AP Calculus AB Summer Assignment
Transcript of AP Calculus AB Summer Assignment
Name: _____________________________ Date: ________________ Period: ____
AP Calculus AB Summer Assignment
Congratulations on making it into AP Calculus! I am very pleased and excited to be teaching this course and
canβt wait to challenge you! AP Calculus AB is a fast-paced course that is taught at the college level. There is a
lot of material in the curriculum that must be covered before the AP exam in May. Therefore, we cannot
spend a lot of class time re-teaching prerequisite skills.
The purpose of this assignment is to have you practice the mathematical skills necessary in order for you to be
successful throughout the year and ultimately on the AP exam. I am including some reference sheets at the
end of the packet that may help you, but donβt rely on them too heavily. They are a reference, not a crutch! I
have also included some websites you may want to refer to if you need help with the topics presented in the
packet.
One piece of advice: Donβt fake your way through any of these problems because you will need to understand
everything in this very well. If not, it is possible that you will get calculus problems wrong in the future β not
because you do not understand the calculus concept, but because you do not understand the algebra or
trigonometry behind it.
Each problem should be done in the space provided. For the AP exam, you will not box, circle or underline
your final answer so letβs practice that now! Please do not circle or box your final answer. Just leave it as is.
Remember, NO WORK = NO CREDIT!! (Double exclamation points because itβs that important) I do not care
how good you are with mental math; all work must be provided. You may use additional sheets and please be
neat. Also, do not rely on a calculator. Half of your AP exam next year is taken without a calculator so use
paper and pencil techniques as much as possible.
It is a mistake to decide to do this right as school ends. Let it go until mid-summer. I want these techniques to
be relatively fresh in your mind in the fall, but of course do not wait until last minute. Spend some quality
time with this packet βΊ
In addition to the questions provided, you must memorize the unit circle. You will not be allowed to use the
unit circle on any test or quiz in the class. We use trigonometry often (radians only) so be familiar with it!
You will also need a graphing calculator for this class. It is highly recommended that you purchase the TI-84
calculator and have it the first week of school. Half of the AP test is taken with a graphing calculator. It is too
time consuming to explain how to use a variety of calculators during class so only TI-84 will be taught. If you
cannot purchase one, another version of the calculator will be lent to you for the year. A scientific calculator is
not allowed during any parts of the exam; therefore, it will not be allowed during class.
If you have any questions please do not hesitate to contact me at [email protected]. Good luck
and see you in the fall!
Please refer to the following websites should you need any help:
http://purplemath.com/modules/index.htm
http://khanacademy.org
http://fireflylectures.com/free-algebra/
http://whyu.orghttp://whyu.org
Topic 1: Equations of Lines
Important Things to Remember About Lines
Slope-Intercept Form: π¦ = ππ₯ + π
** Where π is the slope and π is the y-intercept. **
Point-Slope Form: π¦ β π¦1 = π(π₯ β π₯1)
** Where (π₯1, π¦1) is the point going through the line. **
Vertical line: π₯ = π
** Where π is the value of π₯-value that the line passes through and the slope is undefined. **
Horizontal line: π¦ = π
** Where π is the value of π¦-value that the line passes through and the slope is zero. **
Parallel lines have equal slopes.
Perpendicular lines have opposite reciprocal slopes.
Example: Write an equation for each line.
(a) Containing (0,1) and with a slope of β2. (b) Containing points (3, β4) and (9,0).
Solution: (a) The slope, π, is given as β2. The line contains (0,1), so this point is the π¦-intercept, or π, which is 1.
Substituting these numbers into slope-intercept form gives you the equation π¦ = β2π₯ + 1.
(b) First find the slope: π =π¦2βπ¦1
π₯2βπ₯1=
β4β0
3β9=
2
3
β’ Next, substitute the coordinates of one of the given points (does not matter which point you choose to use) into the point-slope equation (since you have the points and the slope) β π¦ +
4 =2
3(π₯ β 3) OR π¦ =
2
3(π₯ β 9).
β’ Both are acceptable answers!
β’ For the purpose of this course, you do not need to go any further. You may leave your equation in point-slope form rather than distributing to get the equation in slope-intercept form.
Now you practice!
1. Determine the equation of a line passing through the point (5,-3) with an undefined slope.
2. Determine the equation of a line passing through the point (-4,2) with a slope of 0.
3. Use point-slope form to find a line passing through the point (2,8) and parallel to the line π¦ =5
6π₯ β 1.
4. Find the equation of the line passing through the points (-3,6) and (1,2).
5. Use point-slope form to find a line perpendicular to π¦ = β2π₯ + 9 and passing through the point (4,7).
Find the x and y intercepts for each equation.
** The π₯-intercepts are found by plugging in π¦ = 0 and solving for π₯. The π¦-intercepts are found by plugging
in π₯ = 0 and solving for π¦. **
6. π¦ = 2π₯ β 5
7. π¦ = π₯2 + π₯ β 2
8. π¦ = π₯β16 β π₯2
9. π¦2 = π₯3 β 4π₯
Topic 2: Points of Intersection
Remember that a system of equations is made up of equations of lines. You can solve a system by using the substitution or elimination method. Your answer is the POINT OF INTERSECTION. Therefore, your answer
must be expressed as an ordered pair.
