2 b 16x2 9

sin 16x29

4x , cos 34x , tan

16x293


Math 140 Lecture 18 Exam 3 covers lectures 13 -18. Study the recommended exercises.

DEFINITION. For any function f: f is even iff f (-x) = f (x). f is odd iff f (-x) = - f (x). Graphically, the left half (the half plane left of the y-axis) of

an even function is the reflection of the right half across

the y-axis. The left half of an odd function is the negativeof this reflection.

f(x) = x2 is even since f (-x) = (-x)

2 = x

2 = f (x). g(x) = x3 is odd since g(-x) = (-x)

3 = -x

3 = -g(x).

xx

23

even odd

Instead of thinking of sin θ as a function of an angle θ,we can think of it as a function sin(t) of a real variable t.

MINUS THEOREM. sin(-t) = -sin(t), cos(-t) = cos(t), tan(-t) =-tan(t). Thus cos(t) is even; sin(t) and tan(t) are odd functions.

Proof.

t

1

-t

sin(t)

sin(-t)=-sin(t)cos(-t)=cos(t)

tan(-t) = sin(-t)/cos(-t) = -sin(t)/cos(t) = -tan(t).

DEFINITION. A function f is periodic with period p iff f (x + p) = f (x)

for all x. p must be the smallest such positive number.

2π THEOREM. sin(t+2π) = sin(t), cos(t+2π) = cos(t), forintegers n, sin(t+2nπ) = sin(t), cos(t+2nπ) = cos(t).

Proof.

tt+ π

(cos t, sin t)=(cos(t+2 ),sin(t+2 ))

(-cos t, -sin t) = (cos(t+ ), sin(t+ ))

t+2π

π π

π π

π THEOREM. sin(t+π) = -sin(t), cos(t+π) = -cos(t), tan(t+π) = tan(t).

For odd n, sin(t+nπ) = -sin(t), cos(t+nπ) = -cos(t).

Proof. See the picture for sin and cos. The terminal sides

of t + π and t have the same slope, τ tan(t + π) = tan(t).

sin and cos have period 2π, but tan has period π.

Simplify: .cos(3� − x)

Since 3 is odd, cos(3� − x) = − cos(−x) = − cos(x)

Previously we found reference angles in [0, π/2]geometrically. Now we can find them algebraically.

v Take out minus signs (use the Minus Theorem).

v Rewrite in terms of the nearest multiple of π.

v Throw out multiples of π (use 2π and π Theorems).

Rewrite with a reference angle in [0, π/2], then evaluate.

sin(-5π/6). sin(-5π/6) = -sin(5π/6).

0 = 0π/6 < 5π/6 < 6π/6 = π. 5π/6 is nearest 6π/6 = π.-sin(5π/6) = -sin(π−π/6) = sin(-π/6) = -sin(π/6) = -1/2.

cos(4π/3).

π = 3π/3 < 4π/3 < 6π/3. 4π/3 is nearest 3π/3 = π.cos(4π/3)=cos(3π/3+π/3)=cos(π+π/3) = -cos(π/3) = -1/2.

tan(-7π/4). tan(-7π/4) = -tan(7π/4).

π = 4π/4 < 7π/4 < 8π/4 = 2π. 7π/4 is nearest 8π/4 = 2π-tan(7π/4) = -tan(8π/4−π/4) = -tan(2π−π/4) = -tan(-π/4)

= tan(π/4) = sin(π/4)/cos(π/4) = 1.

sin(21π/2). 10π = 20π/2 < 21π/2 < 22π/2 = 11π.sin(21π/2)=sin(20π/2+π/2) = sin(10π+π/2) = sin(π/2) = 1.

cos(-31π/6) = cos(31π/6) = cos(30π/6 + π/6) =

cos(5π+π/6) = cos(π+4π+π/6) = -cos(π/6) = - . 3 /2

PYTHAGOREAN IDENTITIES. sin2t + cos2t = 1, tan2t +1 = sec2t,cot2t +1 = csc2t.

Proof. We’ve proved the first. To prove the second,

multiply both sides by cos2t. For the third, multiply bothsides by sin2t.

Simplify csc t+csc t cot2t

sec2t−tan2t

=csc t [1+cot2t]

(1+tan2t)−tan2t=

csc t(csc2t)

1 = csc3t

Simplify sin2t−cos2t

sin4t−cos4t

=sin2t−cos2t

(sin2t−cos2t)(sin2t+cos2t)= 1

Prove cot � + tan � + 1 =cot�

1−tan � +tan�

1−cot�Let x = tan θ, then cot θ = 1/tan θ = 1/x.Thus we must prove1x + x + 1 =

1/x1−x +

x

1−1/x

iff 1x + x + 1 =

1/x1−x −

x2

1−x

iff 1 + x2 + x = [1 − x3]/(1 − x)

iff iff (x2 + x + 1)(1 − x) = (1 − x3) 1 − x3 = 1 − x3

iff true

11(2). q is acute and . Find , sin(-q), and cos(-q) .tan 5/3 tan tan tan tan 5/3

q5

3

x

x 9 25 34sin sin 5

34sin 5

34 cos cos 3

34cos 3

34

12(6). Prove: 11 sec x 1

1 sec x 2cot2x 0Any acceptable proof.

iff clear the denominators, multiply by 11 sec x 1

1 sec x 2 cot2x 1 secx1 secx

iff 1 secx 1 sec x 2cot2x1 sec2x

iff 2 2cot2x1 1 tan2x 1

iff 2 2cot2x1 tan2x 1

iff 2 2 cot2x tan2x

iff ,2 2

14(a)(4). Rewrite using a reference angle in [0, p/2]. Write using sin or cos, don't give the decimal or radical answer.sin 4011

, 4p is the nearest multiple of p. sin 4011 3 33

11 4011 44

11 4

sin 4411 4

11 sin4 411 sin 4

11 sin 411

(b)(4). Rewrite using a reference angle, then find the exact answer: cos(-19p/6).

, 3p is the nearest multiple of p. cos 196 3 18

6 196 24

6 4

cos18/6 /6 cos3 /6 cos(/6) 32

PYTHAGOREAN IDENTITIES. sin2t + cos2t = 1, tan2t +1 = sec2t, cot2t +1 = csc2t.

15(6). Simplify to a number: cot2tsec2t 1sec2t tan2t 1

1/2cot2ttan2 1 1tan2t 1 tan2t 1 cot2t tan2t

2


11. q is acute and . tan 3/10

q

x

Label the sides of the triangle in a way that represents the equation . Find the third side, x = tan 3/10

cos(- q) =7 symb, chk=11.

sin(q+p) =7 symb, chk=13.

cos(p-q) =8 symb, chk=11.

14a. Rewrite using a reference angle in [0, p/2]. Exact answer with sin, p, no decimal answer. 6 or 8 symb, chk=7.sin 227

First take care of the minus sign "-".

Find the nearest multiples of p above and below . Of these two multiples of p, circle the nearest one.227

Rewrite as the nearest multiple of p plus or minus an acute positive angle. This acute angle is the reference227

angle.

14b. Rewrite using a reference angle in [0, p/2]. Exact answer with cos, p, no decimal answer. 8 or 10 symb, chk=10.cos 187


§6.3 496:9-32. Exam 3 next.For 1-12: (a) Rewrite in terms of an angle in [0,p/2]. (b) Find theexact answer, no decimals. See the example. 1/2 point each.

In 1-15, 4 answers are integers, 2 others are rational.

Example: sin(15p/4) = -sin(p/4) = -1/l2

1(½). cos(2p/3) = 4 symb,chk=3

2(½). cos(-2p/3) = 4 symb,chk=3

3(½). sin(2p/3) = 4 symb,chk=5

4(½). sin(-2p/3) = 5 symb,chk=5

5(½). cos(-13p/4) = 5 symb,chk=3

6(½). sin(13p/4) = 5 symb,chk=3

7(½). cos(9p/4) = 4 symb,chk=3

8(½). sin(-9p/4) = 5 symb,chk=3

9(½). tan(7p/4) = 2 symb,chk=1

10(½). sin(17p/4) = 4 symb,chk=3

11(½). cos(11p) = 2 symb,chk=1

12(½). cos(53p/4) = 5 symb,chk=3

PYTHAGOREAN IDENTITIES. sin2t + cos2t = 1, tan2t +1 = sec2t,cot2t +1 = csc2t.

13(1). Simplify to an integer: sec2t 1tan2t

14(1). Simplify to an integer: csc4 cot4csc2 cot2

§7.1 533:51-86.

15(2). Prove: sin2t cos2t 1 cot2t1 cot2t

16(2). Prove: 1 tan s1 tan s sec2s 2 tan s

2 sec2s

Math 140 Hw 18 Name ______________________________ Score____/ 12 Exam 3

For A-D: (a) Rewrite in terms of an angle in [0,p/2]. (b) Find theexact answer, no decimals. See the example. ½ point each.

A(a). cos11/6

A(b). cos11/6

A(c). sin11/6

A(d). sin11/6

B(a). cos5/4

B(b). sin5/4

C(a). sec5/3

C(b). csc5/3

C(c). tan5/3

C(d). cot5/3

D(a). cos 4 2

D(b). sin 3 2

D(c). sin 2 6

E. Simplify: sin2tcos2t

tan2t1

F. Simplify: sec2tan2

1cot2

G. Prove: csc t sin t cot t cos t

H. Prove: 1

1sec s 1

1sec s 2cot2s

Answers

Aabcd. 32 ,

32 , 1

2 , 12

Bab. 12

, 12

Cabcd. 2, 23

, 3 , 13

Dabc. 12

,32 , 1

E. cos2t

F. sin2

G. sin t cot t cos t sin t cos tsin t cos t

sin2tsin t

cos2tsin t sin2tcos2t

sin t 1sin t csc t

H. 1

1sec s 1

1sec s 1sec s1sec s1sec s1sec s

21sec2s

2tan2s 2cot2s


(a) Rewrite in terms of an angle in [0,p/2]. (b) Find the exact answer, no decimals.

A(a). cos11/6cos11/6 cos12/6 /6 cos2 /6 cos/6 cos/6

32

A(d). sin11/6sin11/6 sin11/6 sin12/6 /6 sin2/6 sin/6 sin/6 1/2

B(a). cos5/4cos5/4 cos /4 cos/4 1/ 2

B(b). sin5/4sin5/4 sin /4 sin/4 1/ 2

C(a). sec5/3sec5/3 sec6/3/3 sec/3 sec/3 2

C(c). tan5/3tan5/3 tan6/3/3 tan/3

tan/3 3 /21/2 3

E. Simplify sin2tcos2t

tan2t1

1sec2t cos2t

F. Simplify sec2tan2

1cot2

tan21tan t2

csc2 1csc2 sin2

G. Prove: csc t sin t cot t cos tProof.sin t cot t cos t sin t cos t/ sin tcos t

sin t cos2t/ sin t sin2tcos2tsin t 1

sin t csc t

Alternate proof.sin t cot t cos t sin t cos t

sin t cos t

sin2tsin t

cos2tsin t sin2tcos2t

sin t 1sin t csc t

H. Prove: 1

1sec s 1

1sec s 2 cot2s

11sec s

11sec s

1sec s1sec s1sec s1sec s

21sec2s

2tan2s 2cot2s


Math 140 Lecture 19Graphs of sin and cosRECALL. sin and cos have period 2p:

sin(x +2p) = sin(x), cos(x +2p) = cos(x). We often graph periodic functions only over one period, e.g., [0, 2p]. Before and after this interval, they repeat.

Graph sin(x).

p/2 p 3p/2 2p-p/2

1

-1

sin(x)

y=x

0

min root max root min rootx-intercept: x = ..., -2p, -p, 0, p, 2p, 3p, ... Max value: 1 at x = ..., -3p/2, p/2, 5p/2, ...Min value: -1 at x = ..., -5p/2, -p/2, 3p/2, ...Amplitude: A = 1 (see definition below) Period: p = 2p

At x = 0, the line y = x is tangent to the graph of sin(x).

Graph cos(x). ... Done similarly ...

DEFINITION. The amplitude of a function f is half the difference between the max and min values of f.

Like periods, amplitudes are always positive.

`Find the amplitude and period of f, g, and h. f g h

-11

2

3

2

-2

1

1

-1

2

3

(3,3)

amplitude = 2 amplitude = 1 amplitude = 1

period = 4 period = 4 period = 3

`Graph y = 2sin(x) over one period.

p/2 p 3p/2 2p

1

-1

sin(x)

2sin(x)

-2

2

amplitude = 2 period = 2p. increases on [0, p/2] and [3p/2, 2p]

For A > 0, y = + Asin(x) has amplitude A. Reflect the graph across the x-axis for -Asin(x).

`Graph y = sin(2x) over one period.

Note sin(x) = 0 iff x = ..., 0, p, 2p, ... . Thus sin(2x) = 0 iff 2x = ..., 0, p, 2p, ... iff x = 0, p/2, p, ... .

p/2 p 3p/2 2p

sin(x)-1

1

sin(2x)

Amplitude = 1, period = p. Increases on [0, p/4] and [3p/4, p].

For B > 0, y = sin(Bx) has period is 2p/B. Reason: sin(Bx) repeats at Bx = 2p. Solving for x gives x = 2p/B.

B is the compression factor.THEOREM. For y = + Asin(Bx) & y = + Acos(Bx) with A,B >0,v amplitude : Av period : p = 2p/B. Hence also, .B 2/p

To graph, mark on the y-axis, 0, p on the x-axis.ADivide into four parts.0, p

`Graph y = -3cos(px) over one period. List the amplitude, period, x-intercepts and the intervalsin the period on which the function increases. y = -3cos(px) = - Acos(Bx)A = 3, B = p amplitude: A = 3 period: p = 2p/B = 2p/p = 2 Draw a one-period box of amplitude A and length p.

Divide the period into four parts.

3

-3

1/2 1 3/2 2

increases on: [0,1]. `Find an equation for the graph. Write it in the form

y = + Asin(Bx) or y = + Acos(Bx) with A, B > 0.

(2,2)

The graph has the shape of y = -Acos(Bx). amplitude: A = 2; period: p = 4.B: B 2

p 24

2Equation: y 2 cos 2 x

1(7). For the given graph, find the amplitude, period, and an equation of the form y=Asin(Bx) or y=Acos(Bx). 2 Amplitude

6 Period Coefficient B/3

Equationy 2cosx/3

To graph a trigonometric function over one period, draw a box. The length is the period p. The height above andbelow the x-axis is the amplitude A. Divide the box into four pieces. Label the divisions on the x-axis. Fill in the graph.

2. (a)(5) Graph over one period beginning at the phase shift. List the period and amplitude.2 cos3x

2 Amplitude Period2/3

Graph

2

-2

2p/3p/3

p/6

(b)(5) Graph over one period beginning at the phase shift. List the period and amplitude.3 sin5x

3 Amplitude Period2/5 2/5 Graph

3

-3

2/51/5

1/10


(-3,2)

1. For the graph below, find the amplitude, period, and an equation of the form y = +Asin(Bx) or y = +Acos(Bx).Warning, the graph pictured includes more than one period. The period is not 6.

Amplitude: A = must be positive

Basic shape:

Period: p =

Coefficient B: B 2p

3 sym, chk=2

Equation: chk=4

`Example: Graph over one period.y 5sinx/3

-5

6p

5

3p3p/2 9p/2

2. Graph over one period. y 3cos2x

Amplitude (must be positive) A

Compression coefficient B

Period p 2B

chk=0

Draw a box (see example above). The length is the period p. The height above and below the x-axis is the amplitude A. Divide the box into four pieces. Label the divisions on the x-axis. Fill in the graph.


