Math, Physics, Calculus Worksheets

Math 261 - Calculus 1

Worksheets

Pierce College

Instructor: Bob Martinez

Linear Functions Review Worksheet

Show all work on your paper as described in class. Video links are included throughout for instruction on how to do the

various types of problems. Important: Work the problems to match everything that was shown in the videos. For example:

Suppose a video shows 3 ways to do a problem, (such as algebraically, graphically, and numerically), then your work should

show these 3 ways also. That is , each video is a model for the work I want to see on your paper.

TI-84 Calculator video - shows some common things we do with the calculator in our class:

http://youtu.be/I-BV9tQk9TsWatch and make sure you can do the methods shown in the video.

ESSAY.

1) Go to http://youtu.be/ZV4xnYXO88g and watch and take notes on the video "What is Calculus?"

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

2) The thickness of a glacier, in meters, is given by T = 50 - 0.4x, where x is the number of years since 1960.

Interpret the slope of the graph as a rate of change.

A) The glacier will disappear in 125 years

B) The glacier's thickness is decreasing by 0.4 meters per year.

C) The glacier's thickness decreases by 1 meter every 0.4 years

D) The glacier was 50 meters thick in 1960

3) In 1912, the glacier on Mount Kilimanjaro in Africa covered 5 acres. By 2002, this glacier melted to only 1 acre.

Assuming this glacier melted at a constant rate each year, write a linear equation that gives the acres, A, of this

glacier t years after 1912. http://youtu.be/HNJtksdcgsE

A) A = - 2

45t + 5 B) A = 5t + 1 C) A = -22.5t + 1912 D) A = .04t - 555

4) The thickness of a glacier, in meters, is given by T = 60 - 0.2x, where x is the number of years since 1960.

Interpret the slope of the graph as a rate of change.

A) The glacier's thickness decreases by 1 meter every 0.2 years

B) The glacier was 60 meters thick in 1960

C) The glacier's thickness is decreasing by 0.2 meters per year.

D) The glacier will disappear in 300 years

5) The fish population, P, in Crystal Lake t years after the year 2000 is given by P = 430 - 60t. What does the slope

of the graph tell you?

A) The fish population is 60.

B) The fish population is decreasing by 60 fish per year.

C) The fish population is 430.

D) The fish population is increasing by 60 fish per year.

6) Find an equation of the line that passes through (1, -7) and (-8, 8)

http://youtu.be/bhQbaQGZHW8

A) y = 3

5x -

38

5B) y =

5

3x +

26

3C) y = -

5

3x -

16

3D) y =-

3

5x -

32

5

1

7) Find the slope of the line joining (1, -3) and (-5, -8)

A)5

6B)

6

5C) -

5

6D)

11

4

8) Find the slope of the line joining (-2, 6) and (5, -6)

A) 0 B) - 7

12C)

12

7D) -

12

7

Find the slope-intercept form for the line satisfying the conditions.

9) Parallel to y = 3x - 2, passing through (1, -4)

A) y = -3x - 1 B) y = 3x + 7 C) y = -3x + 1 D) y = 3x - 7

10) Find an equation for the line that is perpendicular to y = 2

5x + 2,passing through (8, -6)

http://youtu.be/Yff1R6Oyxfo

A) y = - 5

2x + 14 B) y = -

2

5x + 16 C) y =

5

2x - 14 D) y = -

5

2x - 6

11) Find an equation of a line perpendicular to y = 4

5x + 3, passing through (16, -7)

A) y = 5

2x - 26 B) y = -

4

5x + 15 C) y = -

5

4x + 13 D) y = -

5

4x + 2

12) Solve the equation: -8b - 4 = -7 - 4b

A)3

4B)

4

3C) -

4

3D)

12

11

Solve the problem.

13) Decide whether the points in the table lie on a line. If they do, find the slope-intercept form of the line.

x -2 -1 0 1

f(x) -4 1 6 11

A) Yes; f(x) = 4x + 6 B) No C) Yes; f(x) = 5x + 8 D) Yes; f(x) = 5x + 6

The points in the table lie on a line. Find the equation of the line.

14) x 2 3 4 5

y -4 -7 -10 -13

A) y = -3x + 2 B) y = -4x + 3 C) y = -3x + 1 D) y = 3x - 8

2

Give the slope-intercept form of the line shown.

15)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

http://youtu.be/En3NtcW2BCI

A) y = 2

3x - 2 B) y = -

3

2x - 3 C) y = -

2

3x - 2 D) y = -

3

2x - 2

16)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) y = -4x - 3 B) y = 1

4x - 3 C) y =

1

5x - 3 D) y = -

1

4x - 4

17) Find an equation for the line that has a slope of - 3

4 and passes through (4, 2)

A) y = 3

4x - 5 B) y = -

3

4x + 5 C) y = -

3

4x + 4 D) y =

3

4x + 4

18) Find an equation for the line that has a slope of - 2

5 and passes through (4, 3)

A) y = - 2

5x +

23

5B) y =

2

5x -

23

5C) y =

2

5x + 4 D) y = -

2

5x + 4

3

ESSAY.

19) Office Outlet has 600 boxes of printer paper to sell. After 7 days, they had 460 boxes left. Suppose they continue

selling boxes at the same rate.

(a) Write a linear equation for the number of boxes, B, left after t days.

(b) Graph the equation for B. Remember to label the axes.

(c) State the slope of the graph, including units. What does the slope mean in this problem?

(d) What does the t-intercept tell us about this situation?

20) In 1960, a glacier in Arctic Ocean was 60 meters thick. The thickness of the glacier has been decreasing at a

constant rate of 0.2 meter per year.

a) Write an equation for the thickness, T, of the glacier in terms of x, the number of years since 1960.

b) How long will it take the glacier to be 45 meters thick?

c) Predict what year the glacier will disappear?

21) The fish population, P, in Crystal Lake t years after the year 2000 is shown on the graph below.

a) Find the t-intercept of the graph. What does it tell you about the problem?

b) Find the P-intercept of the graph. What does it tell you about the problem?

c) Find the slope of the line shown, including units.

d) What does the slope tell you about the problem?

e) Write P as a function of t.

t2 4 6 8 10 12 14 16

P

400

300

200

100

t2 4 6 8 10 12 14 16

P

400

300

200

100

4

22) A person is driving a car along a highway. The graph below shows the distance d in miles that the driver is

from home after t hours.

a) Find the t-intercept of the graph. What does it tell you about the problem?

b) Find the d-intercept of the graph. What does it tell you about the problem?

c) Find the slope of the line shown, including units.

d) What does the slope tell you about the problem?

e) Write d as a function of t.

x1 2 3 4 5 6

y350

300

250

200

150

100

50

x1 2 3 4 5 6

y350

300

250

200

150

100

50

23) At the Pierce Farm customers can pick their own tomatoes. There is a $3.00 entrance fee and the tomatoes cost

$1.50 per pound. At Winnetka Ranch, the entrance fee is only $1.00, but the tomatoes cost $2.00 per pound.

a) Write an equation for the total cost of buying x pounds of tomatoes at the Pierce Farm.

b) Write an equation for the total cost of buying x pounds of tomatoes at Winnetka Ranch.

c) How many pounds of tomatoes must be purchased in order for the total costs to be the same?

24) The number of female officers in the Marine Corps has been increasing at a constant rate since 1995. In 1995,

there were 690 female officers, and by 2003, this number had increased to 1,090.

a) Find a linear function that relates the number, N , of female officers in the Marine Corps to the number of

years, t, since 1995.

b) Estimate the number of female officers in the year 2000.

c) Interpret the slope of the linear function in part (a).

d) Interpret the N intercept.