Example:
Solve the system {3π₯ + 5π¦ = 112π₯ + 3π¦ = 7
using the
elimination method.
Example:
Solve the system {π₯2 β π¦ = 3π₯ β π¦ = 1
using the substitution
method.
Solution:
β’ First decide which variable you want to eliminate (does not matter which). For this example, I am choosing to eliminate the π₯.
β’ In order to eliminate the variable, their coefficients must be equal and different signs so that when I add the equations together, they will cancel out. To do this, I will multiply the top equation by β2 and the bottom equation by 3 (although you could have also multiplied the top equation by 2 and the bottom equation by β3; it does not matter.
β’ As a result, I end up with this system:
{β6π₯ β 10π¦ = β22
6π₯ + 9π¦ = 21
β’ Add the equations together and you get β βπ¦ = β1 β π¦ = 1.
β’ Now that you have the value for π¦, substitute it into either original equation to solve for π₯.
β’ Final answer: (2,1)
Solution:
β’ Solve one of the equations for one variable. In this case, it will be easier to solve for π¦ in either equation. I will choose to solve for π¦ in the second equation β π¦ = π₯ β 1.
β’ Plug in (substitute) what you got for π¦ in the first equation. β π₯2 β (π₯ β 1) = 3.
β’ Distribute and set the equation equal to zero so you may factor. β π₯2 β π₯ + 1 = 3 β π₯2 β π₯ β 2 = 0 β (π₯ β 2)(π₯ + 1) = 0 β π₯ = 2, β1
β’ Plug in both π₯-values independently into either original equation to solve for π¦.
β’ When π₯ = 2 β¦ β 2 β π¦ = 1 β π¦ = 1
β’ When π₯ = β1 β¦ β β1 β π¦ = 1 β π¦ = β2
β’ Final answers: (2,1) and (β1, β2)
Find the point(s) of intersection of the graphs for the given equations.
10. {π₯ + π¦ = 8
4π₯ β π¦ = 7
11. {π₯2 + π¦ = 6π₯ + π¦ = 4
12. {2π₯ + 7π¦ = 4
3π₯ + 5π¦ = β5
13. {π₯ = 3 β π¦2
π¦ = π₯ β 1
Topic 3: Rational and Negative Exponents
Rules of Exponents
Generalization Explanation
Product Rule
π₯π β π₯π = π₯π+π When multiplying like bases, add
the exponents
Quotient Rule
π₯π
π₯π= π₯πβπ
When dividing like bases, subtract the exponents
Power Rule
(π₯π)π = π₯ππ
When raising a term with an exponent with another exponent,
multiply exponents.
Expanded Power Rule
(π₯π¦)π = π₯ππ¦π
AND
(ππ₯
ππ¦)
π
=πππ₯π
πππ¦π
When raising a product of terms to an exponent, raise each
individual term to the product (this is not the same as when you have a sum or difference of terms
raised to an exponent).
Negative Powers
π₯βπ =1
π₯π OR 1
π₯βπ = π₯π
Negative exponents mean that factors remain but they are in the
denominator if the factor was originally in the numerator and
vice versa.
Zero Powers
π₯0 = 1
Any non-zero π₯-values raised to an exponent of zero is always
equal to 1
Rational Exponents
π₯ππ = βπ₯ππ
The denominator of a rational exponent is the root (index of the radical) and the numerator is the
power.
Simplify using only positive exponents.
14. 3π₯β3
π₯β2
15. (9π₯2π¦β4)β1
2
16. (16π₯2π¦)3
4
17. (π₯2)
3π₯
π₯7
18. β4π₯β16
β(π₯β4)34
19. βπ₯ β βπ₯3 β π₯1
6
20. [(π₯β1π¦1
3) (π₯β4
3π¦2)]2
21. (π₯
12π¦β2
π¦π₯β
74
)
4
Topic 4: Factoring
When factoring a trinomial in the form ππ₯2 + ππ₯ + π, remember that you are looking for the two numbers that
multiply to give you π and also add to give you π. If there π β 1, check to see if the coefficient can be factored out first before you continue to factor.
Example: 2π₯2 β 4π₯ β 30 = 2(π₯2 β 2π₯ β 15) = 2(π₯ β 5)(π₯ + 3)
Special Factorization Methods
General Example(s)
Perfect Square Trinomials
π2 + 2ππ + π2 = (π + π)2 π2 β 2ππ + π2 = (π β π)2
β’ π₯2 + 6π₯ + 9 = (π₯ + 3)2
β’ π₯2 β 8π₯ + 16 = (π₯ β 4)2
Difference of Squares
π2 β π2 = (π + π)(π β π)
β’ 4π₯2 β 9 = (2π₯ + 3)(2π₯ β 3)
AC Method
ππ₯2 + ππ₯ + π where π β 1 and π cannot be factored.