(3,-2)

§5.3 431:80(a)(d). Find the period and amplitude. ½ point

per part. 5 answers are integers, one is a fraction. All are positive.

1(1). period= amplitude=

3

−3

4

8

12


5 1015

−5

−10

−15

−5 20


1 2−1

(0,2/3)

§5.3 429:15-26. One period is an integer, the rest involve π.

Graph over one period. List the amplitude, period,

x-intercepts and the intervals (in the period) on which the

function increases. graph 1 point, ½ point for other parts.

4(3). y = 3 sin xamplitude=

period=

x-intercepts:

increases on:

5(3). y = sin 3x

amplitude=

period=

x-intercepts:

increases on:

6(3). y = −2 cos(x4 )

amplitude=

period=

x-intercepts:

increases on:

7(3). y = −2 cos(�x4

)

amplitude=

period=

x-intercepts:

increases on:


§5.3 429:41-48. Find an equation for the graph of the form y = Asin(Bx)or y = Acos(Bx) with B>0. 1 B involves π, two are fractions.

8(2). chk=5 y =

(3π, −2)

9(2). chk=3 y =

3

−3

(1,0)

10(2). chk=10 y =

( 5 π , 4 )

Math 140 Hw 19 Name _______________________________ Score ____/ 21

1

−1

1

2

−1

−2

1

2

−1

−2

Find the period and amplitude. A. period= amplitude=

6

-6

1

35

2 4

B. period= amplitude=

4

6

2

2 4-2

C. period= amplitude=

3 6-3

-6

-3

Graph over one period. List the amplitude, period,x-intercepts and the intervals (in the period) on which thefunction increases.

D. y sin2xamplitude=

period=

x-intercepts:

increases on:

E. y 2 cos 2x

amplitude=

period=

x-intercepts:

increases on:

F. y 3sin (x/2amplitude=

period=

x-intercepts:

increases on:

G. y cos 2xamplitude=

period=

x-intercepts:

increases on:

Find an equation for the graph of the form y = + Asin(Bx)or y = + Acos(Bx) with B>0. 2 B's involve p, all are fractions.

H. y = (p/3,3/2)

I. y =

(5,1)

J. y =

(4,-p)


Find the period and amplitude. A. period= 4 amplitude= 6

6

-6

1

35

2 4

B. period = 4 amplitude = 2

4

6

2

2 4-2

C. period = 6 amplitude = 3/2

3 6-3

-6

-3

Graph over one period. List the amplitude, period,x-intercepts and the intervals (in the period) on which thefunction increases. Use the amplitude and period to get the box.

D. y sin2xamplitude= 1

period= p

x-intercepts: 0, p/2, p

increases on: [p/4, 3p/4]

E. y 2cos 2x amplitude= 2

period= p

x-intercepts: p/4, 3p/4

increases on: [p/2, p]

F. y 3sin (x/2amplitude= 3

period= 4

x-intercepts: 0, 2, 4

increases on [0, 1], [3, 4]

Find an equation for the graph of the form y = + Asin(Bx)or y = + Acos(Bx) with B > 0. 2 B's involve p, all are fractions.

H. y 32 sin3x/2

(p/3,3/2)

Since p/3 is at the one quarter point, the endpoint of theperiod is at 4p/3. Thus the period p=4p/3. Since 3/2 is thehigh point, the amplitude is A=3/2. Now B = 2p/p =2p/(4p/3) = 2p(3/4p) = 3/2. Since the graph starts at 0 and goes up, it is sin rather thancos, -sin or -cos.

I. y cos2x/5(5,1)

J. y cosx/4

(4,-p)

Note, the amplitude A is p, not -p. Amplitudes andperiods are always positive. 4 occurs halfway through theperiod, hence the period is 8 and B = 2p/p = 2p/8 = p/4.


p/2

1

2p/2

3

2 40

Math 140 Lecture 20 RECALL. For C > 0, the graph of f (x + C) is the graph of f (x) shifted C units to the left. f (x- C) is the graph of f (x) shifted C units to the right. The phase shift is the amount C subtracted from x.f (x + C) = f (x-(-C)) has phase shift -C.f (2x-C) = f ( 2(x-C/2) ) has phase shift C/2.

`Graph y = sin(x- p/2). Phase shift = p/2.

sin(x) sin(x- /2)p

1

-1

1

-1

p 2pp/2p/2-p/2

p 2p 5p/2

To graph y = Asin(Bx+D): Acos(Bx+D) is similar.v Factor out B, and rewrite in the form Asin[B(x-C)]:

Asin[Bx+D] ƒ Asin[B(x+(D/B))] ƒ Asin[B(x-C)]. v The period is . Also find p/4. We assume A, B > 0.p 2/Bv Draw a box with amplitude A, period p, shifted by C.v Draw the +sin or +cos shape in the box.

`Graph y = sin(2x- p) over one period. sin(2x- p) = sin[2(x- p/2)]. Phase shift = p/2, p = p.

sin(2x) sin(2(x- /2))p

pp/2p/2

p 3p/21

-1

1

p is not the shift for sin(2x-p). In sin(2x-p), p is subtracted from 2x.But a phase shift must be subtracted from x. Factoring sin(2x-p) tosin(2(x-p/2)) shows that the phase shift subtracted from x is p/2.

`Graph -2cos(2px- p/2) over one period. -2cos(2px- p/2) = -2cos[2p(x- 1/4)] � phase shift form

t amplitude = 2, period = 2p/B = 1, phase shift = 1/4.

-2cos(2 (x-1/4)) p

5/41/4

1/2

2

-2

13/4

`Graph y = tan(x). tan(x) = sin(x)/cos(x). tan(x) = 0 iff sin(x) = 0 iff x = ..., 0, p, 2p, ... .

x-intercepts: ..., 0, p, 2p, ... . tan(x) undef. iff cos(x) = 0 iff x = ..., -p/2, p/2, 3p/2, ...

vert. asymptotes: ..., x = -p/2, x = p/2, x = 3p/2, ... .

p/2p 2p0

-p/2

tan(x)

`Graph y = -tan(3x + p/2). -tan(3x + p/2) = -tan[3(x- (- p/6))]. Phase shift = -p/6.

Period = p/3. Draw a strip centered at -p/6 with width p/3(half of the width is before the center -p/6, half is after).

Then fill in -tan(x). Add additional copies of this strip.

0

-tan[3(x+ /6)]

p/3

2p/3

-p/3

p/6-p/6

p

x-intercepts: ..., -p/6, p/6, ... Period = p/3. vert. asymptotes: ..., x = -p/3, x = 0, x = p/3, x = 2p/3, ... .Note: tan(x) has period p, thus tan(Bx) has period p/B.

`Graph y = csc(x). csc(x) = 1/sin(x). To graph csc(x), graph sin(x) and then invert its values.sin(x) = ½ ˆ csc(x) = 1/(½) = 2. sin(x) = 0 ˆ csc(x) = 1/0 = undef.

0 p/2p

2p

csc(x)

1

-1

x-intercepts: none vert. asymptotes: ..., x = 0, x = p, x = 2p, ...

`Graph y = -csc(3x + p) over one period. csc3x csc3x /3 csc3x /3Graph first. Then invert as with y sin3x cscx

above.

3(7). Graph over one period. List the period and phase shift. Draw the asymptotes with dotted lines..y tan 4 x 8

tan 4 x 1

2 tan

4 x 12

Period p=p B

/4 4, p/2 2Phase shift C =C 1/2

-5/23/2-1/2

2(12). Graph over one period beginning at the phase shift. List the period and phase shift. y 2 sin x4

2

Get the box from the amplitude, period and shift.Amplitude A = 2 sin 1

4 (x 2), Period p= p 2

1/4 8 p/4 2Phase shift C =C 2

2p 4p 8p 10p-2p

6p

2

-2

Except for an incorrect shift, the following would also be acceptable.

2p 4p-2p

6p

2

-2


sin cos 2. Graph over one period. 3 cos 6 x 3

Inside the parentheses, factor out the coefficient of x. Write in the form AcosBx C

3cos

Amplitude A = Amplitudes must be positive. Compression coefficient B = 3 symb, chk=6

Period p = = Phase shift C = 2 symb, chk=22B

p4

2 symb, chk=3

Now draw a box which starts at C, has length p and height A above and below the x-axis.Divide the box horizontally into 4 pieces. Then fill in the graph.

3. Graph over one period. Draw the strip with phase shift C at the center. y tan 14 x 8

Again, inside the parentheses, factor out the coefficient of x. Write in the form A tanBx CRecall that . Recall the graph of tan(x): chks for two asymptotes are 7 and 5.p B

p/2p 2p0

-p/2

Math 140 Classwork 20 Score _____/6

1

-1 2p

1

-1

2p

AC

p

§5.3 429:27-40. (1) Graph over one period (start the period atthe phase shift). (2) List the x-intercepts. (3) List both coordinates ofthe highest and lowest points. (4) Give the period and phase shift.

Hint: one phase shift is positive, one is negative.

1(5). First rewrite in the form Asin(B(x+C).y sin2x 2

y

x-intercepts: max point: min point:

2(5). Rewrite: y = y 3 sin 12 x

6

x-intercepts: max points:

min point:

(1) Graph over one period. (2) List the x-intercepts. (3) List the vertical asymptotes. In 3 and 4 choose the periodto include the origin and to be between the asymptotes.

3(3). Center the period at the phase shift.y tanx 3

x-intercept:

vertical asymptotes:

§5.4 441: 1-52.

4(3). y 2 tanx

x-intercept:


5(3) Put (a) and (b) on the same graph. First rewrite with the argument in the form: B(x+C)

(a) y cos3x 3 y

(b) y sec3x 3 y

cos(3x+p/3): x-intercepts:

sec(3x+p/3): vertical asymptotes:



(1) Graph over one period (period doesn't have to start at 0).(2) List the x-intercepts. (3) List both coordinates of thehighest and lowest points.A. First rewrite in the form Asin(B(x-C)).y sin3x

2 y

x-intercepts: period: max point: min point: phase shift:

B. y cosx 2

x-intercepts: period:

max points:

min point: phase shift:

(1) Graph over one period. (2) List the x-intercepts. (3) List the vertical asymptotes.

C. y tanx 4

x-intercept: vertical asymptotes:

D. y 12 tanx/2

x-intercept:


E. Put (a) and (b) on the same graph. First rewrite with the argument in the form: B(x+C)

(a) y 3 cos2x 4 y

(b) y 3 sec2x 4 y

: x-intercepts: 3 cos2x 4

: vertical asymptotes:3 sec2x 4

For answers, see Hw 20 worked examples.


(1) Graph over one period (period doesn't have to start at 0). (2) List the x-intercepts. (3) List both coordinates of thehighest and lowest points. Get the box from the amplitude, period and phase shift.

A. First rewrite in the form Asin(B(x-C)).y sin3x 2

y sin3x 6

-p/6

p/2

1

-1

x-intercepts: -p/6, p/6, p/2 (answers depend on choice of interval)

max point: (0, 1) period: 2p/3

min point: (p/3, -1) phase shift: -p/6

B. y cosx 2

p/2p

2p

1

-1

x-intercepts: p, 2p (answer varies with choice of interval) y= sin(x).

max points: (p/2, 1), (5p/2,1) period: 2p min point: (3p/2,-1) shift: p/2


Continued on next page

(1) Graph over one period. (2) List the x-intercepts. (3) List the vertical asymptotes.

C. y tanx 4

p/4-3p/4 -p/4

x-intercept: -p/4vertical asymptotes: x = -3p/4, x = p/4

E. Put (a) and (b) on the same graph. First rewrite with the argument in the form: B(x+C)

(a) , p = 1y 3 cos2x 4 y 3 cos2x 1

8

(b) y 3 sec2x 4 y 3 sec2x 1

8

1/8

-3sec...

-3cos...5/8

9/8

3

x=3/8x=7/8

-3

You can start the period at 1/8 or anywhere you like.-3cos(2px-p/4): x-intercepts: -1/8, 3/8, 7/8, ...

-3sec(2px-p/4): vertical asymptotes: x = -1/8, x = 3/8, ...

Math 140 Hw 20 Worked examples continued.

Continued from previous page

Math 140 Lecture 21

RECALL. sin2+ cos2 = 1, sin(-x) = -sin(x), cos(-x) = cos(x).cos(p/2-x) = sin(x), sin(p/2-x) = cos(x).

TRIGONOMETRIC ADDITION FORMULAS. sins t sin s cos t cos s sin tcoss t cos s cos t sin s sin tcoss t cos s cos t sin s sin ttans t tan s tan t

1 tan s tan t

tans t tan s tan t1 tan s tan t

Proof of .coss t cos s cos t sin s sin t

st

s+t

t

-s

(cos(s+t), sin(s+t))

(1,0)

(cos(t), sin(t))

(cos(-s), sin(-s))

dd

1

1

1

1

=(cos(s), -sin(s))

In both pictures the obtuse angle is s + t, the short sidesare radii of length 1, and the long sides have somecommon length d. Use the distance formula to calculated.

In the first,d coss t 12 sins t 02

td2 cos2s t 2coss t 1 sin2s tt ,d2 2 2 coss t

since .cos2s t sin2s t 1In the second,

d cos t cos s2 sin t sin s2

td2 cos2t 2 cos t cos s cos2s sin2t 2 sin t sin s sin2s

t ,d2 2 2 cos t cos s 2sin t sin ssince cos2t sin2t 1, cos2s sin2s 1,coss coss, sins sins.

Equate the two terms which equal d 2:

2 2coss t 2 2cos t cos s 2sin t sin st2coss t 2 cos t cos s 2sin t sin st .coss t cos s cos t sin s sin t

Proof of .coss t cos t cos s sin t sin scoss t coss t

cos s cost sin s sint. cos s cos t sin s sin t

Proof of .sins t sin s cos t cos s sin tsins t cos 2 s t cos 2 s t

cos 2 scos t sin 2 s sin t. sin s cos t cos s sin t

The other proofs are similar.

In each problem below and in homework: (1) Apply an addition formula. (first point)(2) Simplify. (second point)

`cos 5 cos 30 sin

5 sin 30

= first pointcos 5 30

second pointcos 630

30 cos 5

30 cos 6 3 /2

`sinx ycos y cosx y sin y

= = sinx y y sinx

`sina b sina b

= sinacos b cos a sinb sinacos b cos a sinb

= 2sinacos b

`cosa b cosa b

= cos acos b sina sinb cos acos b sina sinb

= 2sina sin b

`tan

3 tan 6

1 tan 3 tan

6

= tan 3 6 tan 6

13

( = )tan 6 sin/6cos/6

1/23 /2 1

3

` 2 tan 8

1 tan2 8

= tan 8 8 tan 4 1

`Given . tan s 13 , tan t 2

3 , find tans t

tans t tan s tan t1 tan s tan t

= 13

23

1 13

23 1

1 29

17/9 9

7

`Use an addition formula to find tan(p/12). p/12 = 4p/12 - 3p/12 = p/3 - p/4

tan 12 tan 3 4

= tan/3 tan/41 tan/3 tan/4

= = 3 11 3 1

3 11 3

1 3

1 3

= 2 3 13

13 2 3 42 2 3

sins t sin scos t cos s sin t tans t tan s tan t1 tan s tan t

coss t cos scos t sin s sin tcoss t cos scos t sin s sin t tans t tan s tan t

1 tan s tan t

5(5). Write with no products and no p: at most 7 symbols. cos 8 cosx 38 sin 8 sinx 3

8

cos 8 x 38 cos 4

8 x cos 2 x sin x

5(5). Write with no products and no p: at most 7 symbols. sin 58 cosx 3

8 cos 58 sinx 3

8

sin 58 x 3

8

sinx 88 sin(x )

sin x

6(2). Write in terms of and no p: . tanx tanx /4

tanx /4 tan x tan/41 tan x tan/4

tan x 11 tan x

6(2). Find the exact answer. tanx 4, tany z, tanx y

tanx tany

1 tanx tany

4 z1 4z


sins t sin s cos t cos s sin t tans t tan s tan t1 tan s tan t

coss t cos scos t sin s sin tcoss t cos scos t sin s sin t tans t tan s tan t

1 tan s tan t

5. Write with no products and no p: 4 or 6 symbols. 4 symb, chk=0 cos8 cosx 58 sin8 sinx 5

8

6. Find the exact answer, 6 symbols. 6 symb, chk=13 tanx 5, tany 6, tanx y

6. Write in terms of and no p: tanx tanx 6 Answers involving fractions of fractions must be simplified to a single fraction. Leave , in the denominator. 16 symbols, 20 with ( )s.3

Math 140 Classwork 21 Score_____/5

§7.2 539:13-18. (a) Apply an addition formula (show the step immediately after

the addition formula). First point.