25) The table below shows the amount of water in a tank w, in gallons, after t minutes.

t (minutes) 4 8 12 16

w (gallons) 200 160 120 80

(a) Write a linear equation for the amount of water in the tank, w, in terms of the time t.

(b) State the slope of the equation in part (a). What are the units of the slope? What does the slope mean in the

context of this problem?

(c) State the t-intercept of the equation in part (a). What does the t-intercept tell us about this situation?

5

Answer KeyTestname: LINEAR FUNCTIONS REVIEW WORKSHEET

1) watch the video

2) B

3) A

4) C

5) B

6) C

7) A

8) D

9) D

10) A

11) C

12) A

13) D

14) A

15) C

16) B

17) B

18) A

19) a) B=600-20t b) the graph c) -20 boxes/day means the number of boxes is decreasing at

20 boxes per day d) The number of days when there will be no boxes left (B = 0 at t = 30 days)

20) a) T = -0.2x + 60 b) 75 years (in the year 2035) c) 300 years (in the year 2260)

21) a) (14,0) in 14 years after the year 2000 (in 2014) there will be no fish left in the lake. b) (0,420) In the year 2000 there

were 420 fish in the lake. c) -30 fish/year d) The fish population in the lake is decreasing at the rate of 30 fish per

year. e) P = -30t + 420

22) a) (6,0) In 6 hours the driver will be home b) (0,300) At time 0 the driver was 300 miles away from home. c) -50

miles/hour d) The driver gets 50 miles closer to home every hour e) d = -50t + 300

23) a) P = 1.5x + 3 b) W = 2x + 1 c) 4 pounds

24) a) N = 50t + 690 b) 940 female officers c) slope = 50 female officers per year. The number of female officers in the

Marine Corps is increasing at the rate of 50 female officers per year. d) The N intercept is (0,690) In 1995 there were

690 female officers in the Marine Corps.

25) a) w = -10t + 240 b) -10 gallons/minute The amount of water in the tank is dropping at a rate of 10 gallons per minute.

c) (24,0) In 24 minutes all the water will be gone.

6

Quadratic Functions Review Worksheet






Solve the problem.

1) A projectile is thrown upward so that its distance above the ground after t seconds is h = -13t2 + 416t. After how

many seconds does it reach its maximum height? http://youtu.be/ZdCbfjfdP6MA) 32 sec B) 8 sec C) 24 sec D) 16 sec

2) A window washer accidentally drops a bucket from the top of a 144-foot building. The height h of the bucket

after t seconds is given by h = -16t2 + 144. When will the bucket hit the ground?

A) -3 sec B) 16 sec C) 3 sec D) 12 sec

Solve the problem.

3) An object is thrown upward with an initial velocity of 11 ft per second. Its height is given by h = -11t2 + 66t at

time t seconds. After how many seconds does it hit the ground?

A) 7 sec B) 3 sec C) 6 sec D) 9 sec

4) If an object is thrown upward with an initial velocity of 15 ft per second, its height is given by h = -15t2 + 90t

after time t seconds. What is its maximum height?

A) 45 ft B) 135 ft C) 225 ft D) 228 ft

Use the graph of y = ax2 + bx + c to solve the quadratic equation or inequality.

5) ax2 + bx + c < 0

x-10 10

y

10

-10

x-10 10

y

10

-10

http://youtu.be/-96ycSNVzdgA) x < -4 or x > 2 B) -4 ≤ x ≤ 2 C) -4 < x < 2 D) 2 < x < 4

1

6) ax2 + bx + c ≤ 0

x-10 10

y

10

-10

x-10 10

y

10

-10

A) x ≤ 2 or x ≥ 5 B) 2 < x < 5 or x > 5 C) 2 ≤ x ≤ 5 D) x < -2 or x < 5

Solve the quadratic inequality using your graphing calculator. Write your answer in interval notation.

7) x2 + 5x - 14 > 0 http://youtu.be/fWAqsyZnLioA) (-∞, -7) ∪ (2, ∞) B) (2, ∞) C) (-∞, -7) D) (-7, 2)

8) x2 - 2x - 3 < 0

A) (-∞, -1) B) (-∞, -1) ∪ (3, ∞) C) (-1, 3) D) (3, ∞)

9) Solve: x2 - 2x - 8 < 0

A) -4 < x < 2 B) x > -2 or x > 4 C) -2 < x < 4 D) x < -2 or x > 4

Solve the problem.

10) A flare fired from the bottom of a gorge is visible only when the flare is above the rim. If it is fired with an initial

velocity of 160 ft/sec, and the gorge is 336 ft deep, during what interval can the flare be seen?

(h = -16t2 + vot + ho.) http://youtu.be/-af_Bsd3Bb8A) 0 < t < 3 B) 6 < t < 10 C) 9 < t < 13 D) 3 < t < 7

11) A coin is tossed upward from a balcony 200 ft high with an initial velocity of 48 ft/sec. During what interval of

time will the coin be at a height of at least 40 ft?

(h = -16t2 + vot + ho.)

A) 0 ≤ t ≤ 1 B) 0 ≤ t ≤ 5 C) 5 ≤ t ≤ 10 D) 4 ≤ t ≤ 5

12) If a rocket is propelled upward from ground level, its height in meters after t seconds is given by

h = -9.8t2 + 78.4t. During what interval of time will the rocket be higher than 147 m?

A) 0 < t < 3 B) 3 < t < 5 C) 6 < t < 8 D) 5 < t < 6

2

Answer KeyTestname: QUADRATIC FUNCTIONS REVIEW WORKSHEET

1) D

2) C

3) C

4) B

5) C

6) A

7) A

8) C

9) C

10) D

11) B

12) B

3

Polynomial Functions Review Worksheet






Evaluate the expression graphically.

1) f(2)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) -3 B) -4 C) -2 D) -1

2) f(-1)

x-5 -4 -3 -2 -1 1 2 3 4 5

y10987654321

-1-2-3-4-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y10987654321

-1-2-3-4-5

A) -3 B) -4 C) -2 D) -5

1

3) f(0)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) -3 B) -2 C) -1 D) -4

Evaluate f(x) at the given value of x.

4) f(x) = 2x3 + 3x2 - x + 26 for x = -2

A) 26 B) 24 C) 14 D) 12

Solve the problem.

5) A(x) = -0.015x3 + 1.05x gives the alcohol level in an average person's blood x hrs after drinking 8 oz of

100-proof whiskey. If the level exceeds 1.5 units, a person is legally drunk. Would a person be drunk after 7

hours?

A) Yes B) No

6) The following polynomial approximates the shark population in a particular area.

R(x) = -0.021x5 + 3.785x4 + 300, where x is the number of years from 1985. Use a graphing calculator to describe

the shark population from the years 1985 to 2010.

A) The population remains stable.

B) The population increases.

C) The population decreases.

7) The polynomial function I(t) = -0.1t2 + 1.6t represents the yearly income (or loss) from a real estate investment,

where t is time in years. After how many years does income begin to decline?

A) 10.7 yr B) 8 yr C) 16 yr D) 7 yr

Solve the equation.

8) 6x4 - 3x3 = 0

A) 0, - 1

2B) 0,

1

3C) 0,

1

2D) 0, -

1

3

Factor completely.

9) x2 - x - 20

A) (x + 1)(x - 9) B) (x + 5)(x - 4) C) (x + 4)(x - 5) D) (x - 5)(x + 5)

10) x2 - 6x - 40

A) (x - 4)(x - 10) B) (x - 4)(x + 1) C) (x - 4)(x + 10) D) (x + 4)(x - 10)

2

Solve the equation.