β’ Find ac.
β’ Find factors of ac that add up to b.
β’ Write the original expression, but rewrite bx using your new factors.
β’ Factor by grouping.
β’ 2π₯2 β 5π₯ β 3
β’ ππ = β6, π = β6,1
β’ 2π₯2 β 6π₯ + π₯ β 3
β’ 2π₯(π₯ β 3) + 1(π₯ β 3)
= (π₯ β 3)(2π₯ + 1)
Factoring by Grouping
Only use this method if the polynomial you are trying to
factor has four terms:
ππ + ππ β ππ β ππ = π(π + π) β π(π + π)
= (π + π)(π β π)
β’ Factor the GCF of the first two factors and factor the GCF of the second two factors.
β’ What is in parenthesis will be one of your final factors.
β’ What is left outside the parenthesis is the other factor.
β’ π₯3 + 3π₯2 β 2π₯ β 6
β’ (π₯3 + 3π₯2) + (β2π₯ β 6)
β’ π₯2(π₯ + 3) β 2(π₯ + 3)
= (π₯ + 3)(π₯2 β 2)
** Although there are many other factoring methods, these are the ones that we will mostly be using
throughout the course. Factoring is SUPER important! I will not have time to reteach factoring throughout the
year so make sure that this is a skill that you have mastered. **
Factor each of the following polynomials completely.
22. π₯2 β 20π₯ + 36
23. 2π₯3 β 2π₯2 β 4π₯
24. π₯3 β 5π₯2 β π₯ + 5
25. 2π₯2 + π₯ β 6
26. 9π₯2 β 4π¦4
27. 3π₯2 + 3π₯ β 60
28. 5π₯2 + 29π₯ + 20
29. π₯6 + 2π₯4 β 16π₯2 β 32
Topic 5: Solving Polynomials by Factoring
To solve a polynomial by factoring simply means to set the polynomial equal to zero first, factor and then set each individual factor equal to zero and solve for π₯.
Solve by factoring.
30. 4π₯2 + π₯ β 36 = π₯2 + 4π₯ 31. 9π₯2 + 2π₯ β 4 = 2π₯ β 3
Topic 6: The Quadratic Formula
The solutions to a quadratic equation written in the form ππ₯2 + ππ₯ + π = 0, where π β 0, can be found by using the quadratic formula:
π₯ =βπ Β± βπ2 β 4ππ
2π
** Where the values of π₯ represent the zeros or roots or π₯-intercepts of the quadratic equation. **
Example: Use the quadratic formula to solve: 4π₯2 + 4π₯ β 8 = 1
Solution:
β’ First, you have to set the quadratic equal to zero in order to use the quadratic formula: β 4π₯2 + 4π₯ β 9 = 0
β’ In this case, π = 4, π = 4, and π = β9. Plug in these values into the quadratic formula.
β’ π₯ =β4Β±β42β4(4)(β9)
2(4)=
β4Β±β16+144
8=
β4Β±β160
8
β’ Simplify the square root, if possible.
β’ π₯ =β4Β±4β10
8β Simplify your final answer
β’ Final answer: β1Β±β10
2
Use the quadratic formula to solve each equation.
32. 4π₯2 + 3 = 8π₯
33. 3π₯2 + π₯ = 2π₯2 + 6π₯ β 4
Topic 7: Functions
In function notation, the symbol π(π₯) is read π of π₯ and interpreted as the function value of π at π₯. In other words, π¦ = π(π₯).
** To find a function value, simply plug in the value into every π₯ and simplify. **
Example: If π(π₯) = π₯2 + 3π₯, find π(β5).
Solution:
β’ π(β5) means to replace π₯ with the value β5 and evaluate for π(π₯).
β’ π(β5) = (β5)2 + 3(β5) = 25 β 15 = 10
β’ Final answer: π(β5) = 10
Evaluate each function for the given function value.
Let π(π₯) = 5 β2π₯
3 and π(π₯) =
1
2π₯2 + 3π₯
34. π (1
2)
35. β3π(β2)
36. π(6) + 2π(1)
37. π(4) β π(5)
Topic 8: Operations with Functions
To write a sum, difference, product, or quotient of two functions, π and π, write the sum difference, product, or quotient of the expressions that define π and π, then simplify. The domain of the combined
function is restricted to the domains of each original function.
Example: Let π(π₯) = π₯2 + 3π₯ + 2 and π(π₯) = 5π₯ β 1. Write an expression for each function.