(b) Simplify. Second point.

Examples: `sinxcosy x cosx siny x

4 symbols sinx y x siny`cos 4

cos 4 cos sin 4 sin 1

2cos sin

`Given .tanx 3, tany 5: Find tanx y4 symbols, chk=9 35

135 18

1(2). sin 6 cos 3 cos 6 sin

31 symbol, chk=1

2(2). cos 2u cos 3u sin2u sin3u4 symbols

3(2). coss t cos t sins t sin t4 symbols

4(2). cos 32

4 symbols

5(2). sint 6 sint

6 4 symbols

6(2). cos 4 cos

4 6 symbols, chk=2

7(4). Given tan s 12 , tan t 1

3 Find tans t1 symbol, chk=1

Find tans t3 symbols, chk=8

8(2). tan/5 tan/30

1 tan/5 tan/304 symbols, chk=4

9(2). 2 tan/12

1 tan2/124 symbols, chk=4

Math 140 Hw 21 Name _________________________________ Score ____/20

(1) Simplify using an addition formula (show the stepimmediately after the addition formula). First point.

(2) Complete the simplification. Second point.


sinx y x siny`cos 4


2cos sin

`Given .tanx 3, tany 5: Find tanx y 35

135 18

A. sin3cos cos3sin

B. cos 2ucos 3u sin2usin3u

C. sinABcos A cosAB sinA

D. sin 32

E. sin 4 s sin 4 s

F. cos 3 cos 3

G. Given :tan s 2, tan t 3 Find tans t

Find tans t

H. tanttan2t

1tant tan2t

I. tanxytany

1tanxy tany

AnswersA. B. sin2 cos 5uC. D. cos qsinBE. F. 2 sin s 3 sinG. tans t 1, tans t 1/7H. I. tan3t tanx


(1) Simplify using an addition formula (show the stepimmediately after the addition formula). First point.

(2) Complete the simplification. Second point.


sinx y x siny`cos 4


2cos sin

`Given .tanx 3, tany 5: Find tanx y 35

135 18

A. sin3cos cos3sin

sin3 sin2Answer: sin2

B. cos 2ucos 3u sin2usin3u

cos2u 3u cos5uAnswer: cos 5u

C. sinABcosA cosABsinA

sinAB A sinBAnswer: sinB

D. sin 32

sin cos 32 cos sin 3

2 sin0 cos1 cosAnswer: cos q

E. sin 4 s sin 4 s

sin 4 cos s cos 4 sins sin

4 cos s cos 4 sins

2 cos 4 sin s 2 12

sin s 222 sin s

Answer: 2 sins

F. cos 3 cos 3 cos

3 cos sin 3 sin

cos 3 cos sin

3 sin

2 sin 3 sin 2

32 sin

Answer: 3 sin

G. Given tan s 2, tan t 3 Find tans t

tan stan t1tan s tan t

23123

55 1

Answer: tans t 1

Find tans t

tan stan t1tan s tan t

23123

17

Answer: tans t 1/7

H. tanttan2t

1tant tan2t

tant 2t tan3tAnswer: tan3t

I. tanxytany

1tanxy tany

tanx y y tanxAnswer: tanx


Math 140 Lecture 22DOUBLE-ANGLE FORMULAS.

sin2 2sincos cos 2 cos2 sin2

tan 2 2 tan1tan2

Proof.sin2 sin

, sincos cossin 2sincos

cos 2 cos , coscos sinsin cos2 sin2

tan2 tan

, tantan1tan tan

2 tan1tan2

HALF-ANGLE FORMULAS.

+ if q/2 i quadrant I or II, - if not.sin 2 1cos

2

+ if q/2 i quadrant I or IV, - if not.cos 2 1cos 2

tan 2 sin

1cos Proof.

. Hencecos2 sin2 1 andcos2 1 sin2

sin2 1 cos2

cos 2 cos2 sin2 1 sin2 sin2 1 2sin2

t 2sin2 1 cos2

t . Replacing q by q/2 gives sin 1cos 22

,sin 2 1cos

2

cos 2 cos2 sin2 cos2 1 cos2 2cos2 1

t 2 cos2 1 cos 2

t . Replacing q by q/2 gives cos 1cos 22

,cos 2 1cos 2

tan 2 sin

2 / cos 2 1cos 2 / 1cos

2

1cos 1cos 1cos 1cos

1cos 1cos

, 1cos21cos 2 sin2

1cos 2 sin 1cos

` , . cos 13

32 2

Evaluate sin(2q), cos(2q), sin(q/2), cos(q/2). First find sin(q). The inequality ˆquadrant IVˆsin q<0.

sin 1 cos2 1 19 8

9 23 2

sin 2 2 sincos 2 23 2 1

3 4 2

9

cos 2 cos2 sin2 19

89 7

9

ˆ 32 2 3

4 2

ˆ is II quadrant ˆ 2 sin

2 0, cos 2 0

sin 2 1cos

2 11/32 31

6 1/ 3

cos 2 1cos 2 11/3

2 316 2

3

`Evaluate using half-angle formulas. = = sin

12 sin /62

1cos 6

2 1 3 /2

2 2 3

4 12 2 3

= = cos 12 cos /6

2

1cos 6

2 1 3 /2

2 2 3

4 12 2 3

= = tan 12 tan /6

2

sin

6

1cos 6 1/2

1 3 /2 1

2 3

2 3

2 3 2 3 2 3

`For the angle t in the picture, find cos(2t) and sin(t/2).

t2

3

First find the missing side x, then find sin(t ) and cos(t).x2 4 9, x 9 4 5

sin t 5 /3, cos t 2/3 cos 2t cos2t sin2t 4

9 59 1

9

sin t2 1cos t

2 12/32 1/3

2 16

`Express in a form which does not use powers ofsin4trigonometric functions.

sin4 sin22 1cos 22

2 12 cos 2cos22

4

12 cos 2 1

2 1cos 44

18 3 4 cos 2 cos 4

The calculation for is similar. cos4

+ if q/2 i quadrant I or II, - if not.sin2 2 sincos sin 2 1 cos 2

+ if q/2 i quadrant I or IV, - if not.cos 2 cos2 sin2 cos 2 1 cos 2

tan 2 2 tan1 tan2

tan 2 sin 1 cos

7. Given: . x tan 1, 32

tan 1/x

q

2

1x +1l

xx 0sin 1/ x2 1cos x / x2 1

(a)(2) Find . Draw the triangle, locate the angle p/2- q. is the reciprocal of and has the same sign.tan 2 tan 2 tan

x1 x

(b)(5) Find . cos2

cos2 sin2

x/ x2 12 1/ x2 1

2

x2 1x2 1

(c)(5) Find . sin 2

Divide by 2: 32

quad. II2 2 3

4

1 cos 2

1 x

x2 12


+ if q/2 i quadrant I or II, - if not.sin2 2 sincos sin 2 1cos

2

+ if q/2 i quadrant I or IV, - if not.cos 2 cos2 sin2 cos 2 1cos2

7. Given: . tan 3/2, 3/2Draw a triangle with q and the opposite side marked 3 and the adjacent side marked 2 and find the third side.

Find . In the triangle above, locate the angle p/2- q, it is the complementary angle. 3 symbols, chk=5. tan 2

Find . Use to determine the sign +. 6 or 8 symbols, chk=7.sin 3/2

Find . 6 symbols, chk=6.cos

Find . 5 symbols. Use the double angle formula above.sin2

Find . Use the half angle formula above.cos/2

To determine the sign +, find the quadrant for q/2.To find the quadrant for q/2, divide by 2. 3/2Your answer should not have a fraction of fractions. You may leave in the denominator. 12 symbols.13


§7.3 548:1-8, 21,22, 35-40. Evaluate the expressions. 4 answers are fractions, one a polynomial. The rest have radicals1(4). . cos 2

532 2

First find sin q. The quadrant determines the sign.

(a) sin28 symbols, chk=14

(b) cos 26 symbols, chk=15

(c) sin 2

5 symbols, chk=4

(d) cos 26 symbols, chk=8

2(3). . sin 110 270 360

First find cos q. The quadrant determines the sign.



(c) sin 2

Double radical

3(3). Evaluate using half-angle formulas.

(a) sin/8Double radical

(b) cos/8Double radical

(c) tan/8 4 or 6 symbols, chk=3, 4

4(2). Picture for 4.

t24

7

Note hypotenuse = 242 72 25(a) sin2t7 symbols, chk=2

(b) cost/25 or 6 symbols, chk=25

5(2). . x 2 cos 0 2



Math 140 Hw 22 Name _________________________________Score ____/14

Evaluate the expressions.

A. . sin 34

2

First find cos q. The quadrant determines the sign.

(a) sin2

(b) cos 2

(c) sin 2

(d) cos 2

B. . cos 13 180 270


(a) sin2

(b) cos 2

(c) sin 2

C. Evaluate using half-angle formulas.

(a) sin/12

(b) cos/12

(c) tan/12

Picture for D and E..

4

5b

q

D. (a) (b) (c) sin2 cos 2 tan2

E. (a) (b) (c) sin/2 cos/2 tan/2

F. x 5sin, 0 2

(a) sin2

(b) cos 2

G. . x 1 2sin, 0 2

(a) sin2

(b) cos 2

Answers

A. (a) (b) (c) (d) 3 7

818

4 78

4 78

B. (a) (b) (c) 4 2

979

63

C. (a) (b) (c) 2 3

2

2 3

2 2 3

D. (a) (b) (c) 2425

725

247

E. (a) (b) (c) 110

310

13

F. (a) (b) 2x 25x2

25252x2

25

G. (a) (b) x1 32xx2

212xx2

2


Evaluate the expressions. A. . sin 3

42

First find cos q. The quadrant determines the sign.2 quadrant II cos 0

cos 1 sin2 1 916 7

16

cos 7

4

(a) sin2

sin 2 2 sin cos 2 34

74

3 78

(b) cos 2

cos 2 cos2 sin2 7

4 2 34 2

cos 2 716

916 2

16 18

2

4 2

2

2 quadrant I sin

2 0, cos 2 0

(c) sin 2

sin 2

1cos2

1 7 /42

4 78

(d) cos 2

cos 2

1cos 2

1 7 /42

4 78

Answers: (a) (b) (c) (d) 3 7

818

4 78

4 78

B. . cos 13 180 270


180 270 quadrant III sin 0sin 1 cos2 1 1/32

sin 1 1/9 8/9 2 2 /3(a) sin2

sin 2 2 cos sin 2 13

2 23

4 29

(b) cos 2

cos 2 cos2 sin2 13 2

2 29 2

19

89 7

9

(c) sin 2

quadrant III /2 quadrant II sin 0

sin 2 1cos

2 11/32 4/3

2 23

Answers: (a) (b) (c) 4 2

979

23

C. Evaluate using half-angle formulas.

(a) sin/12

sin 12 sin /6

2 1cos/62

1 3 /22

2 3

2

(b) cos/12

cos 12 cos /6

2 1cos/62

1 3 /22

2 3

2

(c) tan/12tan

12 tan /62 sin/6

1cos/6 1/21 3 /2

12 3

Answers: (a) (b) (c) 2 3

2

2 3

2 2 3

Picture for 17 and 21.

4

5b

qy

D. (a) (b) (c) sin2 cos 2 tan2

y 52 42 2516 9 3sin y/4 3/5, cos 4/5, tan y/4 3/4

(a) sin 2 2 sin cos 2 35

45

2425

(b) cos 2 cos2 sin2 45 2 3

5 2 16

25 925 7

25

(c) tan 2 2 tan 1tan2

23/413/42 3/2

169/16 837 24

7

Answers: (a) (b) (c) 2425

725

247

E. (a) (b) (c) sin/2 cos/2 tan/2sin/2 14/5

2 110

cos /2 14/52 9

10 310

tan/2 sin1cos

3/514/5 3/5

9/5 13

Answers: (a) (b) (c) 110

310

13

F. x 5 sin , 0 2

quadrant I cos 0sin x/5

cos 1 sin2 1 x/52 25x2

5

(a) sin2

sin 2 2 sin cos 2 x5

25x2

5 2x 25x2

25

(b) cos 2

cos 2 cos2 sin2 25x2

5 2 x5 2

25x2

25 x2

25 252x2

25

Answers: (a) (b) 2x 25x2

25252x2

25


Math 140 Lecture 23 Product-to-sum rules RECALL

cosx y cos xcos y sinx siny cosx y cos xcos y sinx siny

Adding these gives cosx y cosx y 2cos x cos y

Subtracting gives cosx y cosx y 2sinx siny

Similarly sinx y sinx cos y cos x siny sinx y sinx cos y cos x siny

Adding these gives sinx y sinx y 2sinx cos y

Dividing by 2 gives PRODUCT-TO-SUM FORMULA.

sin x sin y 12 [cosx y cosx y]

cos x cos y 12 [cosx y cosx y]

sin x cos y 12 [sinx y sinx y]

Write as a sum or difference of trigonometric functions. ` sin

2 cos 6

12 sin 2

6 sin 2

6

. You could continue and get: . 12 sin

3 sin 23 3 /2

` cos xcos 2x 1

2 cosx 2x cosx 2x. 1

2 cosx cos3x 12 cosx cos3x

Solving trigonometric equations RECALL. . sinx sin x, cos x cosxNote: are called supplementary angles.x and xFind all solutions for each equation. ` sin 1

2

y=1/2p/6p-p/6=5p/6

The two simplest solutions: q = p/6, and q = p-p/6 = 5p/6.Since sin(q) has period 2p, adding a multiple of 2p to

either of these also produces a solution. The general solution consists of the two sets

q = p/6 + 2pn or q = 5p/6 + 2pn for n an integer.

` Here we use x for the angle instead of q.cos x 12

The two simplest solutions are x = p/3 and x = -p/3. The general solution is

x = p/3 + 2pn or x = -p/3 + 2pn for n an integer.

CONVENTION. Assume n is an arbitrary, possibly negative, integer. We’ll omit the phrase “for n an integer”.

` tan x 13

Since tan 6 sin 6 / cos 6

12 /

32 1

3

p/6 is one solution. Since tan(x) has period p, adding amultiple of p also gives a solution. The general solutionis x = p/6 + pn.

For sin, cos, add 2pn to the (usually two) simplest solutions. For tan, add pn to the one simplest solution.`sin2x 1

iff 2x = p/2 + 2pn, Only one simple solution here.

iff x = p/4 + pn.