11) 16t3 - 49t = 0

A)7

4B) -

7

4,

7

4, 0 C) 0 D) ±

7

4

12) 3x3 + 26x2 = -48x

A) 6, 8

3B) -6, -

8

3C) -6, -

8

3, 0 D) 0, 6,

8

3

13) x3 + 14x2 + 59x = -70 (solve by graphing and 2nd calc intersect: y1 = x3 + 14x2 + 59x y2 = -70 )

A) -1, -2, -5 B) 7, 2, 5 C) 0, 7, 2, 5 D) -7, -2, -5

Solve the problem.

14) Ariel, a marine biologist, models a population P of crabs, t days after being left to reproduce, with the function

P(t) = -0.00006t3 + 0.016t2 + 7t + 1200 (solve by graphing and 2nd calc zero)

Assuming that this model continues to be accurate, when will this population become extinct? (Round to the

nearest day.)

A) 1512 days B) 911 days C) 707 days D) 547 days

ESSAY

15) An open top rectangular box is to formed by taking an 8.5" by 11" piece of paper, cutting out an x inch by x inch

square out of each corner, then folding the sides up to form the box. Go tohttp://calculusapplets.com/boxproblem.html to see an example and move the slider on the applet

to see the effect of different size cutouts on the resulting shape and volume of the box.

a) Draw a picture of the sheet of paper with the cutouts illustrated. And draw a 3-d picture of the resulting box.

b) Write the volume of the box, V, as a function of the cutout side, x.

c) What is the domain of V? That is, what is the feasible range of values that x can be?

d) Graph V(x) on its domain and find the maximum volume of the box and the cutout side, x, that produces this

maximum. (Use 2nd calc max to find the maximum.)

16) An open top cylindrical can is to be made such that its surface area (around the side and on the bottom) is 100

cm2 .

a) Write the volume, V, of the can as a function of the radius, r, of the can. Leave the symbol, π, in the equation -

do not approximate.

b) Graph V(r) on its domain (on the domain of feasible values of r).

c) Find the radius that produces the greatest volume for the can.

d) What is the height of the max volume can?

FROM THE BOOK

17) Do section 1.3 #13, 15; section 1.6 #7 http://youtu.be/Bg0eJX1DKc4 , 9, 11; Chapter 1 Review #33, 39

3

Answer KeyTestname: POLYNOMIAL FUNCTIONS REVIEW WORKSHEET

1) A

2) B

3) B

4) B

5) A

6) B

7) B

8) C

9) C

10) D

11) B

12) C

13) D

14) D

15) discuss the answer in class

16) max volume = 108.59 cm3 achieved when the radius is 3.26 cm, the height of the max volume can is h = 3.26 cm

17) see answers in the book

4

Rational, Power, and Root Functions Review Worksheet






What are the equations of the vertical and horizontal asymptotes of the graph of the given equation?

1) y = 8

x

A) Vertical: y = 0; horizontal: x = 0 B) Vertical: x = 0; horizontal: y = 0

C) Vertical: y = 0; horizontal: x = 8 D) Vertical: x = 0; horizontal: y = 8

2) y = 8

x + 9

A) Vertical: x = 0; horizontal: y = 8 B) Vertical: y = 9; horizontal: x = 8

C) Vertical: x = 8; horizontal: y = 9 D) Vertical: x = 0; horizontal: y = 9

3) y = 1

(x - 8)2 + 9 http://youtu.be/6J6UtGKE7Zs

A) Vertical: x = 8; horizontal: y = 9 B) Vertical: x = 8; horizontal: y = 0

C) Vertical: x = 0; horizontal: y = 8 D) Vertical: x = 9; horizontal: y = 8

Graph f(x) by a) finding asymptotes, x intercepts, and y intercept, b) on the graphing calculator to verify

4) f(x) = x + 2

x + 1

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

A)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

B)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

1

C)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

D)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

5) f(x) = x - 3

x - 4

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

A)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

B)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

2

C)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

D)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

6) f(x) = -2x - 3

x + 2

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

A)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

B)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

3

C)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

D)

x-8 -4 4 8

y8

4

-4

-8

x-8 -4 4 8

y8

4

-4

-8

Provide an appropriate response.

7) Explain the behavior of the graph of f(x) as it approaches its vertical asymptote.

f(x) = 1

(x - 9)2

A) Approaches -∞ from the left and ∞ from the right

B) Approaches ∞ from the left and -∞ from the right

C) Approaches -∞ from the left and the right

D) Approaches ∞ from the left and the right


f(x) = 3

x - 9

A) Approaches ∞ from the left and -∞ from the right

B) Approaches -∞ from the left and ∞ from the right

C) Approaches ∞ from the left and the right

D) Approaches -∞ from the left and the right


f(x) = -1

x - 5

A) Approaches -∞ from the left and the right

B) Approaches -∞ from the left and ∞ from the right


D) Approaches ∞ from the left and -∞ from the right


f(x) = -2

(x - 3)2

A) Approaches -∞ from the left and the right

B) Approaches ∞ from the left and -∞ from the right


D) Approaches -∞ from the left and ∞ from the right

4

11) Suppose a friend tells you that the graph of f(x) = x2 - 16

x - 4 has a vertical asymptote with equation x = 4. Is this

correct? If not, describe the behavior of the graph at x = 4.

A) This is incorrect. The graph has an oblique asymptote at x = 4.

B) This is correct.

C) This is incorrect. The graph is actually the graph of y = x - 4 with a "hole" at (-4, -8).

D) This is incorrect. The graph is actually the graph of y = x + 4 with a "hole" at (4, 8).

Solve the problem.

12) At an altitude of h feet above the surface of the earth, the approximate distance in miles that a person can see is

given by

d = 1.2247h1/2.

How far can a person see if he or she is 740 feet above the earth's surface? Round your answer to the nearest

tenth of a mile, if necessary.

A) 39 miles B) 33.3 miles C) 34.4 miles D) 35.7 miles

13) The formula

T = .07D1.5

can be used to approximate the duration of a storm, where T is the time in hours and D is the diameter of the

storm in miles. A storm that is 16.4 miles in diameter is heading toward a city. How long can the residents of the

city expect the storm to last? Round your answer to the nearest hundredth of an hour.

A) 5.88 hr B) 4.65 hr C) 4.08 hr D) 6.18 hr

14) In a manufacturing operation, the cost, c, is related to the manufacturing time, t, by the equation

c = t.

Find the exact value of c when t = 500. Do not concern yourself with units.

A) 500 B) 50 C) 10 5 D) 22

15) A manufacturer's cost is given by

C = 2003

n + 200,

where C is the cost, in dollars, and n is the number of parts produced. Find the cost when 343 parts are

produced.

A) $1600 B) $29 C) $900 D) $3904

16) The distance in miles that can be seen from above the surface of the ocean is given by

d = 1.4 h

where h is the height in feet above the surface of the water. How many feet above the water would a pirate have

to climb to see 17 mi? Round your answer to the nearest foot.

A) 74 ft B) 6 ft C) 570 ft D) 147 ft

17) To model the actual speed s (in miles per hour) in an accident which left a skid mark of l feet, police use the

formula

s = S l

L,

where S is the test-car speed (in miles per hour) and L is the test-skid length (in feet). Find the speed S = 54

mph, l = 150 ft, and L = 100 ft. Round your answer to the nearest unit.

A) 81 mph B) 104 mph C) 66 mph D) 44 mph

5

18) The time T in seconds required for a pendulum of length L feet to make one swing is given by

T = 2π L

32.

How long is a pendulum if it makes one swing in 3.00 sec? Round your answer to the nearest tenth of a foot.

A) 15.3 ft B) 8.0 ft C) 14.6 ft D) 7.3 ft

19) The radius r of a cone of height h (in inches) and volume V (in cubic inches) is given by

r = 3V

πh.

If the height of a cone is 8.9 in. and its volume is 198 in.3, find its radius to the nearest hundredth of an inch.