(a) (π + π)(π₯) (b) (π β π)(π₯) (c) (ππ)(π₯)
(d) (π
π) (π₯)
** Keep in mind that the domain of both π and π is (ββ, β) since both functions are polynomial functions and are continuous everywhere and extend in both directions. **
Solution:
(a) (π + π)(π₯) = π(π₯) + π(π₯) = (π₯2 + 3π₯ + 2) + (5π₯ β 1) = π₯2 β 2π₯ + 3
β’ Domain: (ββ, β)
(b) (π β π)(π₯) = π(π₯) β π(π₯) = (π₯2 + 3π₯ + 2) β (5π₯ β 1) = π₯2 + 3π₯ + 2 β 5π₯ + 1 = π₯2 β 2π₯ + 3
β’ Domain: (ββ, β)
(c) (ππ)(π₯) = π(π₯) β π(π₯) = (π₯2 + 3π₯ + 2)(5π₯ β 1) = 5π₯3 β π₯2 + 15π₯2 β 3π₯ + 10π₯ β 2 = 5π₯3 + 14π₯2 + 7π₯ β 2
β’ Domain: (ββ, β)
(d) (π
π) (π₯) =
π(π₯)
π(π₯)=
π₯2+3π₯+2
5π₯β1, where π₯ β
1
5 (because that is the value that would make the denominator
equal to zero.
β’ Domain: (ββ,1
5) βͺ (
1
5, β)
Find each new function and state the combined functionβs domain.
Let π(π₯) = 3π₯2 + 2, π(π₯) = 2π₯ β 1, and β(π₯) = π₯2 + 5π₯.
38. (π + π)(π₯)
39. (πβ)(π₯)
40. (π β β)(π₯)
41. (π
β) (π₯)
Topic 9: Composition of Functions
A composition of functions denoted as π(π(π₯)) or π[π(π₯)], is read as βπ of π of π₯,β and means to plug in
the inside function (in this case π(π₯)) in for π₯ in the outside function (in this case π(π₯)).
Example:
Given π(π₯) = 2π₯2 + 1 and π(π₯) = π₯ β 4, find π[π(1)]. ** To find a composition at a function value, you first need to find π[π(π₯)] and then you plug in the given π₯-value, in this case π₯ = 1. **
Solution:
β’ π[π(π₯)] = π(π₯ β 4) = 2(π₯ β 4)2 + 1 β’ 2(π₯2 β 8π₯ + 16) + 1 β’ 2π₯2 β 16π₯ + 32 + 1 β’ π[π(π₯)] = 2π₯2 β 16π₯ + 33
βNow you can plug in π₯ = 1.
β’ π[π(1)] = 2(1)2 β 16(1) + 33 β’ Final answer: π[π(1)] = 19
Find the following composition of functions at the indicated function value, if given.
Let π(π₯) = π₯2 β 1, π(π₯) = 3π₯ β π₯2, and β(π₯) = 5 β π₯
42. β[π(π₯)]
43. π[β(2)]
44. π[π(π₯)]
45. π[β(π(1))]
Topic 10: Evaluating for Piecewise Functions
A piecewise function is exactly what it sounds like. It is a function that is defined in βpieces.β It is a function that combines pieces of different equations that are only defined at specified intervals. To evaluate for a function value for a piecewise function, you must first identity the interval where the indicated function
value lies and plug it into the equation that is defined in that interval.
Visual Representation
π(π₯) = {π₯ + 3, π₯ < β1
π₯2 , β 1 β€ π₯ β€ 23, π₯ > 2
To graph a linear piecewise function, make sure you remember how to graph lines by using the slope and y-intercept. Also, keep in mind the endpoints of each βpiece.β If the < or > symbols are used, then that π₯-value is NOT included and would be graphed with an open circle. If the β€ or β₯ symbols are used, the that π₯-value IS included and would be graphed with a solid circle.
Example:
Graph the piecewise function given below and find π(3).
π(π₯) = {1
3π₯ + 2, π₯ β€ 3
βπ₯ + 3, 3 < π₯ β€ 4
Solution:
Although you could find π(3) graphically, if you were to find the function value algebraically, you would have to see where π(3) is defined. The value of π₯ = 3 is only included in the first equation and not in the second. Therefore, to find π(3), you would plug in π₯ = 3 into the first equation.
π(3) =1
3(3) + 2 = 3
Graph each function and find the indicated function value algebraically.
46. π(π₯) = {π₯ + 3, π₯ < 0β2π₯ + 5, π₯ β₯ 0
β’ π(3) = _____________
47. π(π₯) = {1
2π₯, β 4 β€ π₯ β€ 2
2π₯ β 3, π₯ > 2
β’ π(2) = _____________
Topic 11: Transformations of Function Graphs
Transformations of function graphs occur when there are translations, reflections and dilations. The operations performed on a function indicate what transformations occurred. Addition and subtraction
mean that there was a translation. Multiplication and division mean that there was a dilation. A negative (not subtraction) means that the graph was reflected either over the π₯-axis or the π¦-axis.