RECALL. sin(x) and cos(x) are always between -1 and 1. ` cos x 3

iff never. tno solution.

` 2 cos2x cos x 1 2cos2x cos x 1 0

2cos x 1cos x 1 0cos x 1

2 or cos x 1x = p/3 + 2pn or x = -p/3 + 2pn or x = p + 2pn.

`sin2x cos x 1 0 1 cos2x cos x 1 0

cos2x cos x 2 0 cos2x cos x 2 0 cos x 1cos x 2

The second is impossible.cos x 1 or cos x 2x = p + 2pn.

` 2 tan2x 3 tanx sec x 2sec2x 0 2 sin2x 3sinx 2 0

2sinx 1sinx 2 0The second is impossible. sin x 1

2 or sin x 2x = -p/6 + 2pn or x = -5p/6 + 2pn.

RECALL. A function is 1-1 iff no horizontal line crosses itsgraph more than once.

sin, cos and tan are not 1-1. But they are 1-1 on the heavily marked intervals. These are the largest such intervals containing first quadrant angles.

-p/2 p/2

0p

sin

cos

tan -p/2 p/2

For sin, the interval is [-p/2, p/2]. For cos, the interval is [0, p]. For tan, the interval is (-p/2, p/2).

tan(p/6) = 1/l3sin x sin y 12 [cosx y cosx y]

tan(p/4) = 1cos x cos y 12 [cosx y cosx y]

tan(p/3) = l3sin x cos y 12 [sinx y sinx y]

8(5). Write as a sum or difference of trig functions and then simplify. cos3 2y sin3 2y

note, move the sin to the front before applying the product formula sin(3 2y) cos(3 2y) 1

2 sin3 2y (3 2y) sin(3 2y) (3 2y)

12 sin(4y) sin(6)

12 sin6 sin4y

9(5). Find all solutions for . Hint: divide by cos2 q.sin2 3cos2 0

tan2 3 0tan2 3tan 3

, 3 n

3 n

10(10). Find all solutions for . Hint: write in terms of cos. Three sets of solutions. 2 sin2x cosx 1 0

21 cos2x cosx 1 02 2cos2x cosx 1 02cos2x cosx 1 02cos2x cosx 1 0(2cosx 1)(cosx 1) 0cos x 1

2 , cos x 1

,x 23 2n

,x 23 2n, or 4

3 2nx 2n


sin2x cos2x 1 sin2x 1 cos2x cos2x 1 sin2xtan0 0, tan 6

13

, tan 4 1, tan 3 3

9. Solve for q. sin2 2cos2 0Divide by everything by cos2 q.Rewrite in terms of tan q.

Solve for q. (Write “no solutions” if there are none.)

10. Solve for x. Should be three sets of solutions.sin2x cosx 1Rewrite in terms of just sin or just cos, not both. Since , you can solve for sin2 in terms of cos2 or cos2 in terms of sin2 getting sin2x cos2x 1

and sin2x 1 cos2x cos2x 1 sin2x

Use one of these equations to rewrite the given equation above entirely in terms of sin or entirely in terms of cos.

Now solve for x.


§7.3 548:41-46. Write each expression as a sum or differenceof trigonometric functions. Don't calculate the answer.

1(1). cos 18o sin72o

chk=21

2(1). sin 38 sin

8chk=9

3(1). cos 5xcos 2xchk=13

§7.5 568:1-12. Find all solutions (write “no solution” ifnone). Use radian measure. In each answer, write n torepresent an arbitrary integer, e.g., -p/3+2pn, p/6+pn.

One problem has no solutions, one has two sets of solutions, onehas three sets, and two have one set.

4(2). sin 12 0

5(2). tan 13 0

chk=6 or 11

§7.5 568:19-24, 46, 47. For 6, 7, 8, show your work.

6(3). 2 sin2x 3sinx 1 0chk=8, 13, 4

7(2). sin2x sinx 6 0

8(2). cos 2sec 3chk=2

Math 140 Hw 23 Name ________________________________ Score ____/14

Write each expression as a sum or difference oftrigonometric functions. Don't calculate the answer.

A. sin20 cos 10

B. cos 5 cos 45

C. sin 27 sin 5

7

D. sin 712 cos

12

E. sin3xsin4x

F. sin6cos 5

Find all solutions (write “no solution” if none). Useradian measure. In each answer, write n to represent anarbitrary integer, e.g., -p/3+2pn.

G. sin 3 /2

H. sin 1/2

I. cos 1

J. tan 3

K. tanx 0

L. 2 cos2 cos 0

M. cos2t sin t sin t 0

N. 2cos2x sinx 1 0

AnswersA. 1

2 sin 10 14

B. 12 cos 3

5 12

C. 12 cos 3

7 12

D. 12 sin

2 sin 23

E. 12 cos x 1

2 cos 7xF. 1

2 sin 12 sin 11

G. 3 2n or 2

3 2nH. 6 2n or 5

6 2nI. 2nJ.

3 nK. x nL. 2 2n, 2

3 2n, or 23 2n

M. t nN. x

6 2n, x 56 2n, or x 3

2 2n


Write each expression as a sum or difference oftrigonometric functions. Don't calculate the answer.A. sin20 cos 10

12 sin200 100 sin200 100

12 sin100 sin300 1

2 sin100 12

Answer: 12 sin 10 1

4

B. cos 5 cos 4

5

12 cos 5

45 cos 5

45

12 cos 3

5 cos 1

2 cos 35 1

Answer: 12 cos 3

5 12

C. sin 27 sin 5

7

12 cos 2

7 57 cos 2

7 57

12 cos 3

7 cos 1

2 cos 37 1

Answer: 12 cos 3

7 12

Find all solutions (write “no solution” if none). Useradian measure. In each answer, write n to represent anarbitrary integer, e.g., -p/3+2pn.G. sin 3 /2

Two simplest: 3 , 2

3Answer:

3 2n or 23 2n

H. sin 1/2Two simplest:

6 , 56

Answer: or, equivalently, 6 2n, 5

6 2nAnswer: 7

6 2n or 116 2n

I. cos 1One simplest: Answer: 2n

J. tan 3Simplest answer:

3Answer:

3 n

L. 2 cos2 cos 0

2cos2 cos 0cos2cos 1 0cos 0, cos 1/2

when cos 0 2 , 2

when cos 12 2

3 , 23

Answer: 2 2n, 23 2n

or, equivalently, Answer:

2 n, 23 2n, 4

3 2n

M. cos2t sin t sin t 0

sin tcos2t 1 0sin tsin2t 0sin3t 0sin t 0Simplest answers: , t 0 Answer: t 2n, 2n

N. 2cos2x sinx 1 0

21 sin2x sinx 1 02 2sin2x sinx 1 02sin2x sinx 1 02sin2x sinx 1 02sinx1sinx1 0sinx 1/2, sinx 1sin x 1

2 when x 6 , 5

6sin x 1 when x

2

Answer: x 6 2n, x 5

6 2n, or x 2 2n

or, equivalentlyAnswer: x

6 2n, x 56 2n, or x 3

2 2n


Math 140 Lecture 24Exam 4: Lectures 19-24, including this lecture.

Inverse trigonometric functionsRECALL. A function is 1-1 iff no horizontal line crosses its

graph more than once. While not 1-1 in general, sin, cos, and tan are 1-1 on the

first quadrant angles in [0, p/2] and on the largerheavily marked “1-1” intervals.

-p/2 p/2

0p

sin

cos

tan -p/2 p/2

1

-1

DEFINITION. sin -1, cos

-1, tan -1 are the inverses of the above

restrictions of sin, cos, and tan. Thus sin

-1(x) = the q i[-p/2, p/2] such that sin(q) = x, cos

-1(x) = the q i[0, p] such that cos(q) = x, tan

-1(x) = the q i(-p/2, p/2) such that tan(q) = x. sin-1(x) and cos-1(x) have domain [-1, 1], for x2[-1, 1],

they are undefined. tan -1(x) has domain (-5,5).

NOTATION. sin -1(x), cos

-1(x) and tan -1(x) are also written:

arcsin(x), arccos(x), and arctan(x).

Warning, sin -1(x) = 1/sin(x). sin

-1(x) is the inverse; (sin(x))

-1 = 1/sin(x) is the reciprocal.

= p/2cos -1(0)= p/2sin -1(1)

= p/3tan -1(l3)= p/3cos -1(1/2)= p/3sin -1(l3/2)

= p/4tan -1(1)= p/4cos -1(1/l2)= p/4sin -1(1/l2)

= p/6tan -1(1/l3)= p/6cos -1(l3/2)= p/6sin -1(1/2)

= 0tan -1(0)= 0cos -1(1)= 0sin -1(0)

The unit circle pictures are

x

arcsin(x)arccos(x)

x

x

1

-1

-1 1

p/2

-p/2

p 0

The heavily marked half circles are the restricted ranges.

`arcsin(-1/2) = -p/6 arccos(-1/l2) = 3p/4 cos -1(2) = undef since 22[-1, 1] arctan(-1) = -p/4 cos(tan-1(-1)) = cos(-p/4) = cos(p/4) = 1/ 2

Since inverses undo each other, we have

tan(tan -1

x) = x for any xtan-1(tan q) = q if qi(-p/2,p/2)

cos(cos-1 x) = x if xi[-1, 1]cos

-1(cos q) = q if qi[0, p]

sin(sin-1 x) = x if xi[-1, 1]sin-1(sin q) = q if qi[-p/2, p/2]

RECALL. . cosx cos x, sin x sinxWhen a formula involves an angle outside the restricted

range, rewrite it in terms of an angle q in the restrictedrange before applying the equations in the box.

` coscos1 15

15

cos1cos 45

45

�important cos1cos 5 cos1cos 5 5

sin1sin 5 5

�important sin1sin 45 sin1sin 5

5

THEOREM. For xi[-1,1], cossin1x 1 x2

sincos1x 1 x2

Proof of 1st. Recall: . cos2 1 sin2t cos 1 sin2t

cossin1x 1 sinsin1x2

1 x2

The “+” was chosen over the “-” since sin-1(x) i[-p/2, p/2] ˆ cos(sin-1(x)) > 0.

` sinsin1 15 1

5

cossin1 15 1 1

25 2425 46

25 2 65

tansin1 15

sinsin1 15

cossin1 15 1/5

2 6 /5

612

` secsin11 cos11 sec 2 0 1/ cos 2 1/0 undef.

In a triangle with hypotenuse 1 and side x, = angle opposite x, sin1x = angle adjacent to x. cos1x

x

arcsin(x)

arccos(x)

1

Note that sin1x cos1x /2 90o

tan(tan -1

x) = x for any xtan-1(tan q) = q if qi(-p/2,p/2)cos(cos-1

x) = x if xi[-1, 1]cos -1(cos q) = q if qi[0, p]

sin(sin-1 x) = x if xi[-1, 1]sin-1(sin q) = q if qi[-p/2, p/2]

if cossin1x 1 x2 sincos1x 1 x2 x 1, 1

11. Find the exact value (no credit for a decimal).

(a) ___/2cos11/ 2

3/4

(b) ___/2sin11/2

/6

(c) ___/2tan1 3

/3

(d) ___/2cos1cos 47

cos1 cos 47

47

(e) ___/2sin1sin 47

sin1 sin 47

sin1 sin 37

sin1 sin 37

37

(f) ___/4tanarcsin 13

sin sin1 1

3cos sin1 1

3

13

1 19

1389

1383

18


11. Find the exact value (no credit for a decimal). (a) The answer is an angle in the second quadrant. 4 symb, chk=5cos11/2.

(d) cos1cos5/9Warning -5p/9 is not correct. -5p/9 is not in the restricted region [0, p] for the inverse of cos.Rewrite in terms of an angle in the restricted region for cos-1 before canceling the inverses. 4 symb, chk=14

cos1cos5/9 cos1cos

(e) sin1sin5/9Warning is not correct. -5p/9 is not in the restricted region [-p/2, p/2] for the inverse of sin.5/9Rewrite in terms of an angle in the restricted region for sin-1 before canceling the inverses. 5 symb, chk=13

sin1sin5/9 sin1sin

(f) tanarcsin 23

Write tan as . Then use the theorem: . 4 or 5 symb, chk=7 or 12 sincos cossin1x 1 x2


§7.4 557:1-8. Evaluate exactly without a calculator: pand rather than 3.14 or 1.41. Answer may be “undef.” 2

1(1). One symbolcos11

2(1). 4 symbol fraction with p, chk=7arccos 2 /2

3(1). 4 symbol fraction with p, chk=2arcsin1

§7.4 557:13-26. 4(1). Carefulsinarcsin2

5(1). 4 symbol fraction with p, chk=2sin1sin 32

6(2). 3 symbol fraction, chk=7tansin1 45

7(1). 4 symbol fraction with radical, chk=3sintan11

8(2). 4 symbol fraction, chk=8tanarccos 513

9(1). 3 symbol fraction, chk=3cosarctan 3

10(1). 5 symb fraction with radical, chk=7sinarccos 13

11(2). 2 symbol radical, chk=2seccos11/ 2 sin11

Math 140 Hw 24 Name _____________________________ Score ____/14 Exam 4

Evaluate exactly without a calculator: p and rather than23.14 or 1.41. Answer may be “undef.”

A. sin1 3 /2

B. tan1 3

C. arctan1/ 3

D. tan11

E. cos12

F. sinsin1 14

G. coscos1 34

H. arctantan/7

I. arcsinsin/2

J. arccoscos2

K. cosarcsin 27

5 symbol fraction with square root

L. sintan115 symbol fraction with square root

M. cossin1 23


N. sincos1 13


O. tanarcsin 2029

5 symbol rational

P. cscsin1 12 cos1 1

2 Integer

AnswersA. B. C. D. E. undefined/3 /3 /6 /4F. 1/4 G. 3/4 H. I. J. 0/7 /2K. L. M. N. 3 5 /7 1/ 2 5 /3 2 2 /3O. 20/21 P. -2

Math 140 Hw 24 Recommended problems, don’t turn this in.

Evaluate exactly without a calculator: p and rather than23.14 or 1.41. Answer may by “undef.”

A. sin1 3 /2

Answer: /3

B. tan1 3

Answer: /3

C. arctan1/ 3

Answer: /6

D. tan11

Answer: /4

E. cos12 is only defined on [-1, 1].cos1x

2p 6.28 > 1.Answer: undef.

F. sinsin1 14

Answer: 1/4

G. coscos1 34

Answer: 3/4

H. arctantan/7

Answer: /7

I. arcsinsin/2

Answer: /2

J. arccoscos2

since 2p is not in the restrictedarccoscos 2 2range [0,p] for cos. But and 0 is in [0,p].cos 2 cos 0

arccoscos 2 arccoscos0 0.Answer: 0

K. 5 symbol fraction with square rootcosarcsin 27

1 27 2

1 449

3 57

L. 5 symbol fraction with square rootsintan11

sin/4 1/ 2or 2 /2

M. 4 symbol fraction with square rootcossin1 23

1 23

2

1 49

53

N. 5 symbol fraction with square rootsincos1 13

1 13

2

1 19

2 23

O. 5 symbol rationaltanarcsin 2029

sinsin120/29cossin120/29

20/29

1400/292 20

21

P. Integercscsin1 12 cos1 1

2

csc 6 3 1/ sin 6 2


Math 140 Lecture 25Trigonometric word problems(a) Find the exact answer. (b) Find the decimal answer.To test if your calculator is in radian or degree mode, calculate sin(1).

sin(1o) .017, sin(1rad) .84 . Give only exact answers on tests. The bookstore has cheap trig calculators for less than $20.