A) 1.54 in. B) 4.61 in. C) 2.6 in. D) 0.87 in.

20) Fred is in a row boat that is 3 miles from the shore of a lake. He wants to get to his house, which is 17 miles

down the shore, as shown below. He will row to shore and then jog the remaining distance along the shore. He

can row at 3 miles per hour and can jog at 6 miles per hour. At about what point along the shore should he

beach the boat and jog the rest of the way if he wants to get home as soon as possible? Round your answer to the

nearest tenth of a mile, if necessary. http://youtu.be/SI1fTmJ5v8k

68%

3 miles

17 miles

A) 15.3 miles from home B) 10.3 miles from home

C) 14.3 miles from home D) 16.3 miles from home

21) Two vertical poles of lengths 32 feet and 87 feet are situated on level ground 50 feet apart, as shown in the figure

below. A piece of wire is to be strung from the top of the 32-ft pole, to a stake in the ground, to the top of the

87-ft pole. At what distance from the 32-foot pole should the stake be located to minimize the amount of wire

used? Round your answer to the nearest tenth of a foot.

32 ft 87 ft

50 ft

A) 14.5 ft B) 10.1 ft C) 17.6 ft D) 13.7 ft

22) The battleship USS Tennessee is 170 miles due south of the destroyer USS Alaska and is sailing north at 40 mph.

If the USS Alaska is sailing east at 25 mph, how long will it be before the distance between the ships is at a

minimum? Express your answer in hours and minutes, rounded to the nearest minute.

A) 41 minutes B) 2 hours 5 minutes C) 3 hours 4 minutes D) 1 hour 26 minutes

6

23) The battleship USS Tennessee is 170 miles due south of the destroyer USS Alaska and is sailing north at 40 mph.

If the USS Alaska is sailing east at 25 mph, how far apart will the ships be when that distance is at a minimum?

Round your answer to the nearest tenth of a mile.

A) 17.8 miles B) 42.5 miles C) 90.1 miles D) 84.3 miles

24) The phone company needs to install a line from point A on shore to an island that is 8 miles away from point B,

as shown in the drawing below. The cost associated with installing underwater cable is $18,600 per mile, while

the cost associated with installing cable over land is $9000 per mile. For the project to be completed at minimum

cost, how much of the cable should be installed over land? In other words, at what distance from point A should

the company stop laying cable over land? Round your answer to the nearest tenth of a mile, if necessary.

8 miles

10 miles

A) 5.6 miles B) 6.7 miles C) 7.1 miles D) 4.8 miles

25) The phone company needs to install a line from point A on shore to an island that is 3 miles away from point B,

as shown in the drawing. The cost associated with underwater cable is $18,600 per mile, while the cost

associated with cable used over land is $9000 per mile. To the nearest hundred dollars, what is the minimum

cost for which this project can be completed?

3 miles

10 miles

A) $137,600 B) $139,600 C) $138,800 D) $135,400

FROM THE BOOK

26) Do section 1.6 #13, 15, 19, 21 http://youtu.be/wOt-IWgtVro

7

Answer KeyTestname: RATIONAL, POWER, AND ROOT FUNCTIONS WORKSHEET

1) B

2) D

3) A

4) B

5) D

6) C

7) D

8) B

9) D

10) A

11) D

12) B

13) B

14) C

15) A

16) D

17) C

18) D

19) B

20) A

21) D

22) C

23) C

24) A

25) C


8

Exponential and Logarithmic Functions Review Worksheet






Graph the function.

1) f(x) = 5x

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

A)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

B)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

C)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

D)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

1

2) f(x) = 1

5

x

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

A)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

B)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

C)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

D)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

Solve the problem.

3) The amount of particulate matter left in solution during a filtering process decreases by the equation

P(n) = 800(0.5)0.4n, where n is the number of filtering steps. Find the amounts left for n = 0 and n = 5. (Round to

the nearest whole number.)

A) 1600 ; 200 B) 800 ; 200 C) 800 ; 25 D) 800 ; 3200

4) The number of bacteria growing in an incubation culture increases with time according to B(x) = 1500(2)x,

where x is time in days. Find the number of bacteria when x = 0 and x = 3 .

A) 1500 ; 9000 B) 3000 ; 12,000 C) 1500 ; 12,000 D) 1500 ; 6000

2

5) A computer is purchased for $3500. Its value each year is about 75% of the value the preceding year. Its value, in

dollars, after t years is given by the exponential function V(t) = 3500(0.75)t. Find the value of the computer after

7 years. Round to the nearest cent.

A) $18,375.00 B) $350.40 C) $262.80 D) $467.19

6) The half-life of a certain radioactive substance is 7 years. Suppose that at time t = 0 , there are 22 g of the

substance. Then after t years, the number of grams of the substance remaining will be N(t) = 221

2

t/14. How

many grams of the substance will remain after 28 years? Round to the nearest hundredth when necessary.

A) 5.5 g B) 2.75 g C) 0.69 g D) 1.38 g

7) The space in a landfill decreases with time as given by the function F(t) = 260 - 30 log (20t + 5) acres where t is

measured in years. How much space is left when t = 6 ?

A) 197 acres B) 180 acres C) 110 acres D) 320 acres

Solve the equation.

8) 15 x = 22 (Round to the nearest hundredth.) http://youtu.be/0jLz4PByEGgA) 0.88 B) 1.38 C) 1.14 D) 0.17

9) 14 x - 2 = 26 (Round to the nearest hundredth.)

A) 3.35 B) 3.86 C) 3.23 D) 2.81

Solve the problem.

10) Sound levels in decibels can be computed by f(x) = 10log(x/So), where x is the intensity of the sound in watts per

square meter and So = 1.00 x 10-12 watt/m2. A certain noise produces 6.81 x 10-5 watt/m2 of power. What is

the decibel level of this noise? (Round to the nearest decibel)

A) 180 decibels B) 78 decibels C) 68 decibels D) 8 decibels

Solve the equation.

11) 6 3 x = 5 x + 1 (Round to the nearest thousandth.)

A) 1.898 B) 0.898 C) 0.427 D) 8.827

12) 3(1 + 2x) = 27 http://youtu.be/3Igo_ar7dVIA) 9 B) 3 C) 1 D) -1

13) 3x = 1

81

A)1

27B) -4 C)

1

4D) 4

Solve the problem.

14) How long will it take for the population of a certain country to triple if its annual growth rate is 4.4 %? (Round

to the nearest year.) http://youtu.be/RnfHYe5pRdAA) 11 yr B) 1 yr C) 68 yr D) 26 yr

3

15) How long will it take for the population of a certain country to double if its annual growth rate is 7.4 %? (Round

to the nearest year.)

A) 27 yr B) 4 yr C) 9 yr D) 1 yr

16) There are currently 79 million cars in a certain country, decreasing by 1.7 % annually. How many years will it

take for this country to have 57 million cars? (Round to the nearest year.)

A) 4 yr B) 182 yr C) 19 yr D) 13 yr

17) An economist predicts that the buying power B(x) of a dollar x years from now will be given by the formula

B(x) = 0.21x. How much will today's dollar be worth in 6 years? Round the answer to the nearest cent.

A) $1.46 B) $1.26 C) $0.39 D) $0.00

18) Suppose that y = 2 - log (100 - x)

0.40 can be used to calculate the number of years y for x percent of a population of

444 web-footed sparrows to die. Approximate the percentage (to the nearest whole per cent) of web-footed

sparrows that died after 4 years.

A) 97% B) 100% C) 95% D) 99%

19) Suppose f(x) = 34.2 + 1.4log (x + 1) models salinity of ocean water to depths of 1000 meters at a certain latitude. x

is the depth in meters and f(x) is in grams of salt per kilogram of seawater. Approximate the salinity (to the

nearest hundredth) when the depth is 710 meters.