** Anything βoutsideβ affects the π¦-values and does the same as the operation indicates. Anything βinsideβ affects the π₯-values and does the opposite operation indicated. **
Transformation Notation
Verbal Description
π(π₯ + β)
Horizontal shift left β units
π(π₯ β β)
Horizontal shift right β units
π(π₯) + π
Vertical shift up π units
π(π₯) β π
Vertical shift down π units
ππ(π₯)
Vertical stretch with a scale factor of π
1
ππ(π₯)
Vertical shrink with a scale factor of 1
π
π(ππ₯)
Horizontal shrink with a scale factor of 1
π
π (1
ππ₯)
Horizontal stretch with a scale factor of π
βπ(π₯)
Reflection over the π₯-axis
π(βπ₯)
Reflection over the π¦-axis
Identify the parent function for each of the following. Then, describe the transformations of the parent
functions indicated from the functionβs equation.
48. π(π₯) = β2|π₯ + 1| β 4
β’ Parent function: __________________
49. π(π₯) =1
3ββ2π₯ + 3
β’ Parent function: __________________
50. π(π₯) = (1
4(π₯ β 1))
2
β’ Parent function: __________________
51. π(π₯) =3
β(π₯+2)β 7
β’ Parent function: __________________
Topic 12: Domain and Range
Domain: The set of all the π₯-values for which a function is defined (input values). Range: The set of all the π¦-values (output values).
β’ All polynomial functions have a domain of (ββ, β) since their graphs go forever left and right encompassing all possible π₯-values.
** Please refer to the reference sheet at the end of this packet for graphs of all parent functions in order to be able to determine domain and range. **
Below are examples of when the domain is restricted based on the functionβs graph and how to find the domain algebraically.
Function Type
Finding Domain Algebraically
Radical Functions (π(π₯) = βπ₯)
You can never have a negative inside of a square root (because then you would end up with the imaginary unit). Therefore, to
find the domain you set the radicand (inside of the radical) greater than or equal to zero and solve for π₯.
Rational Functions (π(π₯) =1
π₯)
Rational functions are basically fractions (division). Since you can never divide by zero, to find the domain, you set the denominator to βnot equalβ zero and find the restriction.
Trigonometric Functions (π(π₯) = sin π₯ and cos π₯ only )
The graphs of sine and cosine are waves that oscillate forever in both directions. Therefore, the domains of both functions are
always (ββ, β).
Natural Log Functions (π(π₯) = ln π₯)
The graph of natural log has a vertical asymptote at π₯ = 0. Therefore, to find the domain, like the radical function, you set
what is βinsideβ of ππ to be greater than (not equal) to zero.
State the domain and range of each of the following functions.
** To find the domain of the functions, follow the guidelines in the chart on the previous page. To determine
the range, you have to visualize the functionβs behavior based on the vertical shifts. If the parent function
normally has a range of [0, β) but the function was shifted up 5 units, then the range would be [5, β). **
52. π(π₯) = π₯2 β 3
β’ Domain: __________________________
β’ Range: ___________________________
53. π(π₯) = 3 sin π₯
β’ Domain: __________________________
β’ Range: ___________________________
54. π(π₯) = ββπ₯ + 3
β’ Domain: __________________________
β’ Range: ___________________________
55. π(π₯) =2
π₯β1
β’ Domain: __________________________
β’ Range: ___________________________
Topic 13: Inverses of Functions
Functions are considered inverses if and only if their π₯ and π¦ values have switched places. Graphically, this means that inverse functions are reflections along the line π¦ = π₯.
** Inverse function notation is denoted as πβ1(π₯). **
To find the inverse of a function, simply interchange the π₯ and π¦ and solve for π¦. You can also verify your answer by checking that π[πβ1(π₯)] = π₯ and πβ1[π(π₯)] = π₯.
Example:
Find the inverse of the function π(π₯) = βπ₯ + 13
. Then, verify that your answer is the inverse by finding at least one of the compositions π[πβ1(π₯)] = π₯ or πβ1[π(π₯)] = π₯.
Solution:
β’ Rewrite π(π₯) as π¦ β π¦ = βπ₯ + 13
β’ Interchange the π₯ and π¦ β π₯ = βπ¦ + 13
β’ Solve for π¦ β π₯3 = π¦ + 1 β π₯3 β 1 = π¦
β’ Rewrite using appropriate inverse function notation β πβ1(π₯) = π₯3 β 1
Verify your answer by finding one of the compositions indicated.
β’ π[πβ1(π₯)] = β(π₯3 β 1) + 13
= βπ₯3 + 1 β 13
βπ₯33= π₯
Since the composition equals to π₯, the inverse is
correct!
Find the inverse for each function. Verify that your answer is correct through composition.
56. π(π₯) = 2π₯ + 1
57. π(π₯) =π₯2
3
58. π(π₯) =5
π₯β2
59. π(π₯) = β4 β π₯ + 1
Topic 14: Even and Odd Functions
Even functions are functions that are symmetric over the π¦-axis.
β’ To determine algebraically if a function is even, you must find out if π(βπ₯) = π(π₯).
β’ This means that if you plug in βπ₯ and all of the terms remain the same sign, then the function is even.
Odd functions are functions that are symmetric over the origin.
β’ To determine algebraically if a function is odd, you must find out if π(βπ₯) = βπ(π₯).
β’ This means that if you plug in βπ₯ and all of the terms change signs, then the function is odd.