`Point P is level with the base of a 1000 ft high building which is4000 ft away. Find the angle of elevation from P to the top ofthe building. (a) the exact radian answer, (b) to nearest degree.

1000

P 4000

q

Let q be the angle. tan(q) = 1000/4000 = 1/4. (a) q = tan-1(1/4) exact answer. (b) q = 14o to nearest degree.

`An antenna sits atop a 1000 ft high building. From point a Pon the ground, the angle of elevation to the top of the buildingis b, the angle of elevation to the top of the antenna is a.Express the height of the antenna in terms of a and b.

1000

P xb a

h

Let h be the height of the antenna.Let x be the distance between P and the building. Thentan 1000

x x Note: find the secondary variable x first. 1000

tan

tan h1000x

x tan h 1000h x tan 1000h 1000

tan tan 1000Answer: Height is ft. Remember the units.1000 tan

tan 1`From a point P level with the base of a mountain, the angle of

elevation of the mountain is a. From a point Q 1 mile closer tothe mountain’s base, the angle of elevation is b. Express theheight of the mountain in terms of a and b.

a b

1

h

xP Q

Let h be the height. Let x be as pictured. Again, find thesecondary variable x first.

t so tan hx x h

tan so tan h

1x h 1 x tan 1 htan tan

h tan htan tan

h tan tan h tanh tan tan tan h tanh tan h tan tan tanhtan tan tan tan

Answer height = miles.tan tan /tan tan

THEOREM. The area of a triangle with sides a and b andincluded angle q is . 1

2 ab sinPROOF. Case q is acute.

a

bq h

To find the area, we need to know the height h. sin h

a h a sin

Area . 12 hb 1

2 a sinb 12 ab sin

Case q is obtuse. Recall sin(p-q) = sin q.

a

b

qhp-q

sin ha

h a sin a sin

Area (same answer) 12 hb 1

2 a sinb 12 ab sin

`Find the area of an octagon (stop sign) of radius 1 ft.Give the exact answer and the decimal answer to 2 places.

1q1

In each of the 8 triangles, the two sides are radii andhave length 1.

. 28

4The area each triangle 1

2 11 sin 4 .1

2 22

24

The area of the octagon = 8x the area of each triangle . 8 2

4 2 2Exact answer = sq.ft. Decimal answer = 2.83 sq. ft.2 2

2(6). A boat is sighted from two lighthouses which areone mile apart along a straight east-west shoreline. Fromthe first lighthouse, the angle between the boat and dueeast is 35o. From the second lighthouse, the anglebetween the boat and due west is 50o. How far is the boatfrom the shore? Exact answer plus units.

d

1st lighthouse 2nd Lighthouse

Boat

1 mile1-x x

EastWest35 50o o

Let d be the distance between the boat and the nearest point on the shoreline.Let x be the distance between the closest lighthouse and the nearest point on the shore line.

tan50o dx x d

tan 50o

tan 35o d1 x 1 x tan35o d

1 dtan 50o tan 35o d

tan 35o d tan 35o

tan 50o d

tan35o tan 50o d tan 35o d tan50o

tan50o tan 35o d tan35o d tan50o

milesd tan 50o tan 35o

tan 35o tan 50o

6(6). Find the area of a pentagon whose sides each havelength 2. Give the exact answer.

q

2

1 1

h

qq/2

5 isosceles triangles.Inner angle: 2/5Opposite side: 2Divide the triangle in halfwith height h.

tan 5 1

hh 1

tan 5

Area of triangle: 12 bh 1

2 2 1tan

5 1

tan 5

Area of pentagon: 5tan

5

Math 140 Discussion Lecture 25 Exam problems points/time 12pts 17minsNo calculators allowed on any test.

6. Find the area of a regular (all sides are equal) 7 sidedfigure inscribed in a circle of radius 1 inch. Give the exactanswer, no decimals. Do not solve for any triangle's height or width.Don’t use the area formula ½bh. Instead use ½absinq of Lecture 25.

The 7-sided figure divides into 7 triangles each with two sides of length 1. Draw the picture.

What is the included angle of each triangle? chk=9

What is the area of each triangle? chk=11,12Recall .A 1

2 ab sin

What is the total area? chk=18, remember the units

2. A boat is sighted from of two lighthouses which areone mile apart along a straight east-west shoreline. Fromthe first lighthouse, the angle between the boat and dueeast is 55o. From the second lighthouse, the anglebetween the boat and due east is 70o. How far is the boatfrom the shore? Exact answer plus units. This is a variant ofthe problem in Lecture 25.

d

Lighthouse Lighthouse

Boat

1 milex

70o

55 o

Let d be the distance between the boat and the nearest point on the shoreline.Let x be the distance between the closest lighthouse and the nearest point on the shore line.

There are two right triangles in the picture. Apply tangentto the two given angles to get two equations involving xand d. chk=7, chk=11

Solve these two equations for d. chk=34, remember the units


§6.2 484:44. §7.4 558:53-55. 2sina has 5 symbols, 2sin(a) has 7 symbols.

1(3). From a point level with and 1000 ft away from thebase of the Washington Monument, the angle of elevationto the top of the monument is 29o. Find the height of themonument. (a) Draw the picture. (b) Give the exact answer using

tan (remember the units). (c) Find the height to the nearest foot.

(a) Picture.

(b) 10-12 symbols + units, chk=12 Exact height =

(c) 3 symbols + units, chk=14 Nearest ft =

2(3). In triangle DABC, the sides are AC = BC = 8 inchesand AB=4. Find the angles. (a) Give the exact answer. (b) Find

the angle to the nearest degree.Hint. Let P be the midpoint of AB. Then APC is a right triangle.

8

4A B

8

C

VA = VB = = chk=5,13

exact radian answer nearest degree answer

VC = = exact radian answer nearest degree answer

3(2). Find the distance AB if AC = 400m, VC = 90o, VA = 40o. Find AB. (a) Give the exact answer. (b) Find the

3-digit distance to the closest meter.

B A

C

40

?

400

(a) 10-12 symbols + units, chk=8 Exact distance =

(b) 3 symbols + units, chk=9 Nearest meter (m) =

4(4). A surveyor stands 30 yards from the base of abuilding. On top of the building is a vertical radioantenna. Let a be the angle of elevation when thesurveyor sights to the top of the building. Let b be theangle of elevation when the surveyor sights to the top ofthe antenna. (a) Draw the picture. (b) Give the length of the antenna in terms of the angles a and b.(a) Picture.

(b) 13-17 symbols + units Antenna height =

5(3). The radius of the circle in the following figure is 1unit. Express the lengths of OA, AB, and DC in terms oftrigonometric functions of a. (No units are given in this case.)

aO A

B

C

D

1

4-6 symbols OA =

4-6 symbols AB =

4-6 symbols DC =


2sina has 5 symbols, 2sin(a) has 7 symbols.

A. When earth E, Mercury M, and the sun S are lined upso that VEMS is a right angle, VSEM is 21o. Given thatthe distance ES from the earth to the sun is 93 millionmiles, find the distance MS from Mercury to the sun tothe nearest million miles.

B. A contractor must fence a plot with the shape of a righttriangle. One angle is 40o. The hypotenuse is 60 m. Findthe length of fencing required. Round to the nearestmeter.

C. Given: q = 39.4o and x = 43.0 ft.

q q

x xh

(a) Find the height h to one decimal.

(b) Find, to one decimal, the total area of the triangle(the large isosceles triangle, not the two right-triangle pieces).

D. From a point at ground level, the angle of elevation tothe top of a mountain is 38o. 200 m further away from themountain, the angle is 20o. Find the mountain’s height.

E. The arc in the picture is part of the unit circle.

q

A P

BO

(a) Express in terms of q: VBOA, VOAB, VBAP, VBPA.

(b) Express in terms of sin q, and cos q: AO, AP, OB, BP.

Answers

A. 33 million miles B. 145 m C. (a) 27.3 ft. (b) 906.9 ft2 D. 136 m. E. (a) VBOA = 90o

- q, VOAB = q, VBAP = 90o- q,VBPA = q. (b) AO = sin q, AP = cos q, OB = sin2

q, BP = cos2 q.

Math 140 Hw 25 Recommended problems, don’t turn this in.

2sina has 5 symbols, 2sin(a) has 7 symbols.

A. When earth E, Mercury M, and the sun S are lined upso that VEMS is a right angle, VSEM is 21o. Given thatthe distance ES from the earth to the sun is 93 millionmiles, find the distance MS from Mercury to the sun tothe nearest million miles.

M

S

E21o

93x

sin210 x93

x 93sin 210 93.358 33

Answer: 33 million miles

B. A contractor must fence a plot with the shape of a righttriangle. One angle is 40o. The hypotenuse is 60 m. Findthe length of fencing required. Round to the nearestmeter.

40o

60x

y

, sin 400 x60 x 60sin400 60.643 38.567

cos 400 y

60 , y 60 cos 400 60.766 45.963Fence = 60 + 38.567 + 45.963 = 144.530

Answer: 145 m

C. Given: q = 39.4o and x = 43.0 ft.

q q

x xh

b

. sin hx h

43 h 43sin39.40 43.635 27.293cos b

x b43 . b 43 cos39.40 43.773 33.228

(a) Find the height h to one decimal. Answer: 27.3 ft.

(b) Find, to one decimal, the total area of the triangle. Each half has area 1

2 hb 12 27.29333.228 453.446

Answer: 906.9 ft2 D. From a point at ground level, the angle of elevation tothe top of a mountain is 38o. 200 m further away from themountain, the angle is 20o. Find the mountain’s height.

20 38o o

200m

h

x

tan 380 hx , x h

tan 380

tan 200 h200x

h200h/ tan 380

200 htan 380 h

tan 200

htan 200 h

tan 380 200

h 200/1/ tan200 1/ tan 38

1.468] = 136.2h 200/

Answer: 136 m.

E. The arc in the picture is part of the unit circle.

q

A P

BO

(a) Express in terms of q: VBOA, VOAB, VBAP, VBPA.

VBOA = 90o - qVOAB = 900 - VBOA = 90o - (90o - q) = qVBAP = 90o - VOAB = 90o - q VBPA = 90o - VBAP = 90o - (90o - q) = q

(b) Express in terms of sin q, and cos q: AO, AP, OB, BP.

AO = height of P = sin q

AP = x-coordinate of P = cos q

OB = ?sin(VOAB) = OB/OAOB = OA.sin(VOAB) = sinq.sinq = sin2q

BP = 1 OB 1 sin2 cos2


Math 140 Lecture 26CONVENTION. Assume side a is opposite angle A, side b is

opposite angle B and side c is opposite angle C. SINE LAWS. In any triangle, the ratio of one angle’s sine

and its opposite side equals the ratio of any otherangle’s sine and opposite side.

Although written as one, there are 3 equations. Each involves two

sides and two angles.

orsin Aa sin B

b sin Cc

asin A b

sin B csin C

PROOF. Recall that the area of a triangle is half the productof any two sides times the sine of their included angle.

Thus the area of the triangle can be written three ways:

12 bc sin A 1

2 ac sin B 12 ab sin C

multiply by 2bcsinA ac sinB absinC

divide by abc

sin A

a sin Bb sin C

cCOSINE LAWS. For any two sides of a triangle, the sum of

their squares minus twice their product times the cos ofthe included angle equals the square of the third side.

Each involves three sides and one angle.

a2 b2 c2 2bc cos A

b2 a2 c2 2ac cos B

c2 a2 b2 2abcos C

PROOF. We prove the last equality for the case C acute.

C

B

A

c

b

a

x

h

b-x

sin C ha , cos C x

a , so h a sin C, x a cos C c2 b x2 h2

c2 b2 2bx x2 h2

c2 b2 2bacos C a2 cos2C a2 sin2C c2 b2 2abcos C a2cos2C sin2C

c2 a2 b2 2abcos C

STRAIGHT ANGLE SUM FACT. The sum of a triangle’s 3angles is a straight angle.

VA +VB +VC = 180o = pIf you know two angles, you can solve for the third.

Solving for C gives VC = 180o - (VA +VB).

Today’s problems involve 4 quantities; each is a side oran angle whose measure is given or wanted. Usually --

v For 2 sides and 2 angles: use the sine law involving the2 sides (if necessary, get the third angle with theStraight Angle Sum Theorem).

v For 3 sides and 1 angle: use the cosine law involvingthe angle.

b

aC B

A

c

`Given b, VA, VC: find c. 2 side, 2 angle problem. Use the sine law with b and c.

where B = 180o - (A+C) .c

sin C bsin B

`Given a, b, c: find VC. 3 side, 1 angle problem. Use the cosine law with VC.

. Solve for cosC then take cos -1.c2 a2 b2 2abcos C

Give an exact answer and a 2-place decimal answer. `VA = 20o, VB = 30o, c = 40cm. Find a. 2 angle, 2 side problem with sides a, c. Use the sine law for c and a.

where C = 180o - (VA +VB) a

sinA csinC

= 180o - (20o

+ 30o) = 180o - 50o = 130o.

Now solve for a.

a

sinA csinC

cm �Exact answer a c sin Asin C 40 sin 20o

sin 130o

�2-place decimal answer 17.86cmAdditional table-user step:

, so sinx sin x sin180o x. sinC sin130o sin180o 130o sin50o

`VA = 20o, b = 50, c = 60. Find a. 3 side, 1 angle problem with angle VA. Use the cosine law for VA.

a2 b2 c2 2bc cos ASolve for a.

a b2 c2 2bccosA 502 602 26050cos 20o

�Exact answer 6100 6000 cos 20o

�2-place decimal answer 21.49`a = 20, b = 20, c = 30. Find VA. 3 side, 1 angle problem with angle VA. Use the cosine law for VA.

a2 b2 c2 2bc cos ASolve for . cos A

2bc cos A b2 c2 a2

cos A b2 c2 a2/2bc A cos1b2 c2 a2/2bc

cos1900/1200 cos13/4 41.41o

b

aC B

Ac

c

ab

C

A

B

Give exact answers, not decimal answers.

A

ca

bC

B

4(9). a = 3, b = 1, c = 3. Find VB.

3 side, 1 angle problem with angle VB. Use the cosine law for VB.

b2 a2 c2 2ac cos BSolve for . cos B

2accos B a2 c2 b2

cos B a2 c2 b2/2ac

B cos1a2 c2 b2/2ac cos132 32 12/2 3 3 cos117/18

3(9). VC = 40o, b = 3, a = 5. Find c.

3 side, 1 angle problem with angle VC. Use the cosine law for VC.

c2 b2 a2 2ba cos CSolve for c. c b2 a2 2bacosC

32 52 235cos 40o

34 30 cos 40o

5(6). VA = 50o, a = 4, c = 5. Find the two values of VC.

There are two values since the side opposite is < side adjacent.2 side, 2 angle problem.

Use the sine law with sides a, c.

sin Cc sin A

a

sin C c sin Aa 5 sin 500

4

and C sin1 5 sin 500

4

orC 1800 sin1 5 sin 500

4

C sin1 5 sin 500

4

5(10). VA = 60o, c = 10, a = 9. Find b.

3 side, 1 angle problem with angle VA. Use the cosine law for VA

a2 b2 c2 2bc cos A

0 b2 2bc cos 600 c2 a2

b2 2c 12 b c2 a2 0

b2 10b 100 81 0

b2 10b 19 0

b b b2 4ac

2a

10 100 4 19

2 10 24

2 5 6

Math 140 Discussion Lecture 26 Exam problems points/time 34pts 26minsNo calculators allowed on any test. Note, this is longer than usual, watch your time.