A) 98.93 grams of salt per kilogram of seawater B) 96.13 grams of salt per kilogram of seawater

C) 38.19 grams of salt per kilogram of seawater D) 30.21 grams of salt per kilogram of seawater

20) Suppose f(x) = 30.6 + 1.4log (x + 1) models salinity of ocean water to depths of 1000 meters at a certain latitude. x

is the depth in meters and f(x) is in grams of salt per kilogram of seawater. Approximate the depth (to the

nearest tenth of a meter) where the salinity equals 37.

A) 37,276.9 meters B) -0.8 meters C) -1.0 meters D) 37,274.9 meters

FROM THE BOOK

21) Do section1.2 #15, 17, http://youtu.be/sf-mGSj5gCk ,27, 29 http://youtu.be/1swEiISxIuQ ;section 1.4 #47, 53 http://youtu.be/WVCzQXSaCD0

4

Answer KeyTestname: EXPONENTIAL AND LOGARITHMIC FUNCTIONS REVIEW WORKSHEET

1) D

2) A

3) B

4) C

5) D

6) A

7) A

8) C

9) C

10) B

11) C

12) C

13) B

14) D

15) C

16) C

17) D

18) A

19) C

20) D


5

Trigonometric Functions Review Worksheet





ESSAY.

1) Watch and take notes on this Radian Angles video: http://youtu.be/TArUro2zXcg

2) Take notes on these 3 Unit Circle Trig Explanation videos and know for a test:http://youtu.be/FrXpbS6pBlM , http://youtu.be/_9SA2VK4O2Y , http://youtu.be/7pci1NP0Bok

3) Go to http://www.youtube.com/watch?v=YfcIaUF2JqM After watching the 3 videos in the problem above, watch this video and copy down the unit circle with all the

angles and the corresponding points on the unit circle shown.


Find the exact value by drawing the unit circle, the angle, and the terminal point coordinates. Verify with your calculator.

4) sin 2π

A) 0 B)2

2C) 1 D) undefined

Find the exact value. Verify with your calculator.

5) cos 0

A) 0 B) 1 C)2

2D) undefined

6) tan 0

A) 1 B) 0 C)2

2D) undefined

7) tan π

A) 0 B) -1 C) 1 D) undefined

8) cos π

A) 0 B) -1 C) 1 D) undefined

9) sin (38π) http://youtu.be/KSeOYmnDdXwA) -1 B) 0 C) 1 D) undefined

10) sin (- π

2)

A) 1 B) 0 C) -1 D) undefined

11) cos (-π)

A) -1 B) 0 C) 1 D) undefined

1

12) cos π

4

A) -2

2B) 2 C)

3

2D)

2

2

13) cos 45°

A)2

2B) 2 C)

1

2D)

3

2

The figure shows an angle θ in standard position with its terminal side intersecting the unit circle. Evaluate the indicated

circular function value of θ.

14) Find cos θ.

- 5

13,

12

13

http://youtu.be/YXplSLI08Rg

A) - 12

13B) -

5

13C)

12

13D) -

5

12

2

15) Find tan θ.

7

25, -

24

25

A) - 25

7B) -

7

24C)

25

24D) -

24

7

16) Find tan θ.

- 5

13,

12

13

A) - 12

5B) -

13

12C)

12

5D) -

5

12

3

17) Find sin θ.

- 5

13,

12

13

A)12

13B) -

12

13C)

5

12D) -

5

13

Find the exact circular function value.

18) tan -3π

4

A) 3 B) -1 C)3

3D) 1

19) tan 7π

6

A)3

2B) 3 C)

3

3D) - 3

20) cos 2π

A) 1 B)1

2C) 0 D) -1

21) sin 4π

3

A)3

2B) -

1

2C) -

3

2D) -1

22) sin 3π

4

A) - 2

2B) -

1

2C)

2

2D)

1

2

4

23) cos -2π

3 http://youtu.be/Lb1FXS-riE4

A) - 3

2B) -

1

2C) undefined D)

3

2

Find the exact value. Do not use a calculator.

24) sin 405° http://youtu.be/vw7jiLIBeuQ

A) - 1

2B) -

2

2C)

2

2D)

1

2

25) cos 8π

3

A)3

2B) -

1

2C) -

3

2D)

1

2

Use the fact that the trigonometric functions are periodic to find the exact value of the expression. Do not use a calculator.

26) tan 9π

4

A) 1 B)3

3C) -1 D) 3

27) sin 22π

3

A)3

2B) -

3

2C) -1 D) -

1

2

28) cos 20π

3

A)1

2B) -

1

2C)

3

2D) -

3

2

29) tan 720°

A)3

3B) 0 C) 1 D) undefined

30) tan 390°

A)3

2B) 3 C) - 3 D)

3

3

31) sin 495°

A) - 1

2B) -

2

2C)

1

2D)

2

2

Graph the sinusoidal function.

5

32) y = 3 sin (πx) http://youtu.be/1CBtrE-qoaQ

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

A)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

B)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

C)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

D)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

6

33) y = -3 cos (πx)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

A)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

B)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

C)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

D)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

7

34) y = 3 sin (2x)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

A)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

B)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

C)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

D)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

8

35) y = 3 cos (πx)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

A)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

B)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

C)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

D)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

9

36) y = -4 sin (1

4x)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

A)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

B)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

C)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

D)

x-π π 2π 3π

y6

4

2

-2

-4

-6

x-π π 2π 3π

y6

4

2

-2

-4

-6

Solve the problem.

37) A surveyor is measuring the distance across a small lake. He has set up his transit on one side of the lake 150

feet from a piling that is directly across from a pier on the other side of the lake. From his transit, the angle

between the piling and the pier is 55°. What is the distance between the piling and the pier to the nearest foot?

http://youtu.be/gcyA7IobPAUA) 123 ft B) 86 ft C) 214 ft D) 105 ft

10

38) A radio transmission tower is 220 feet tall. How long should a guy wire be if it is to be attached 5 feet from the

top and is to make an angle of 25° with the ground? Give your answer to the nearest tenth of a foot.

A) 508.7 ft B) 242.7 ft C) 237.2 ft D) 520.6 ft

39) A building 180 feet tall casts a 90 foot long shadow. If a person looks down from the top of the building, what is

the measure of the angle between the end of the shadow and the vertical side of the building (to the nearest

degree)? (Assume the person's eyes are level with the top of the building.)

A) 63° B) 30° C) 27° D) 60°

40) A tree casts a shadow of 26 meters when the angle of elevation of the sun is 24°. Find the height of the tree to the

nearest meter.

A) 13 m B) 12 m C) 10 m D) 11 m

41) A twenty-five foot ladder just reaches the top of a house and forms an angle of 41.5° with the wall of the house.

How tall is the house? Round your answer to the nearest 0.1 foot.

A) 18.7 ft B) 18.6 ft C) 19 ft D) 18.8 ft

FROM THE BOOK

42) Do section 1.5 #19, 21 http://youtu.be/d6ziBWJ8rx0 , 23-27 odd

11

Answer KeyTestname: TRIGONOMETRIC FUNCTIONS REVIEW WORKSHEET

1)

2) see the video

3) see the video

4) A

5) B

6) B

7) A

8) B

9) B

10) C

11) A

12) D

13) A

14) B

15) D

16) A

17) A

18) D

19) C

20) A

21) C

22) C

23) B

24) C

25) B

26) A

27) B

28) B

29) B

30) D

31) D

32) B

33) C

34) D

35) C

36) D

37) C

38) A

39) C

40) B

41) A


12

Limits and Continuity Worksheet





ESSAY

1) Go to http://youtu.be/5vSUrN-nqwE and watch and take notes on "The Idea of a Limit" .