** If you plug in βπ₯ into the function and only some terms change signs and others donβt, then the function is neither even nor odd. **
Example of an even function:
π(π₯) = 2π₯2 + 3|π₯|
β π(βπ₯) = 2(βπ₯)2 + 3|βπ₯| = 2π₯2 + 3|π₯|
Since π(βπ₯) looks identical to π(π₯), then π(π₯) is an even function.
Example of an odd function:
π(π₯) = β5π₯3 + 3π₯
β π(βπ₯) = β5(βπ₯)3 + 3(βπ₯) = 5π₯3 β 3π₯
Since all of the terms in π(βπ₯) are the opposite signs of π(π₯), then π(π₯) is an odd function.
Example of a function that is neither even nor odd:
π(π₯) = 2π₯5 + 1
β π(βπ₯) = 2(βπ₯)5 + 1 = β2π₯5 + 1
Since only one term changed signs and the 1 did not, this means that π(π₯) is neither even nor odd.
State whether each of the following functions are even, odd, or neither.
60.
61.
62. π(π₯) = 2π₯4 β 5π₯2
63. π(π₯) = π₯5 β 3π₯3 + π₯
64. π(π₯) = 2π₯2 β 5π₯ + 3
65. π(π₯) = 2 cos π₯
Topic 15: Asymptotes of Rational Functions
Asymptotes are vertical or horizontal lines that the function will get infinitely close to but never touch.
Finding Asymptotes Algebraically
Example
To find vertical asymptotes, set the denominator
equal to zero to find the π₯-value for which the function is undefined (granted that the factor that
makes the denominator equal to zero is not a common factor of the numerator β this is a
removable discontinuity aka a hole).
Find the vertical asymptote of π(π₯) =π₯2β5π₯β14
π₯2β3π₯β28.
β’ Factor both the numerator and denominator
β’ π(π₯) =(π₯β7)(π₯+2)
(π₯β7)(π₯+4)β Simplify
β’ π(π₯) =π₯+2
π₯+4
β’ At π₯ = β4 there is a vertical asymptote where at π₯ = 7 there is a hole.
Horizontal asymptotes can be found depending on the following three cases:
(a) If the degree of the numerator < the degree of the denominator, then there is a horizontal asymptote at π¦ = 0.
(b) If the degree of the numerator = the degree of the denominator, then there is a horizontal asymptote that is equal to the quotient of the leading coefficients.
(c) If the degree of the numerator > the degree of the denominator, then there is NO horizontal asymptote.
Finding the horizontal asymptotes for all three cases:
(a) π(π₯) =3π₯
π₯2+2β HA @ π¦ = 0
(b) π(π₯) =2π₯2+3π₯β1
4π₯2+5β HA @ π¦ =
2
4=
1
2
(c) π(π₯) =5π₯3βπ₯
π₯2+2π₯β No horizontal asymptote
For each function, find the equations of the vertical and horizontal asymptotes as well as any holes that may
be present on the graph (if any).
66. π(π₯) =π₯+4
π₯2+1
67. π(π₯) =π₯2+5π₯β6
π₯2β1
68. π(π₯) =2π₯3
π₯2β4
69. π(π₯) =2π₯2β4π₯+2
π₯2β3π₯β4
Topic 16: Solving Rational Equations
There are two ways of solving rational equations: 1. Cross multiplication β this method only works if you have one rational expression that equals to
another rational expression. 2. Multiply the least common denominator (LCD) to the entire equation β this method will always work
but is most used when you have addition or subtraction of rational functions on at least one side of the equation.
** Keep in mind that rational equations have excluded values which can lead to extraneous solutions. The excluded values are the values of π₯ that make the denominator equal to zero. **
Example of cross multiplication:
β’ π₯β5
π₯+1=
π₯β9
π₯+5β Excluded values: π₯ = β1, β5
β’ Cross multiply
β’ (π₯ β 5)(π₯ + 5) = (π₯ β 9)(π₯ + 1) β Distribute
β’ π₯2 β 25 = π₯2 β 8π₯ β 9 β Solve for π₯.
β’ Although the resulting equation is a quadratic, if you subtract π₯2 from both sides, the term will cancel out.
β’ β16 = 8π₯ β Divide by 8 to both sides.
β’ Final answer: π₯ = β2
** Since π₯ = β2 is not an excluded value, that is the real solution. **
Example of multiplying the LCD:
β’ 12
π₯β1β
8
π₯= 2 β Excluded values: π₯ = 1, 0
β’ LCD: π₯(π₯ β 1)
β’ Multiply the LCD to the entire equation.
β’ π₯(π₯ β 1) [12
π₯β1β
8
π₯= 2]
β’ 12π₯ β 8(π₯ β 1) = 2π₯(π₯ β 1) β Distribute
β’ 12π₯ β 8π₯ + 8 = 2π₯2 β 2π₯ β Since the resulting equation is a quadratic, set it to equal to zero and solve by factoring.