Give exact answers, no decimal answers.Use either the sin law (2 sides, 2 angles) or the cosinelaw (3 sides, 1 angle).

A

ca

bC

B

3. VB = 55o, a = 3, c = 5. Find b. chk=20

5a. VC = 40o, c = 4, a = 5. Find the two values of VA. Recall if q is a solution to , so is 180o- q when the sidesin aopposite is < the side adjacent. chk=13, 22

5b. VC = 60o, c = 3, a = 2. Find the one b.

Use the quadratic formula . Since triangle sidesx b b24ac2a

have positive lengths; discard any negative answers. chk=7

Math 140 Classwork 26 Score ___/9No calculators allowed on any test.

A

ca

bC

B

§6.4 506:11-26. §6.5 513:11-17. (a) Give the unsimplified exact answer obtained using thelaw of sines or cosines or Straight V Thm. E.g.

or .a 82 52 285 cos200 b 5 sin100o

sin50o(b) Give the decimal answer accurate to two places. “sym”means “symbols”including “.”. 11.23o, 11.23cm have “5 sym+units”

A B

C

c

ab

1(2). VA = 30o, VB = 135o, a=4 cm, find b.

Exact b =

Decimal (4sym + units, chk=17) b =

2(2). VA=VB=35o, c=16cm, find a. Hint, 1st find VC.VC=180-(35+35)=110o.

asin A c

sin C , a c sin Asin C

Exact a =

Decimal (4 sym + units, chk=23) a =

3(4)ab. In figure (a) and (b), find x.

3010 cm

4 cmo x

(a)

10 cm4 cm

o150 x

(b)

(a) Exact x =

Decimal (4 sym+units, chk=18) x =

(b) Exact x =

Decimal (5 sym+units, chk=11) x =

4(5. a=33, b=7, c=37. Find VA, VB, VC. An unsimplified exact answer for an angle might be of the form

or A cos1202 52 122/2205 B 180 70 85

A B

C

c

ab

Exact answers may use the cosine law or be calculated as B is.

Exact A =

Decimal (5 sym + unit. chk=17) VA=

Exact C =

Decimal (3 sym + unit, chk=3) VC=

Decimal for VB (4 sym + unit, chk=16) VB=

5(4). Find the perimeter of a regular nine-sided polygoninscribed in a circle of radius 4cm.

1st find q, 2nd find other two two triangle angles. 3rd find x. 4th find the perimeter.

Exact value of perimeter x =

Decimal answer (5 sym + units, chk=18) =


4 xq

(a) Give the unsimplified exact answer obtained using thelaw of sines or cosines or Straight V Thm. E.g.

or .a 82 52 285 cos200 b 5 sin100o

sin50o(b) Give the decimal answer accurate to two places.See text or worked examples for the answers

A B

C

c

ab

A. VA = 60o, VB = 45o, a=12cm, find b.

Exact b =

Decimal b =

B. VB=100o, VC=30o, c=10cm, find a.

Exact a =

Decimal a =

C. In figure (a) and (b), find x.

(a) (b)

60120

5cm

8cm

5cm

8cm

x x

oo

(a) Exact x =

Decimal x =

(b) Exact x =

Decimal x =

D. In figure (a) and (b), find x.

(a) (b)7.3cm

11.5cm

7.3cm

11.5cm40o

140ox

x

(a) Exact x =

Decimal x =

(b) Exact x =

Decimal x =

An unsimplified exact answer for an angle might be of the form or A cos1202 52 122/2205 B 180 70 850


A B

C

c

ab

E. a = 7, b = 8, c = 13. Find VA, VB, VC.

Exact A =

Decimal (5 sym + units) VA=

Exact C =

Decimal (3 sym + units) VC=

Exact B =

Decimal (5 sym + units) VB=

F. a = b = , c = 2. Find VA, VB, VC. 2/ 3

Exact A =

Decimal (2 sym + units) VA=

Exact C =

Decimal (3 sym + units) VC=

Exact B =

Decimal (2 sym + units) VB=

G. Find the perimeter of a regular pentagon inscribed in acircle of radius 1cm.

1

Exact value of a side x =

Decimal answer for perimeter =

Answers

See Hw 26 worked examples.


(a) Give the unsimplified exact answer obtained using thelaw of sines or cosines or Straight V Thm. E.g.

or .a 82 52 285 cos200 b 5 sin100o

sin50o(b) Give the decimal answer accurate to two places.

A B

C

c

ab

A. VA = 60o, VB = 45o, a = 12cm, find b.

Exact b = cm 12 sin 45sin 60 4 6

Decimal b = 9.80 cm

B. VB=100o, VC=30o, c=10cm, find a.

Exact a = cm10 sin 50sin 30

Decimal a = 15.32 cm

C. In figure (a) and (b), find x.

(a) (b)

60120

5cm

8cm

5cm

8cm

x x

oo

(a) Exact x = cm52 82 2 5 8cos 60Decimal x = 7 cm

(b) Exact x = cm52 82 2 5 8cos 120 129Decimal x = 11.36 cm

D. In figure (a) and (b), find x.

(a) (b)7.3cm

11.5cm

7.3cm

11.5cm40o

140oxx

(a) Exact x = cm7.32 11.52 2 7.3 11.5cos 40Decimal x = 7.54 cm

(b)Exact x = cm7.32 11.52 2 7.3 11.5cos 140Decimal x = 17.72 cm

An unsimplified exact answer for an angle might be of the form or A cos1202 52 122/2205 B 180 70 85


A B

C

c

ab

E. a = 7, b = 8, c = 13. Find VA, VB, VC. C = cos1(a2 b2 c2/2ab)

Exact A = cos1 8213272

2813

Decimal (5 sym + units) VA = 27.80o

Exact C = cos1 7282132

287

Decimal (3 sym + units) VC = 120o

Exact B = cos1 1327282

2137

Decimal (5 sym + units) VB = 32.200

F. a = b = , c = 2. Find VA, VB, VC. 2/ 3

Exact A = cos1 222/ 3 22/ 3 2

222/ 3

Decimal (2 sym + units) VA = 30o

Exact C = cos1 2/ 3 22/ 3 222

22/ 3 2/ 3

Decimal (3 sym + units) VC = 120o

Exact B = cos1 222/ 3 22/ 3 2

222/ 3

Decimal (2 sym + units) VB = 30o

G. Find the perimeter of a regular pentagon inscribed in acircle of radius 1cm.

1

1

q x

q =360/5 = 72o x = 12 12 2 1 1cos 720

Value of x = 1.18 cm

Decimal answer for perimeter = 5x = 5.88 cm


Math 140 Lecture 27A triangle is determined uniquely up to congruence (1) by two sides and an included angle, and also,(2) by two angles and an included side.

side-angle-side angle-side-angle two sides, nonincluded

However, given two sides and a nonincluded angle, theremay be 0, 1, or 2 triangles.

`Suppose VA = 30o, b = 2. Then there is no triangle witha =.5, one triangle with a =1, two triangles with a =1.5,and one triangle with a = 2.5.

A A A A

a=1a=1.5 a=2.5b=2 b=2 b=2 b=2

none one two solutions one solution

a=.5

30 30 30

Suppose two sides and a nonincluded angle are known. When solving for the third side using a cosine law, you

may get an answer of the form . Then there is: s tv No solution if t < 0 or both . s t 0v Two solutions if t > 0 and both . s t 0v One solution otherwise. When solving for a second angle using a sine law, you

may get an answer of the form q = sin-1t, t > 0. There is: v No solution if t >1. v Two solutions if t <1 and the larger of the two sides is adjacent to the given angle (q is one angle, p-q = the other).v One solution otherwise. Ìn a triangle, sin A = . What are the possible angles,1/2

in degrees, for A? One is A = 30o, the other is 180-30 = 150o. Try doing this and the next problem by drawing accurate pictures.

Ìs there a triangle in which a = 2, b = 3, and VA = 60o ? Such a triangle exists iff the third side c exists. Solving for c. a2 b2 c2 2bc cos A4 9 c2 23c cos 60o

0 c2 23c 12 5

c2 3c 5 0

c 3 32415

21 3 11

2 undefinedSince c is undefined, the triangle does not exist.

à = 2, b = 2, V A = 30o. Find VC if VB is acute. First find VB using the sin law, then find VC.

Thus one answer issin Bb sin A

a , sin B ba sin A.

= 30o. TheB sin1 ba sin A sin1 2

2 sin 30o sin1 12

other answer is 180o-30o = 150o. The acute angle is 30o.VC = 180o-(VA +VB) = 180o-(30o

+30o) = 120o .

Polar coordinates DEFINITION. The polar coordinates (r, q) of a point P are

its distance r (radius) from the origin and the angle qbetween the positive x-axis and the line from (0,0) to P.

The usual (x, y) are the rectangular coordinates.

x

yr

q

Note, thisis not theunit circle;it has radius r.

(0,0)

P quad I

quad IVquad III

quad II

`Plot the polar coordinates points (1, p/4), (2,- p/2).

From the picture we have, for positive r

tan1 yx if II, III tan1 y

x if I, IV

r x2 y2x r cosy r sin

tan yxr2 x2 y2cos x

rsin yr

THEOREM. If a point has rectangular coordinates (x, y) andpolar coordinates (r, q) with r > 0, then:

and x, y r cos, r sin if (x, y) is in quadrant I, IV,r, x2 y2 , arctan

yx

if (x, y) in quad. II, III.r, x2 y2 , arctanyx

NEGATIVE r. is the point on the oppositer, r, side of the origin (0,0) as . r,

(r, )

(-r, )q

q

(0,0)q

`Convert from polar coordinates to rectangular: . 7, 6 r = 7, q = p/6, (x, y) = (rcosq, rsinq) =

�answer 7 cos 6 , 7 sin 6

7 32 , 7

2 `Convert from rectangular coordinates to polar:(-1, ).3

x = -1, y = . Point is in quadrant II, tadd p. 3(r, q) = x2 y2 , tan1 y

x

1 3 , tan1 31 2,

3 2, 23

�answer `Convert the polar equation to a rectangular equation:

. r sin 2cos 0 ƒ y 2x/r 0 y 2 x

x2y2 0

� answer (hw simplifies more) y x2 y2 2x 0`Convert the rectangular equation to a polar equation:

2x y2 0 � answer 2r cos r2 sin2 0

7(a)(2) Find the rectangular coordinates of the point withpolar coordinates . r, 4, 5/4

x, y (rcos, r sin) 4cos(5/4),4sin(5/4) 4cos /4,4 sin /4 4cos/4, 4 sin/4

4 22 , 4 2

2 2 2 , 2 2

7(b)(3) Find the polar coordinates of the point withrectangular coordinates . x,y 2, 2

r x2 y2

22 22

2 2 tan1y/x tan11 /4

r, 2 2 , 3/4

8(a)(3) Convert the polar equation to a rectangularequation: , . cotcsc r r 0

cossin

1sin r

cos r sin2xr r

yr

2

x y2

8(d)(2) Convert the rectangular equation to a polarequation: . x y2 y 1

rcos r2 sin2 r sin 1

Math 140 Quiz Discussion Lecture 27 Exam problems points/time 10pts 8mins

7(a) Find the rectangular coordinates of the point withpolar coordinates . Note: this is in quad. II.r, 3,/4

chk=10 or 14 x, y (r cos, r sin)

7(b) Find the polar coordinates of the point withrectangular coordinates x, y 1, 3

What is the quadrant for this point? I?, II?, III?, IV?

Find r using= chk=2r x2 y2

Find q.If q is in quadrant I or IV, tan1y/xOtherwise use the formula chk=7 tan1y/x

Now write your answer as an ordered pair: chk=9

r,

8(a) Convert the polar equation to a rectangular equation: , . sin r r 0

First replace sin q with y/r.

Multiply by r to clear the fraction.

Now replace r with chk=4x2 y2

8(b) Convert the rectangular equation to a polar equation: . Recall: .x2 y2 1 x r cos, y r sin


A B

C

c

ab

1(2). sinB = , 3 /2What are the possible values, in degrees, for VB? 2 and 3 symbols + units, chk=6,3

2(1). cosB = , 3 /2What is the possible value, in degrees, for VB?3 symbols + units, chk=6

3(3). Is there any triangle in which a = 2, b = 3, and VA = 41o? One point for the answer, two for showing your work.

4(4). a = 30, b = 36, VA = 20o. There are two possibletriangles with these sides and angles. Decimal answer only.

Find the areas of these two triangles. Show your work.Either find c or find VC. Then use Lecture 25's area formula.One answer is almost 40, the other is between 360 and 390.

The smaller area to 2 decimal places: __ __ . __ __Decimal answer only, chk=23.

The larger area to 2 decimal places: __ __ __ . __ __Decimal answer only.

§8.1 587:25-40. Convert the polar coordinates torectangular coordinates. Give exact answers. Both involve

the square root of 2.

5(1). 5, 4

6(1). 5, 4

Convert the rectangular coordinates to polarcoordinates. Give exact answers with r >0. Both involve p.

7(1). 3, 3 9 symbols, chk=11

8(1). is undefined, draw the picture instead.0,2 tan1 yx

8 symbols, chk=4

§8.1 587:41-60. Convert the polar equation to arectangular. Simplify any radicals. E.g. x2

+ y2 = 9 insteadof x2 y2 3

9(1). 2 sin 3cos r11 or 13 symbols, chk=9

10(1). r 48 symbols, chk=11

Convert the rectangular equation to a polar equation.11(1). x2 y2 253, 4 or 5 symbols, chk=5 or 9

12(1). y x2

13 symbols, chk=4


A B

C

c

ab

A(a). sinB = , 1/ 2What are the possible values, in degrees, for VB?

(b). cosB = , 1/ 2What is the possible value, in degrees, for VB?

(c). sinA 1/4What is the possible value, in degrees, for VA?

(d). cos A 2/3What is the possible value, in degrees, for VA?

B. Find the lengths a, b, and c in the following picture.

d

ab

c

A B C2cm

VA = 50o, VB = 110o, VC = 95o.

a =

b =

c =

Convert the polar coordinates to rectangular coordinates.Give exact answers. Both involve the square root of 2.

C. 3, 23

D. 4, 11/6

E. 4,/6

Convert the rectangular coordinates to polar coordinates. Give exact answers with r>0.

F. 1,1

E.g., x2+2x+y2+3y = 0. Simplify any radicals. E.g. x2

+ y2 = 9 instead of x2 y2 3

G. r 2cos

H. r tan

I. r 3cos 2

Convert the rectangular equation to a polar equation.Assume r = 0.

J. 3x 4y 2

K. y2 x3

L. 2xy 1

AnswersA. (a) 45o,135o (b) 45o (c) 14.5o,165.5o (d) 131.8o B. a 2 sin 700

sin 200 cm, b 2 sin 500

sin 200 cm,C. , D. , E. 3/2, 3 3 /2 2 3 ,2 2 3 ,2F. G. 2 , 5/4 x 12 y2 1H. I. x4 x2y2 y2 0 x2 y23 9x2 y22

J. K. r 2/3cos4sin r tan2secL. r2 csc 2


A B

C

c

ab

A(a). sinB = , 1/ 2What are the possible values, in degrees, for VB?B sin1 1

2 450, or 1800 450 1350

(b). cosB = , 1/ 2What is the possible value, in degrees, for VB?B cos1 1

2 450

For sin, there may be two values q and p-q. For cos, there's only one.

(c). sinA 1/4What is the possible value, in degrees, for VA?