FROM THE BOOK

2) Read and take notes on section 1.8 Limits: The idea of a limit; Definition of a limit; Properties of Limits; One

and two sided limits; Limits at infinity; Continuity

3) Do section 1.8 #1, 2

ESSAY

4) Go to http://calculusapplets.com/tablelimits.html scroll down and hit "launch presentation" to

make the applet larger.

a) From the drop down menu arrow in the upper right corner choose 1. A nice line. Move the slider (in the lower

left corner) so the x values go toward c =1 from the left and write what number the y values approach. This is

limx → 1-

.5x Then move the slider so the x values go toward c =1 from the right and write what number the y

values approach. This is limx → 1+

.5x Are the left and right limits equal? So, what is the value of limx →1

.5x ?

b) From the drop down menu choose 2. A line with a displaced point. Move the slider the same way you did in

(a) and write down the left limit and the right limit.

c) From the drop down menu choose 5. A jump discontinuity. Move the slider the same way you did in (a) and

write down the left limit and the right limit. So, what is the value of limx →1

f(x) ?


Solve the problem.

5) What conditions, when present, are sufficient to conclude that a function f(x) has a limit as x approaches some

value of a?

A) f(a) exists, the limit of f(x) as x→a from the left exists, and the limit of f(x) as x→a from the right exists.

B) The limit of f(x) as x→a from the left exists, the limit of f(x) as x→a from the right exists, and these two

limits are the same.

C) Either the limit of f(x) as x→a from the left exists or the limit of f(x) as x→a from the right exists

D) The limit of f(x) as x→a from the left exists, the limit of f(x) as x→a from the right exists, and at least one of

these limits is the same as f(a).

1

Use the graph to evaluate the limit.

6) limx→-1

f(x)

x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y

1

-1

x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y

1

-1

A) -1 B)1

2C) ∞ D) -

1

2

7) limx→0

f(x)

x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

A) 1 B) does not exist C) -1 D) 0

2

8) limx→0

f(x)

x-2 -1 1 2 3 4 5

y12

10

8

6

4

2

-2

-4

x-2 -1 1 2 3 4 5

y12

10

8

6

4

2

-2

-4

A) 0 B) 6 C) does not exist D) -1

9) limx→0

f(x)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

http://youtu.be/5Im6jcaoiAwA) 1 B) ∞ C) does not exist D) -1

3

10) limx→0

f(x)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

A) does not exist B) ∞ C) -1 D) 1

11) limx→0

f(x)

x-4 -3 -2 -1 1 2 3 4

y

4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y

4

3

2

1

-1

-2

-3

-4

A) -2 B) does not exist C) 0 D) 2

4

12) limx→0

f(x)

x-4 -3 -2 -1 1 2 3 4

y

4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y

4

3

2

1

-1

-2

-3

-4

A) 0 B) -2 C) 1 D) does not exist

13) limx→0

f(x)

x-4 -3 -2 -1 1 2 3 4

y

4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y

4

3

2

1

-1

-2

-3

-4

A) 2 B) -2 C) -1 D) does not exist

5

14) Find limx→(-1)-

f(x) and limx→(-1)+

f(x)

x-4 -2 2 4

y

2

-2

-4

-6

x-4 -2 2 4

y

2

-2

-4

-6

A) -5; -2 B) -7; -2 C) -7; -5 D) -2; -7

15) limx→0

f(x)

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4

x-4 -3 -2 -1 1 2 3 4

y4

3

2

1

-1

-2

-3

-4


6

Use the table of values of f to estimate the limit.

16) Let f(x) = x2 + 8x - 2, find limx→2

f(x). http://youtu.be/YHiZPKZ_gfM

x 1.9 1.99 1.999 2.001 2.01 2.1

f(x)

A)

x 1.9 1.99 1.999 2.001 2.01 2.1

f(x) 16.692 17.592 17.689 17.710 17.808 18.789 ; limit = 17.70

B)

x 1.9 1.99 1.999 2.001 2.01 2.1

f(x) 5.043 5.364 5.396 5.404 5.436 5.763 ; limit = 5.40

C)

x 1.9 1.99 1.999 2.001 2.01 2.1

f(x) 16.810 17.880 17.988 18.012 18.120 19.210 ; limit = 18.0

D)

x 1.9 1.99 1.999 2.001 2.01 2.1

f(x) 5.043 5.364 5.396 5.404 5.436 5.763 ; limit = ∞

17) Let f(x) = x - 4

x - 2, find lim

x→4f(x).

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x)

A)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 5.07736 5.09775 5.09978 5.10022 5.10225 5.12236 ; limit = 5.10

B)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 ; limit = 1.20

C)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 ; limit = ∞

D)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 3.97484 3.99750 3.99975 4.00025 4.00250 4.02485 ; limit = 4.0

7

18) Let f(x) = x2 - 5, find limx→0

f(x).

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x)

A)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -2.9910 -2.9999 -3.0000 -3.0000 -2.9999 -2.9910 ; limit = -3.0

B)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -4.9900 -4.9999 -5.0000 -5.0000 -4.9999 -4.9900 ; limit = -5.0

C)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -1.4970 -1.4999 -1.5000 -1.5000 -1.4999 -1.4970 ; limit = -15.0

D)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -1.4970 -1.4999 -1.5000 -1.5000 -1.4999 -1.4970 ; limit = ∞

19) Let f(x) = x - 4

x2 - 5x + 4, find lim

x→4f(x).

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x)

A)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 0.2448 0.2344 0.2334 0.2332 0.2322 0.2226 ; limit = 0.2333

B)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) -0.3448 -0.3344 -0.3334 -0.3332 -0.3322 -0.3226 ; limit = -0.3333

C)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 0.4448 0.4344 0.4334 0.4332 0.4322 0.4226 ; limit = 0.4333

D)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 0.3448 0.3344 0.3334 0.3332 0.3322 0.3226 ; limit = 0.3333

8

20) Let f(x) = x2 + 2x - 15

x2 - 2x - 3, find lim

x→3f(x).

x 2.9 2.99 2.999 3.001 3.01 3.1

f(x)

A)

x 2.9 2.99 2.999 3.001 3.01 3.1

f(x) 2.0256 2.0025 2.0003 1.9998 1.9975 1.9756 ; limit = 2

B)

x 2.9 2.99 2.999 3.001 3.01 3.1

f(x) 1.9256 1.9025 1.9003 1.8998 1.8975 1.8756 ; limit = 1.9

C)

x 2.9 2.99 2.999 3.001 3.01 3.1

f(x) 2.1256 2.1025 2.1003 2.0998 2.0975 2.0756 ; limit = 2.1

D)

x 2.9 2.99 2.999 3.001 3.01 3.1

f(x) -0.9048 -0.9900 -0.9990 -1.0010 -1.0101 -1.1053 ; limit = -1

21) Let f(x) = sin(5x)

x, find lim

x→0f(x).

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) 4.99791693 4.99791693

A) limit does not exist B) limit = 0 C) limit = 4.5 D) limit = 5

22) Let f(θ) = cos (5θ)

θ, find lim

θ→0f(θ).

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(θ) -8.7758256 8.7758256

A) limit = 5 B) limit does not exist C) limit = 0 D) limit = 8.7758256

Find the limit.

23) limx→18

2

A) 18 B) 3 2 C) 2 D) 2

24) limx→-9

(6x - 10)

A) -64 B) 44 C) 64 D) -44

Give an appropriate answer.

25) Let limx → -3

f(x) = 1 and limx → -3

g(x) = -10. Find limx → -3

[f(x) - g(x)]. http://youtu.be/E9fF0kbgShg

A) -9 B) -3 C) 11 D) 1

9

26) Let limx → 7

f(x) = 4 and limx → 7

g(x) = 5. Find limx → 7

[f(x) · g(x)].