β’ 2π₯2 β 6π₯ β 8 = 2(π₯2 β 3π₯ β 4) = 2(π₯ β 4)(π₯ + 1) = 0
β’ Final answer: π₯ = 4, β1
** Since π₯ = 4, β1 are none of the excluded values, both are the real solutions of the equation. **
Solve for π₯ for each rational equation:
70. π₯β5
π₯+1=
3
5
71. π₯+3
π₯β4=
π₯+4
π₯β10
72. 1
π₯β2+
1
π₯2β7π₯+10=
6
π₯β2
73. 2π₯+3
π₯β1=
10
π₯2β1+
2π₯β3
π₯+1
Topic 17: Logarithms and Exponential Functions
For π₯ > 0 and π > 0, π β 1 β¦ π¦ = logπ π₯ is equivalent to ππ¦ = π₯,
β¦where π is the base and π¦ is the exponent.
** Basically, a logarithmic function is the inverse of an exponential function. **
*** The common logarithm, log10 π₯, is usually written as log π₯. ***
The logarithmic function can have any base as long as the base is not zero or negative. However, the natural logarithmic function (ln π₯) has a base of π, where π is the exponential base and is
equal to the irrational number 2.718281 β¦
Throughout the course, we will be working with logarithmic functions, more commonly the natural logarithmic function.
Below is a list of properties you should familiarize yourself with regarding the properties of log. These properties can be applied to both logarithmic and natural logarithmic ( ln π₯) functions.
Property
Definition
Example
Product
logπ ππ = logπ π + logπ π
β’ log3 9π₯ = log3 9 + log3 π₯ β’ ln π₯(π₯ + 1) = ln π₯ + ln π₯ + 1
Quotient
logπ
π
π= logπ π β logπ π
β’ log2 (4π₯+1
π₯+2)
= log2(4π₯ + 1) β log2(π₯ + 2)
β’ ln (8
π₯) = ln 8 β ln π₯
Power
logπ ππ = π β logπ π
β’ log2 8π₯ = π₯ β log2 8 β’ ln π₯2 = 2 ln π₯
Equality
logπ π = logπ π , then π = π
β’ log3(3π₯ β 4) = log3(5π₯ + 2), then 3π₯ β 4 = 5π₯ + 2
β’ This property does not apply to the natural log function.
Logarithmic Exponent with Equal Bases
πlogπ π₯
β’ 4log4(π₯β1) = π₯ β 1 β’ πln 2π₯ = 2π₯
β’ This is because the base of ln π₯ is π so they would cancel each other out.
Expand the following logarithmic functions using the properties of logarithms.
74. ln(π₯π¦)6 75. log[(π₯ + 1)(π₯2)] 76. ln (5π₯+3
π¦2 )
Topic 18: Natural Log and Exponential Functions with Base π
Since the natural logarithmic function has a base of π, this means that ln π₯ and π are reciprocal functions and would cancel each other out if they are applied to one another.
When π has an exponent of ln π₯:
πln π₯ = π₯
Example:
πln(2π₯+4) = 2π₯ + 4
When π is βinsideβ of ln π₯:
ln ππ₯ = π₯
Example:
ln ππ₯2+1 = π₯2 + 1
To solve for π₯ when given an exponential equation with base π, apply the function ππ to both sides of the equation to cancel out the π.
Example: Solve for π₯ given that π2π₯ = 5
Solution:
β’ Apply ππ to both sides β ln π2π₯ = ln 5
β’ Simplify β 2π₯ = ln 5
β’ Divide by 2 β π₯ =ln 5
2
** Unless this was a calculator question, you are not expected to know the ln 5, so you would leave your
final answer as π₯ =ln 5
2. **
To solve for π₯ when given a natural logarithmic equation, apply the function π to both sides of the equation to cancel out the ππ.
Example: Solve for π₯ given that ln(4π₯ + 1) = 3
Solution:
β’ Apply π to both sides β πln(4π₯+1) = π3
β’ Simplify β 4π₯ + 1 = π3
β’ Isolate the π₯ β π₯ =π3β1
4
** Just like in the previous example, you would leave your answer as is. **
Solve for π₯ for each of the given equations.
77. ln π2π₯ = 5π₯ β 3
78. 5 ln π3π₯ = 30
79. 8ππ₯ = 51
80. ln(3π₯ β 2) = 7
Topic 19: The Unit Circle
Remember that you are expected to know the Unit Circle! However, you only need to remember the first quadrant. The values of sine and cosine never change as you revolve around the Unit Circle, only the signs.
** I have provided a copy of the Unit Circle at the end of this packet. **
β’ All trigonometric function values are positive in
quadrant 1.
β’ Only sine and cosecant function values are
positive in quadrant 2.
β’ Only tangent and cotangent function values are
positive in quadrant 3.
β’ Only cosine and secant function values are
positive in quadrant 4.
On the Unit Circle, the π₯ = cos π and π¦ = sin π.
β’ Tangent is found by dividing sine by cosine
β tan π =sin π
cos π.
β’ Secant is the reciprocal of cosine.