, or A sin1 14 14.50 1800 14.50 165.50

(d). cos A 2/3What is the possible value, in degrees, for VA?

=131.81oA cos1 23 1800 cos1 2

3 1800 48.190

The intermediate step is needed only for table users, calculators cando = 130.81o directly. cos1 2

3

B. Find the lengths a, b, and c in the following picture.

ab

c

A

B

C2cm

E

VA = 50o, VB = 110o, VC = 95o. E = 180o - (50o+110o) = 20o

, , a = cmasin B 2

sin Ea

sin 110 2sin 20 2 sin 1100

sin 200 2 sin 700

sin 200

, cmbsin A 2

sin E , bsin 50 2

sin 20 b 2 sin 500

sin 200

c = cmcsin A a

sin C , csin 50

2sin 70/ sin 20sin 95

2 sin 500 sin 700

sin 200 sin 950

Convert the polar coordinates to rectangular coordinates.Give exact answer. Both involve the square root of 2.

C. 3, 23

x, y r cos , r sin 3 cos 23 , 3 sin 2

3

= 3 12 , 3

32 3

2 ,3 3

2

D. 4, 11/6x, y r cos, r sin 4cos

6 , 4 sin 6

= 4 3 /2, 41/2 2 3 ,2E. 4,/6x, y r cos, r sin 4cos

6 , 4 sin 6

=4 3 /2, 41/2 2 3 ,2

Convert the rectangular coordinates to polar coordinates. Give exact answer with r > 0.

F. 1,1r x2 y2 12 12 2

tan1 yx tan1 1

1 tan11 4

tan-1 is correct only up to a multiple of p. Since (-1,-1)is in quadrant III, we must add p to p/4.Answer: 2 , 5/4

Convert the polar equation to a rectangular equation with0 on the right, everything else on the left. E.g.,x2+2x+y2+3y = 0. Simplify any radicals. E.g. x2+y2

= 9instead of x2 y2 3

G. r 2cos. Clear the denominator.x2 y2 2 x

x2y2

or, converting to a circle equation,x2 y2 2xAnswer: x 12 y2 1

H. r tan. Square both sidesx2 y2 y

x

x2 y2 y2/x2

x4 y2x2 y2

Answer: x4 x2y2 y2 0

I. r 3cos 2x2 y2 3cos2 sin2

x2 y2 3 x2

x2y2 y2

x2y2

x2 y2 9x2y2

x2y2 2

Answer: x2 y23 9x2 y22

Convert the rectangular equation to a polar equation.J. 3x 4y 2

3r cos 4r sin 2r3cos 4sin 2Answer: r 2/3cos 4sin

K. y2 x3

r sin2 r cos3

r2 sin2 r3 cos3r sin2/ cos3Answer: r tan2 sec


Math 140 Lecture 28 (Omitted on short semesters)The graph of y = ax2+bx+c is a vertical parabola with a

vertical axis of symmetry. Other parabolas havehorizontal or slanted axes.

axis

focus

directrix

parabola

vertex

DEFINITION. A parabola consists of all points equidistantbetween a given focus point and a given directrix line.The axis is the line through the focus and perpendicularto the directrix. The vertex is the intersection of theparabola and the axis.

In a parabola, light rays parallel to the axis are reflected tothe focus. Telescope mirrors and satellite antennas havethis shape. The vertex lies halfway between the focusand the directrix.

`Find the equation for the parabola with focus (0, p) anddirectrix y = -p.

(0,p) = focus

y=-p

(x,y)

directrix

(0,0) = vertex

axis

y

p

For any point (x, y), The distance between (x, y) and the directrix y = -p is y+p.The distance between (x, y) and the focus (0, p) is

. x 02 y p2 x2 y2 2py p2

(x, y) is on the parabola iff the distances are equal iff y p x2 y2 2py p2

iff y p2 x 2 y2 2py p 2

iff y2 2py p2 x2 y2 2py p2

iff iff 2py x2 2py 4py x2

iff iff x2 4py x2 ky where k 4p and p k/4

VERTICAL PARABOLA THEOREM. For k = 0, the graph of is the vertical parabola with focus (0, p) andx2 ky

directrix y = -p where p = k/4. The axis is the y-axis; thevertex is (0, 0).

Exchanging x and y gives ¥ HORIZONTAL PARABOLA THEOREM. For k = 0, the graph of

is a horizontal parabola with focus (p, 0) and y2 kxdirectrix x = -p where p = k/4. The axis is the x-axis; thevertex is (0, 0).

(p,0) = focus

x=-pdirectrix

axis(0,0) vertex

focal-width point

x2-parabolas such as y = x2 are vertical; y

2-parabolas are horizontal.

`Find the focus, directrix and graph of y = -x2/8. x2

= -8y. Parabola is vertical. p = k/4 = -8/4 = -2. Focus : (0, p) ƒ (0, -2). Directrix: y = -p ƒ y = 2. Graph ... .

Hint, first mark the two focal-width points on a line through thefocus, || to the directrix and equidistant from focus and directrix.

`Find the focus, directrix and graph of . 3y2 4xWrite it in horizontal-parabola form: y

2 = kx.

. y2 43 x, p k/4 4

3 /4 13

Focus (p, 0): (ƒ, 0). Directrix x = -p: x = -ƒ. Graph ... .

THEOREM. In any equation, replacing each

down b unitsshifts the graphy by y+b

up b unitsshifts the graphy by y-b

left by a unitsshifts the graphx by x+a

right by a unitsshifts the graphx by x-a

When a parabola is shifted, so are its focus, directrix,vertex, and axis.

To graph a parabola, get the squared variable on the left,the rest on the right. Complete the square if needed.Write the equation in either the: vertical parabola form: orx a2 ky b horizontal parabola form: y b2 kx a

`Find the focus, directrix and graph of .x2 2x 9 8y 0, complete the squarex2 2x 8y 9, x2 2x 1 8y 8

, . A vertical parabola.x 12 8y 1 p k /4 8/4 2For : vertex = (0,0), focus = (0,2), directrix y = -2. x2 8yTo get , shift right 1 unit and up 1 unit.x 12 8y 1Vertex: (0,0) shifted right 1 and up 1 ƒ (1,1). Focus: (0,2) shifted right 1, up 1 ƒ (1,3). Directrix: y = -2 shifted right 1, up 1 ƒ y = -1.

directrix: y=-1

focus= (1,3)

vertex= (1,1)

y

24(15). Find the axes/directrix and the focal point(s). Draw the graph of y2 2y 4x 3 0.

y2 2y 4x 3 0y2 2y 4x 3y2 2y 1 4x 3 1y 12 4x 4y 12 4x 1

This is the horizontal parabola shifted right 1 and up 1 to get y2 4x y 12 4x 1k 4 p k/4 4/4 1Vertex: 0, 0 1, 1Focus: p, 0 1, 0 0, 1Directrix: x p, x 1 x 2(a)(3) Directrix: x 2(b)(3) Focal point(s): 0, 1(c)(9) Graph

x=2

focus (0,1)vertex=(1,1)


Here’s an example, your problem is in the next column.Example. Find the axes/directrix and the focal point(s). Draw the graph of y2 2y 4x 3 0.

Write in parabolic form y b2 kx ay2 2y 4x 3 0y2 2y 4x 3y2 2y 1 4x 3 1y 12 4x 4y 12 4x 1

This is the horizontal parabola y2 4xshifted right 1 and up 1 to get y 12 4x 1k 4 p k/4 4/4 1Vertex: 0, 0 1, 1Focus: p, 0 1, 0 0, 1Directrix: x p, x 1 x 2(a)(3) Directrix: x 2(b)(3) Focal point(s): 0, 1(c)(9) Graph

x=2

focus (0,1)vertex=(1,1)

focal width point

focal width point

24. Find the axes/directrix and the focal point(s). Draw the graph of x2 2x 4y 3 0.

Write in parabolic form x b2 ky a

This is the vertical parabola __________________

shifted ________ and ________ to get

Vertex: chk=20, 0

Focus: chk=10, p ____, ____ ____, ____

Directrix: chk=2y p, ________ _________

Graph: Mark the focal width points on the graph.


Graph the parabola. On the graph, mark and give thecoordinates for focus and vertex. Draw the directrix witha dotted line. First locate the focus and vertex and draw the directrix. Then markthe two “focal-width points” where the line through the focus whichis parallel to the directrix intersects the circle around the focus whichtouches the directrix.

A(3). x2 4y

B(3). y2 8x

C(4). y2 6y 4x 17 0

D(4). x2 8x y 18 0

AnswersFor C and D, see Hw 28 worked examples.

For `A and B, see the next page.

Math 140 Hw 28 Practice only, do not turn in.


A. x2 4y

B. y2 8x

C. y2 6y 4x 17 0

D. x2 8x y 18 0

AnswersFor C and D, see Hw 28 worked examples.

For `A and B, see the next page.


A. x2 4yvertex: (0, 0)focus: (0, 1)directrix: y 1

directrix: y=-1

focus= (0,1)

vertex= (0,0)

focal width points

B. y2 8xvertex: (0, 0)focus: (-2, 0)directrix: x 2

directrix: x=2

focus= (-2,0) vertex= (0,0)



C. y2 6y 4x 17 0

y2 implies a horizontal parabola.Complete the square to put the equation in the form:y b2 kx a

y2 6y 4x 17. Add 9 to both sides.6/22 32 9

y2 6y 9 4x 8y 32 4x 2

Thus p k4 4

4 1

For : vertex = (0,0), focus = (1,0), directrix x = -1. y2 4xTo get , shift right 2 and up 3.y 32 4x 2

Shifting (0,0) right 2 and up 3 gives the vertex: (2,3).Shifting (1,0) gives the focus: (3,3).Shifting x = -1 gives the directrix: x = 1.

directrix: x=1

focus= (3,3)vertex= (2,3)

D. x2 8x y 18 0

implies a vertical parabola.x2

Complete the square to put the equation in the form:x a2 ky b

x2 8x y 18. Add 16 to both sides.8/22 42 16

x2 8x 16 y 2x 42 1y 2

Thus p k4 1

4

For : x2 1yvertex = (0,0), focus = (0,1/4), directrix y = -1/4. To get , shift right 4 unit and up 2 unit.x 42 1y 2

Shifting (0,0) right 4 and up 2 gives the vertex: (4,2).Shifting (0,1/4) gives the focus: (4,9/4).Shifting y = -1/4 gives the directrix: y = 7/4.

directrix: y=7/4

focus= (4,9/4)

vertex= (4,2)


Math 140 Lecture 29RECALL. A circle is the set of all points such that the

distance to a center point is some constant r.

focifoci

minor axis

major axis

vertices

center

DEFINITION. An ellipse is the set of all points such thatthe sum of the distances to two focus points is the distancebetween the vertices. The vertices are the two pointsfarthest apart. The major axis goes from a vertex at oneend of the ellipse through the two foci to the vertex at theopposite end. The minor axis is a perpendicular bisectorof the major axis.

A light ray emitted from one focus point is reflected tothe opposite focus point. Planetary orbits are ellipses.

Leta = major radius = ½ the major axis length. b = minor radius = ½ the minor axis length.c = focal radius = ½ the distance between the foci.

THEOREM. . t .a2 b2 c2 c a2 b2

The graph of is a circle with center (0, 0)x2 y2 r2

and radius r. This can be written as .x2

r2 y2

r2 1

THEOREM. For a > b > 0, the graphs of and x2

a2 y2

b2 1 are ellipses. (0, 0) is the center. a, b, c are they2

a2 x2

b2 1major, minor and focal radii where .c a2 b2

v is a horizontal ellipse. x2

a2 y2

b2 1 Foci:(-c, 0) and (c, 0). Major axis: the line segment (-a,0)(¯ ¯ a,0).

Minor axis: (0,¯ -b)(0¯¯ ,b). Here x has the bigger radius.

v is a vertical ellipse. y2

a2 x2

b2 1 Foci: (0, -c) and (0, c). Major axis: (0,-¯¯a)(0,¯ ¯ a).

Minor axis: (-b,0)(¯ ¯ b,0). Here y has the bigger radius.

To graph, complete the squares if some variable isn’t asquare. Get 1 on the right. Write the equation in one ofthe forms above.

`Find the major and minor axes and foci; draw the graph: .4x2 y2 4

x2 y2

4 1, x2

12 y2

22 1, y2

22 x2

12 1a 2, b 1, c 4 1 3Vertical ellipse (y has the bigger radius).

Major axis: (0,-2)(0,2)¯¯¯¯¯¯¯¯. Write your axes like this on the final.Minor axis: (-1,0)(1,0)¯¯¯¯¯¯¯¯.Foci: or .0, 3 , 0, 3 0, 3

Set your compass to a major radius. Put the point at the end of aminor radius. Draw an arc. It intersects the major axisat the two foci.

`Find the major and minor axes and foci; draw the graph: .16x2 96x 25y2 256

complete square for x16x2 6x 25y2 25616x2 6x 9 25y2 256 16 9

divide by 400, get 1 on right16x 32 25y2 400x32

25 y2

16 1x32

52 y2

42 1a 5, b 4, c 25 16 9 3Horizontal ellipse (x has the bigger radius).

This is shifted right 3 units.x2

52 y2

42 1Major axis: (-5,0)(5,0)¯¯¯¯¯¯¯¯ shifted right 3 ƒ (-2,0)(8,0)¯¯¯¯¯¯¯¯.Minor axis: (0,-4)(0,4)¯¯¯¯¯¯¯¯ shifted right 3 ƒ (3,-4)(3,4)¯¯¯¯¯¯¯¯.Foci: (-3,0), (3,0) shifted right 3 ƒ (0,0), (6,0).

(3,4)

(3,-4)

(-2,0) (8,0)(0,0) (6,0)

focus focus

(1,0)(-1,0)

(0,2)focus

focus(0,-2)

25(15). Find the axes/directrix and the focal points. Draw the graph of x2 2x 4y2 8y 1.

x2 2x 4y2 8y 1x2 2x 4y2 8y 1x2 2x 1 4y2 2y 1 1x 12 4y2 2y 1 1 1 4x 12 4y 12 4

x 12

4 y 12

1 1

x 12

22 y 12

12 1

Horizontal ellipse ƒ shifted left 1, up 1 to x2

22 y2

12 1x 12

22 y 12

12 1

a 2, b 1, c a2 b2 4 1 3

Center: 0, 0 1, 1Major axis: (-2,1)(2,1)¯¯¯¯¯¯¯¯ ƒ (-3,1)(1,1)¯¯¯¯¯¯¯¯Minor axis: (0,0)(0,2)¯¯¯¯¯¯¯ ƒ (-1,0)(-1,2)¯¯¯¯¯¯¯¯¯Focal points: 3 , 0 1 3 , 0

(a)(3) Axes: 3, 11, 1, 1, 01, 2(b)(3) Focal points: 1 3 , 1(c)(9) Graph

1

2

1-1-3


Here’s an example, your problem is in the next column.Example. Find the axes and the foci. Draw the graph of

x2 2x 4y2 8y 1.

x2 2x 4y2 8y 1

x2 2x 4y2 8y 1

x2 2x 1 4y2 2y 1 1

x 12 4y2 2y 1 1 1 4

x 12 4y 12 4

x12

4 y12

1 1

must have squares on the botomx12

22 y12

12 1

Horizontal ellipse ƒ x2

22 y2

12 1

shifted left 1, up 1 to x12

22 y12

12 1

a 2, b 1, c a2 b2 4 1 3

Center: 0, 0 1, 1

Major axis: (-2,0)(2,0)¯¯¯¯¯¯¯¯ ƒ (-3,1)(1,1)¯¯¯¯¯¯¯¯

Minor axis: (0,-1)(0,1)¯¯¯¯¯¯¯¯ ƒ (-1,0)(-1,2)¯¯¯¯¯¯¯¯¯

Focal points: 3 , 0 1 3 , 1

(a)(3) Axes: (-3,1)(1,1¯¯¯¯¯¯¯), (-1,0)(-1,2¯¯¯¯¯¯¯) (b)(3) Focal points: 1 3 , 1(c)(9) Graph

1

2

1-1-3

Problem. Find the axes and the foci. Draw the graph of 4x2 8x y2 2y 1.