A) 9 B) 5 C) 7 D) 20

27) Let limx → -8

f(x) = -7 and limx → -8

g(x) = -4. Find limx → -8

f(x)

g(x).

A)4

7B)

7

4C) -8 D) -3

Find the limit.

28) limx→2

(x3 + 5x2 - 7x + 1)

A) 15 B) does not exist C) 0 D) 29

29) limx→0

x3 - 6x + 8

x - 2

A) 4 B) 0 C) -4 D) Does not exist

30) limx→0

1 + x - 1

x http://youtu.be/tzUSjMBzBuk

A) 1/4 B) Does not exist C) 0 D) 1/2

Determine the limit by sketching an appropriate graph.

31) limx → 6-

f(x), where f(x) = -2x - 6 for x < 6

4x - 5 for x ≥ 6 http://youtu.be/MwTbTOgRSNg

For more info on graphing piecewise defined function in the calculator, seehttp://mathbits.com/mathbits/tisection/precalculus/piecewise.htm

A) -18 B) -4 C) -5 D) 19

32) limx → 6+

f(x), where f(x) = -4x - 3 for x < 6

5x - 2 for x ≥ 6

A) -27 B) 28 C) -2 D) -1

33) limx → 4+

f(x), where f(x) = x2 + 4 for x ≠ 4

0 for x = 4

A) 0 B) 16 C) 12 D) 20

34) lim

x → 5-f(x), where f(x) =

16 - x2 0 ≤ x < 4

4 4 ≤ x < 5

5 x = 5

A) 0 B) 4 C) Does not exist D) 5

35) lim

x → -7+f(x), where f(x) =

x -7 ≤ x < 0, or 0 < x ≤ 3

1 x = 0

0 x < -7 or x > 3

A) Does not exist B) 7 C) -7 D) -0

10

Find the limit, if it exists.

36) limx→0

x3 + 12x2 - 5x

5x

A) 5 B) 0 C) Does not exist D) -1

37) limx→1

x4 - 1

x - 1

A) 4 B) Does not exist C) 0 D) 2

38) limx → 10

x2 - 100

x - 10


39) limx → -9

x2 + 17x + 72

x + 9 http://youtu.be/G-sDRUmTbX0

A) 306 B) -1 C) 17 D) Does not exist

40) limx → 5

x2 + 3x - 40

x - 5


41) limx → 2

x2 + 2x - 8

x2 - 4

A) - 1

2B)

3

2C) Does not exist D) 0

42) limx → 3

x2 - 9

x2 - 7x + 12

A) - 6 B) 0 C) - 3 D) Does not exist

43) limh → 0

(x + h)3 - x3

h

A) 3x2 + 3xh + h2 B) 3x2 C) Does not exist D) 0

44) limx → 9

9 - x

9 - x http://youtu.be/1sF82HMZz4o

A) 0 B) -1 C) Does not exist D) 1

11

Compute the values of f(x) and use them to determine the indicated limit.

45) If f(x) = x4 - 1

x - 1, find lim

x → 1 f(x).

x 0.9 0.99 0.999 1.001 1.01 1.1

f(x)

A)

x 0.9 0.99 0.999 1.001 1.01 1.1

f(x) 1.032 1.182 1.198 1.201 1.218 1.392 ; limit = ∞

B)

x 0.9 0.99 0.999 1.001 1.01 1.1

f(x) 1.032 1.182 1.198 1.201 1.218 1.392 ; limit = 1.210

C)

x 0.9 0.99 0.999 1.001 1.01 1.1

f(x) 4.595 5.046 5.095 5.105 5.154 5.677 ; limit = 5.10

D)

x 0.9 0.99 0.999 1.001 1.01 1.1

f(x) 3.439 3.940 3.994 4.006 4.060 4.641 ; limit = 4.0

46) If f(x) = x3 - 6x + 8

x - 2, find lim

x → 0 f(x).

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x)

A)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -1.22843 -1.20298 -1.20030 -1.19970 -1.19699 -1.16858 ; limit = ∞

B)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -4.09476 -4.00995 -4.00100 -3.99900 -3.98995 -3.89526 ; limit = -4.0

C)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -2.18529 -2.10895 -2.10090 -2.99910 -2.09096 -2.00574 ; limit = -2.10

D)

x -0.1 -0.01 -0.001 0.001 0.01 0.1

f(x) -1.22843 -1.20298 -1.20030 -1.19970 -1.19699 -1.16858 ; limit = -1.20

12

47) If f(x) = x - 4

x - 2, find lim

x → 4 f(x). http://youtu.be/q0OgTtfx0uk

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x)

A)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 3.97484 3.99750 3.99975 4.00025 4.00250 4.02485 ; limit = 4.0

B)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 ; limit = 1.20

C)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 ; limit = ∞

D)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 5.07736 5.09775 5.09978 5.10022 5.10225 5.12236 ; limit = 5.10

48) If f(x) = x - 2, find limx → 4

f(x).

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x)

A)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 3.9000 2.9000 1.9000 2.0000 3.0000 4.0000 ; limit = ∞

B)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 3.9000 2.9000 1.9000 2.0000 3.0000 4.0000 ; limit = 1.95

C)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) 1.47736 1.49775 1.49977 1.50022 1.50225 1.52236 ; limit = 1.50

D)

x 3.9 3.99 3.999 4.001 4.01 4.1

f(x) -0.02516 -0.00250 -0.00025 0.00025 0.00250 0.02485 ; limit = 0

13

For the function f whose graph is given, determine the limit.

49) Find limx→-1-

f(x) and limx→-1+

f(x).

x-6 -4 -2 2 4 6

y

4

2

-2

-4

-6

-8

-10

x-6 -4 -2 2 4 6

y

4

2

-2

-4

-6

-8

-10

A) -5; -2 B) -7; -5 C) -2; -7 D) -7; -2

50) Find limx→2-

f(x) and limx→2+

f(x).

x-4 -3 -2 -1 1 2 3 4

y

8

6

4

2

-2

-4

-6

-8

x-4 -3 -2 -1 1 2 3 4

y

8

6

4

2

-2

-4

-6

-8

A) does not exist; does not exist B) 1; 1

C) -4; 3 D) 3; -4

14

51) Find limx→2+

f(x).

x-2 -1 1 2 3 4 5 6 7

f(x)

8

7

6

5

4

3

2

1

-1x-2 -1 1 2 3 4 5 6 7

f(x)

8

7

6

5

4

3

2

1

-1

A) 5 B) 1.3 C) -1 D) 4

52) Find limx→1-

f(x).

x-5 -4 -3 -2 -1 1 2 3 4 5

f(x)5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

f(x)5

4

3

2

1

-1

-2

-3

-4

-5

A) 2 B)1

2C) -1 D) does not exist

15

53) Find limx→1+

f(x).

x-5 -4 -3 -2 -1 1 2 3 4 5

f(x)5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

f(x)5

4

3

2

1

-1

-2

-3

-4

-5

A) 31

2B) does not exist C) 3 D) 4

54) Find limx→0

f(x).

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5


16

55) Find limx→0

f(x).

x-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

y87654321

-1-2-3-4-5-6-7-8

x-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

y87654321

-1-2-3-4-5-6-7-8

A) 0 B) does not exist C) 2 D) -2

56) Find limx→-1

f(x).

x-4 -2 2 4

y

4

2

-2

-4

A

x-4 -2 2 4

y

4

2

-2

-4

A

A) -1 B) - 2

3C) does not exist D)

2

3

17

57) Find limx→∞

f(x).

x-5 -4 -3 -2 -1 1 2 3 4 5

f(x)5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

f(x)5

4

3

2

1

-1

-2

-3

-4

-5

A) -2 B) ∞ C) 0 D) does not exist

Note: The previous problem brings up an issue: We can say limx→a

f(x) = ∞ indicating that as

x approaches the value a, the f(x) values get larger without bound. But the limit in this caseis still technically "does not exist" since infinity is not a number. So think of it as that infinityis just a special way the limit is failing to exist and writing infinity gives us more information(as opposed to failing because the limit from the left does not equal the limit from the right).