β’ Cosecant is the reciprocal of sine.
β’ Cotangent is the reciprocal of tangent.
Examples of finding a trigonometric function value:
(a) secπ
3β Since secant is the reciprocal of cosine,
we would first need to find cosπ
3 and then its
reciprocal.
β’ cosπ
3=
1
2β Therefore sec
π
3= 2
(b) tanπ
3=
sinπ
3
cosπ
3
=β3
21
2
β’ To divide the fractions, βcopy, change, flip.β
β’ β3
2(
2
1) = β3
β’ Therefore, tanπ
3= β3
Example of finding the inverse of a trigonometric function value:
To evaluate for the inverse of a trigonometric function means to find the value of the angle, π, that would give the value inside the parenthesis.
(a) cosβ1 (ββ2
2) =
3π
4,
5π
4
β’ Both angles are the solutions because the
cosine value for both angles is equal to ββ2
2.
(b) Evaluate sin (cosβ1 (1
2)) on the interval
0 < π <π
2.
β’ The interval provided lets you know what quadrant of the Unit Circle that the angle is located in. According to the interval given, the angle is in quadrant 1.
β’ First, we would have to evaluate for the inverse cosine and then find the sine value at the angle that you found.
β’ sin (cosβ1 (1
2)) = sin (
π
3) =
β3
2
Evaluate the following trigonometric and inverse trigonometric function values.
81. sec11π
6
82. cot π 83. sin5π
4
84. sinβ1(β1)
85. csc (sinβ1 (β2
2) ) on the
interval π
2< π₯ < π
86. tan (sinβ1 (β1
2)) on the
interval π < π₯ <3π
2
Topic 20: Solving Trigonometric Equations
To solve a trigonometric equation follows the same logical pathways as solving any other equation. The goal is to isolate the trigonometric function and then applying the inverse of the trigonometric function to solve for π₯. Depending on how the equation is presented, you may isolate the trigonometric function from
the beginning or (if given a squared trigonometric function) you may have to square root or factor.
Examples:
Solve the following trigonometric equations on the interval [0,2π).
** Side note: Remember that intervals could be written in either set notation or interval notation. The interval given indicates that π₯ = 0 is included and could be a solution but π₯ = 2π is not included and therefore will not be a solution. The interval simply means that you will state all of the solutions in one full revolution around the Unit Circle. **
(a) 4 sin π₯ = 2 sin π₯ + β2
(b) 4 cos2 π₯ + 1 = 4
(c) 2 sin2 π₯ + 3 sin π₯ = β1
Solutions:
(a) Isolate sin π₯ by first subtracting 2 sin π₯ from both sides of the equation
β 2 sin π₯ = β2
β’ Divide by 2 β sin π₯ =β2
2
β’ Apply the inverse of sine to isolate the π₯ β π₯ = sinβ1 (β2
2)
β’ Final answer: π₯ =π
4,
3π
4
(b) Isolate cos2 π₯ by subtracting the 1 and then dividing the 4
β cos2 π₯ =3
4
β’ Square root both sides to eliminate the exponent.
β βcos2 π₯ = β3
4=
β3
β4=
β3
2β cos π₯ =
β3
2
β’ Apply the inverse of cosine to isolate the π₯ β π₯ = cosβ1 (β3
2)
β’ Final answer: π₯ =π
6,
11π
6
(c) Since this equation is quadratic, first set it equal to zero so we can factor β 2 sin2 π₯ + 3 sin π₯ + 1 = 0
β’ Just so that it is easier to see how this can be factored, I will be replacing sin π₯ with π’ during the factoring process.
β’ If π’ = sin π₯, then 2 sin2 π₯ + 3 sin π₯ + 1 = 2π’2 + 3π’ + 1
β’ We can factor the quadratic using the ππ method. In this case, ππ = 2 and π = 2,1. Thereforeβ¦
β’ 2π’2 + 2π’ + π’ + 1 = 2π’(π’ + 1) + 1(π’ + 1) = (2π’ + 1)(π’ + 1) β’ Now replace π’ back with sin π₯.
β’ (2 sin π₯ + 1)(sin π₯ + 1) = 0 β’ Set each individual factor equal to zero and isolate sin π₯.
β’ 2 sin π₯ + 1 = 0 β sin π₯ = β1
2 and sin π₯ + 1 = 0 β sin π₯ = β1
β’ Apply the inverse of sine to solve for π₯.
β’ π₯ = sinβ1 (β1
2) =
7π
6,
11π
6 and π₯ = sinβ1(β1) =
3π
2
β’ Final answer: π₯ =7π
6,
11π
6,
3π
2
Solve the following trigonometric equations on the interval [0,2π).
87. 4 sin2 π₯ β 1 = 0
88. cos2 π₯ = cos π₯
89. 2 cos π₯ + β3 = 0
90. 1 β sin2 π₯ = 2 + 2 sin π₯
** Thatβs all folks! I hope you found this packet a helpful review. If you have any questions and would like
to speak to me, feel free to message me via Remind. See you in the fall! **