Complete the square. Then write the equation in ellipse

form. In this case with squaresy ?2

a2 x ?2

b2 1

on the bottom.

a = ? b = ? c = ? chk=3

Major axis: chk=6

Minor axis: chk=4

Focal points: chk=10

Graph:


Graph the ellipse. On the graph, mark and give thecoordinates for endpoints of the major and minor axesand the foci. A. 4x2 9y2 36

B. x52

52 y12

32 1

C. 3x2 4y2 6x 16y 7 0

D. 5x2 3y2 40x 36y 188 0

AnswersFor C and D, see Hw 29 worked examples.For A and B, see the next page.


Graph the ellipse. On the graph, mark and give thecoordinates for endpoints of the major and minor axesand the foci.

C. 3x2 4y2 6x 16y 7 03x2 6x 4y2 16y 73x2 2x 4y2 4y 73x2 2x 1 4y2 4y 4 7 3 163x 12 4y 22 12x12

4 y22

3 1x12

22 y22

3 2 1

a 2, b 3 , c 4 3 1 1Horizontal ellipse (x has the bigger denominator).

This is shifted right 1, down 2 units.x2

22 y2

32 1

Shifting (-2,0)(2,0)¯¯¯¯¯¯¯¯ right 1 gives major axis: (-1,-2)(3,-2)¯¯¯¯¯¯¯¯¯.Shifting (0,-¯ ¯ l3)(0,¯ ¯ ¯ l3)¯ gives minor axis: (1,-2-¯¯¯¯l3)(1,¯ ¯ ¯ -2+l3)¯ .

Shifting (-1,0), (1,0) gives foci: (0,-2), (2,-2).

(0,-2) (2,-2)(3,-2)(-1,-2)

(1,-3.73)

(1,-.27)

D. 5x2 3y2 40x 36y 188 0

5x2 40x 3y2 36y 188

5x2 8x 3y2 12y 188

5x2 8x 16 3y2 12y 36 188 5 16 3 36

5x 42 3y 62 0

This is true iff x = 4 and y = 6. Thus it is a degenerateellipse, it is just the single point (4,6).

(4, 6)


Math 140 Lecture 30

--------

minor axis

-----------major axis

----verte

xvertex---- ---focal axis---focal point

focal point---

-

----asy

mptote

center


DEFINITION. A hyperbola is the set of all points such thatthe difference of the distances to two focal points is thedistance between the vertices. The focal axis through thetwo foci intersects the hyperbola at its two vertices. Themajor axis is the segment between the vertices, the minoraxis is a perpendicular bisector. The ends of the axes arethe midpoints of a box whose diagonals are asymptotes.The box corners and the foci are equidistant from thebox’s center. To get the foci, draw an arc around the boxcenter from the box corner to the focal axis.

The path of a stone thrown upward (in a vertical gravitationalfield) is a parabola. The path of an orbiting comet is an ellipse withthe sun at one focus. The path of a comet passing through the solarsystem is one piece of a hyperbola with the sun at one focal point.

Leta = the major radius = ½ major axis length. b = the minor radius = ½ the minor axis length. c = the diagonal radius = ½ the hyperbola box diagonal = the focal radius = ½ the distance between the foci.

THEOREM. c2 a2 b2, thus c a2 b2 .THEOREM. The graphs of and x2

a2 y2

b2 1 y2

a2 x2

b2 1are hyperbolas (a must be below the positive square) witha = the major radius, b = the minor radius, and

= the focal radius.c a2 b2

v is a horizontal hyperbola with x2

a2 y2

b2 1

(0,¯ -b) (0, b)minor axis: (-a, 0) (a, 0)major axis: (-c, 0) and (c, 0)foci:

v is a vertical hyperbola with y2

a2 x2

b2 1

(-b, 0) (b, 0)minor axis: (0,¯ -a) (0, a)major axis: (0, -c) and (0, c)foci:

To graph, complete the squares if some variable isn’t asquare, then write the equation in one of the above forms.

`Find the axes and foci; draw the graph and asymptotes: . 4y2 x2 4

y2 x2

4 1, y2

12 x2

22 1a 1, b 2, c 1 4 5

Vertical hyperbola (the y 2 is positive).

Major axis: (0,-1)(0,1¯¯¯¯¯¯¯), minor axis: (-2,0)(2,0¯¯¯¯¯¯¯), foci: (0,l5), (0, -l5). On the final, write the axes as above.

focal point

(2,0)

(0,1)

focal point

(0,-1)

(-2,0)

`Find the axes and foci; draw the graph and asymptotes: . 16x2 96x 25y2 256

complete square for x16x2 6x 25y2 25616x2 6x 9 25y2 256 16 9

divide by 400, get 1 on right16x 32 25y2 400x32

25 y2

16 1x32

52 y2

42 1a 5, b 4, c 25 16 41

Horizontal hyperbola (the x2 is positive).This is shifted right 3 units.x2

52 y2

42 1Shifting (-5,0)(5,0)¯¯¯¯¯¯¯¯ right 3 gives the major axis: (-2,0)(8,0)¯¯¯¯¯¯¯¯.Shifting (0,-4)(0,4)¯¯¯¯¯¯¯¯ right 3 gives the minor axis: (3,-4)(3,4)¯¯¯¯¯¯¯¯.Shifting (-l41¯¯,0), (l41¯¯,0) gives the foci: (3-l41¯¯,0), (3+l41¯¯,0).

(3,4)

(3,-4)

(-2,0) (8,0)focal point focal point

26(15). Find the axes/directrix and the focal points. Draw the graph of x2 2x 4y2 8y 7.x2 2x 4y2 8y 7x2 2x 4y2 8y 7x2 2x 1 4y2 2y 7 1x 12 4y2 2y 1 7 1 4x 12 4y 12 4x 12

4 y 12

1 1

x 12

22 y 12

12 1

Horizontal hyperbola ƒ shifted left 1, up 1 to x2

22 y2

12 1 x 12

22 y 12

12 1

a 2, b 1, c a2 b2 4 1 5Center: 0, 0 1, 1Major axis: (-2,0)(2,0)¯¯¯¯¯¯¯¯ ƒ (-3,1)(1,1)Minor axis: (0,-1)(0,1)¯¯¯¯¯¯¯¯ ƒ (-1,0)(-1,2)¯¯¯¯¯¯¯¯¯

Focal points: 5 , 0 1 5 , 0(a)(3) Axes: 3, 11, 1, 1, 01, 2(b)(3) Focal points: 1 5 , 1(c)(9) Graph

1

2

1-1-3


Here’s an example, your problem in is the next column.

26 Example. Find the axes/directrix and the focal points. Draw the graph of x2 2x 4y2 8y 7.x2 2x 4y2 8y 7x2 2x 4y2 8y 7x2 2x 1 4y2 2y 7 1x 12 4y2 2y 1 7 1 4x 12 4y 12 4x 12

4 y 12

1 1

x 12

22 y 12

12 1

Horizontal hyperbola ƒ x2

22 y2

12 1

shifted left 1, up 1 to x 12

22 y 12

12 1

a 2,b 1,c a2 b2 41 5

Major axis: (-2,0)(2,0)¯¯¯¯¯¯¯¯ ƒ (-3,1)(1,1)¯¯¯¯¯¯¯¯ Note the line over the top.

Minor axis: (0,0)(0,2)¯¯¯¯¯¯¯ ƒ (-1,0)(-1,2)¯¯¯¯¯¯¯¯¯

Focal points: 5 , 0, 5 , 0 1 5 , 01 5 , 0

(a)(3) Axes: (-3,1)(1,1¯¯¯¯¯¯¯), (-1,0)(-1,2¯¯¯¯¯¯¯)

(b)(3) Focal points: 1 5 , 1, 1 5 , 1

(c)(9) Graph

1

2

1-1-3

26 Problem. Find the axes/directrix and the focal points. Draw the graph of 4y2 8y x2 2x 1.Complete the square. Write the equation in hyperbola form. In this case

is wrongy y02

a2 x x02

b2 1 x 1

a = ? chk=1b = ? chk=2c = ? chk=5

Major axis: chk=4Line on top.

Minor axis: chk=6

Focal points: chk=7

Graph:


Graph the hyperbola. On the graph, mark and give thecoordinates for endpoints of the major and minor axesand the foci. Draw the asymptotes with a dotted line.A. x2 4y2 4

B. x52

52 y12

32 1

C. y22

22 x12

12 1

D. x2 y2 2y 5 0

E. y2 25x2 8y 9 0

AnswersFor D and E, see Hw 30 worked examples.For A, B and C, see the next page.


A. x2 4y2 4Major axis: (-2,0)(2,0)¯¯¯¯¯¯¯¯.Minor axis: (0,-1)(0,1)¯¯¯¯¯¯¯¯.Foci: (-l5, 0), (l5, 0)

(0,1)

focal point focal point(-2.23,0) (2.23,0)

(0,-1)

(2,0)(-2,0)

B. x52

52 y12

32 1Major axis: (0, -1)(10, -1)¯¯¯¯¯¯¯¯¯¯¯.Minor axis: (5,-4)(5,2)¯¯¯¯¯¯¯¯.Foci: (5-l34¯¯, -1), (5+l34¯¯, -1).

(5, 2)

focal point focal point

(5, -4)

(10, -1)(0, -1)

C.y22

22 x12

12 1Major axis: (1,0)(1,4)¯¯¯¯¯¯¯.Minor axis: (0,1)(2,1)¯¯¯¯¯¯¯.Foci: (1, 2-l5), (1, 2+l5).

(1, 4)

(1, 0)

(0, 1) (2, 1)

Math 140 Hw 30 Recommended problems, answers.

Graph the hyperbola. On the graph, mark and give thecoordinates for endpoints of the major and minor axesand the foci. Draw the asymptotes with a dotted line.

D. x2 y2 2y 5 0

x2 y2 2y 5

x2 y2 2y 1 5 1

x2 y 12 4

x2

22 y12

22 1

2.83a 2,b 2,c 22 22 8 2 2

Horizontal hyperbola (the x 2 is positive).

This is shifted up 1 unit. x2

22 y2

22 1Shifting (-2,0)(2,0¯¯¯¯¯¯¯) up 1 gives major axis: (-2,1)(2,1¯¯¯¯¯¯¯).

Shifting (0,-2)(0,2¯¯¯¯¯¯¯) up 1 gives minor axis: (0,-1)(0,3¯¯¯¯¯¯¯).Shifting (-2.83,0), (2.83,0) up 1 gives foci: (-2.83,1), (2.83,1).

or precisely, .2 2 , 1, 2 2 , 1

(0,3)

focal point focal point(-2.83,1) (2.83,1)

(0,-1)

(2,1)(-2,1)

E. y2 25x2 8y 9 0

y2 8y 16 25x2 9 16

y 42 25x2 25

y42

52 x2

12 1

5.10 a 5, b 1, c 25 1 26

Vertical hyperbola (the y 2 is positive).

This is shifted down 4 units.y2

52 x2

12 1

Shifting (0,-5)(0,5¯¯¯¯¯¯¯) down 4 gives major axis: (0,-9)(0,1¯¯¯¯¯¯¯).

Shifting (-1,0)(1,0¯¯¯¯¯¯¯) down 4 gives minor axis: (-1,-4)(1,-4)¯¯¯¯¯¯¯¯¯.Shifting (0,-5.10), (0,5.10) down 4 gives foci: (0,-9.10), (0,1.10).

or precisely, 0,4 26 , 0,4 26

(0, 1)

(0, -9)

(-1, -4) (1, -4)


Math 140 Lecture 31 (omitted)Note the difference between parabolas, ellipses, and

hyperbolas. v If only one variable is squared, it is a parabola.

Write with the squared variable on the left.Vertical parabola like y = x2

or y =x a2 ky b-x2

Horizontal parabola.y b2 kx a

Suppose both variables are squared. If necessary,complete and replace by x and by y. x c y d

Write with 1 on the right, squares in the denominators.

v Horizontal ellipse: with a > b > 0. Bigger first.x2

a2 y2

b2 1

v Vertical ellipse: with a > b > 0. y2

a2 x2

b2 1

v Horizontal hyperbola: . Positive first.x2

a2 y2

b2 1

v Vertical hyperbola: . y2

a2 x2

b2 1

There is clearly a missing case,

, but this is impossible since the left side x2

a2 y2

b2 1is negative and thus cannot equal 1 which is positive.

Second degree equations such as whichx2 xy y2 1involve a product of x and y are also parabolas, ellipsesand hyperbolas but their axes may be slanted ratherthan horizontal or vertical. The simplest example is

. Solving for y gives . The graph is xy 1 y 1x

In this case the focal axis of the hyperbola is the major diagonal . Except for this graph you won’t be y xasked to graph equations with a slanted axis. Such graphs can be quite difficult and require techniques which aren’t covered until upper division linearalgebra.

Identify each of the following as a horizontal/vertical parabola/ellipse/hyperbola. Graph.

Check your graphs by calculating the x and y intercepts.

` . x squared, y not: vertical parabola. x2 y 2If we didn’t have to find the focus and directrix wecould easily graph this by rewriting it as: . y 2 x2

x2 y 2 x2 y 2

This is shifted up 2 units. x2 yk = -1, p = k/4 = -1/4. center: (0, 0) shifts to (0, 2). focus: (0, p) = (0, -1/4) shifts to (0, -1/4 + 2) = (0, 7/4).

directrix: y = -p ƒ y =1/4 shifts to y = 1/4 + 2 = 9/4.

1 2 3 4 5 6 7-1-2-3-4-5-6-7

1

2

3

4

5

-1

-2

-3

-4

-5

x

y

` y2 4x2 8x 0 y2 4x2 2x 0

y2 4x2 2x 1 1 4 y2 4x 12 4

y2

4 x 12 1

Both positive, y 2 has the larger

y2

22 x12

12 1denominator, ta vertical ellipse.

This is shifted right 1 unit. y2

22 x2

12 1Center (0,0) shifts to (1,0). a = major radius = 2. b = minor radius = 1. c = focal radius = . a2 b2 22 12 3Graph

To locate the foci, set yourcompass to the distance betweenthe center and a vertex. Then put the compass point at an endpoint of the minor axis.

The arc of the pencil end willintersect the major axis at thetwo foci.

major axis: (1,2)(1,-2¯¯¯¯¯¯¯) minor axis: (0,0)(2,0¯¯¯¯¯¯)foci: 1, 3 , 1, 3

` ƒ . x2 y2 1 x2

12 y2

12 1x2 is positive, thorizontal hyperbola.

Final. The final is cumulative, review using practice exams 1-5. Bring

scratch paper. There is not much room for work on the final.

Come 5 minutes early to fill out a teacher evaluation for me.

`Find the area of a triangle with sides 2, 3, 4.Let q be the angle opposite side 4. Then 42 22 32 223 cost12cos 4 9 16 3, cos11/4Area = .1

2 23 sincos1 14 3 1 1

16 3 15

4

verticalellipse

(0,0) (2,0)

(1,2)

(1,-2)

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