Find the limit.

58) limx→-2

1

x + 2 http://youtu.be/nTfty6Q4wNw

A) Does not exist B) -∞ C) 1/2 D) ∞

59) limx → 7+

1

(x - 7)2

A) -∞ B) 0 C) ∞ D) -1

60) limx → -4-

5

x2 - 16

A) ∞ B) -1 C) 0 D) -∞

61) limx→(π/2)+

tan x http://youtu.be/VyGOMJ-O0l4

A) 0 B) -∞ C) 1 D) ∞

62) limx→(-π/2)-

sec x

A) -∞ B) 1 C) ∞ D) 0

18

63) limx → 1-

x2 - 4x + 3

x3 - x

A) - 1 B) 0 C) ∞ D) -∞

64) limx → 4+

x2 - 6x + 8

x3 - 4x

A) 0 B) Does not exist C) ∞ D) -∞

65) limx → 3+

2

x2 - 9

A) 0 B) -∞ C) 1 D) ∞

66) limx→∞

7

x - 1 http://youtu.be/110FiQsvSfI

A) -1 B) 1 C) -8 D) 6

67) limx→-∞

5

5 - (9/x2)

A) - 5

4B) -∞ C) 1 D) 5

68) limx→-∞

-5 + (5/x)

7 - (1/x2)

A)5

7B) -

5

7C) ∞ D) -∞

69) limx→∞

x2 - 7x + 9

x3 - 6x2 + 14

A)9

14B) 0 C) 1 D) ∞

70) limx→-∞

-4x2 - 3x + 6

-18x2 - 4x + 9

A)2

9B) ∞ C) 1 D)

2

3

71) limx→-∞

cos 4x

x

A) 1 B) 0 C) 4 D) -∞

19

72) limx→ - ∞

2x3 + 4x2

x - 6x2

A) -∞ B) ∞ C) - 2

3D) 2

73) limx→∞

9x3 - 5x2 + 3x

-x3 - 2x + 6

A) ∞ B)3

2C) -9 D) 9

74) limx→∞

2x + 1

15x - 7

A)2

15B) 0 C) ∞ D) -

1

7

ESSAY

75) Definition of Continuity: Go tohttp://www.youtube.com/watch?v=hlorAjS0xWE&feature=topics and watch and take notes

on everything, and know for a test.

76) In the book, Continuity is discussed in section 1.8 - Read and take notes


Find all points where the function is discontinuous.

77)

A) x = 4 B) None C) x = 2 D) x = 4, x = 2

78)

A) x = -2, x = 1 B) x = 1 C) None D) x = -2

20

79)

A) x = 0, x = 2 B) x = 2 C) x = -2, x = 0, x = 2 D) x = -2, x = 0

80)

A) x = -2, x = 6 B) x = 6 C) None D) x = -2

81)

A) None B) x = 1, x = 4, x = 5 C) x = 4 D) x = 1, x = 5

82)

A) x = 1 B) x = 0, x = 1 C) x = 0 D) None

83)

A) None B) x = 3 C) x = 0 D) x = 0, x = 3

84)

A) None B) x = -2 C) x = 2 D) x = -2, x = 2

21

85)

A) x = -2, x = 0, x = 2 B) None C) x = 0 D) x = -2, x = 2


86) Is f continuous at f(1), that is, at x = 1?

f(x) =

-x2 + 1,

4x,

-2,

-4x + 8

4,

-1 ≤ x < 0

0 < x < 1

x = 1

1 < x < 3

3 < x < 5

t-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

d6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

(1, -2)

t-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

d6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

(1, -2)

A) Yes B) No

87) Is f continuous at f(3), that is, at x = 3?

f(x) =

-x2 + 1,

3x,

-4,

-3x + 6

2,

-1 ≤ x < 0

0 < x < 1

x = 1

1 < x < 3

3 < x < 5

t-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

d6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

(1, -4)

t-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

d6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

(1, -4)

A) No B) Yes

22

88) Is f continuous at x = 0?

f(x) =

x3,

-3x,

5,

0,

-2 < x ≤ 0

0 ≤ x < 2

2 < x ≤ 4

x = 2

t-5 -4 -3 -2 -1 1 2 3 4 5

d10

8

6

4

2

-2

-4

-6

-8

-10

(2, 0)

t-5 -4 -3 -2 -1 1 2 3 4 5

d10

8

6

4

2

-2

-4

-6

-8

-10

(2, 0)

A) Yes B) No

89) Is f continuous at x = 4?

f(x) =

x3,

-2x,

6,

0,

-2 < x ≤ 0

0 ≤ x < 2

2 < x ≤ 4

x = 2

t-5 -4 -3 -2 -1 1 2 3 4 5

d10

8

6

4

2

-2

-4

-6

-8

-10

(2, 0)

t-5 -4 -3 -2 -1 1 2 3 4 5

d10

8

6

4

2

-2

-4

-6

-8

-10

(2, 0)

A) Yes B) No

90) Is f continuous on (-2, 4]? Note: A function is continuous on an interval if it is continuous atevery point in the interval

t-5 -4 -3 -2 -1 1 2 3 4 5

d10

8

6

4

2

-2

-4

-6

-8

-10

(2, 0)

t-5 -4 -3 -2 -1 1 2 3 4 5

d10

8

6

4

2

-2

-4

-6

-8

-10

(2, 0)

A) Yes B) No

23

ESSAY

91) Go to http://www.youtube.com/watch?v=Lgr-1ZKPnR4 Watch and take notes on the

Intermediate Value theorem. Use the Intermediate Value Theorem to prove that 6x3 + 5x2 + 4x + 7 = 0 has a

solution between -2 and -1.

92) Use the Intermediate Value Theorem to prove that 2x4 + 10x3 - 6x - 6 = 0 has a solution between -5 and -4.


Find numbers a and b, or k, so that f is continuous at every point.

93)

f(x) =

-7,

ax + b,

21,

x < -4

-4 ≤ x ≤ 3

x > 3

http://youtu.be/6d0WlFdk2QoA) a = 4, b = 33 B) a = -7, b = 21 C) a = 4, b = 9 D) Impossible

94)

f(x) =

x2,

ax + b,

x + 6,

x < -5

-5 ≤ x ≤ -2

x > -2

A) a = 7, b = -10 B) a = -7, b = 10 C) a = -7, b = -10 D) Impossible

95)

f(x) =

3x + 8,

kx + 4,

if x < -1

if x ≥ -1

A) k = - 1 B) k = - 4 C) k = 7 D) k = 4

96)

f(x) =

x2,

x + k,

if x ≤ 4

if x > 4

A) k = 20 B) k = 12 C) k = -4 D) Impossible

97)

f(x) =

x2,

kx,

if x ≤ 2

if x > 2

A) k = 4 B) k = 2 C) k = 1

2D) Impossible

ESSAY

98) Formal Definition of a Limit (also called the "epsilon - delta definition"): Go to

http://www.youtube.com/watch?v=NAAhJ5ucYJg . Watch and take notes on the definition and

the proof of the problem shown in the video. In the book, the formal definition of the limit is in section 1.8

- Read!

24


Solve the problem.

99) Select the correct statement for the definition of the limit: limx→x0

f(x) = L

means that __________________

A) if given any number ε > 0, there exists a number δ > 0, such that for all x,

0 < x - x0 < ε implies f(x) - L < δ.

B) if given any number ε > 0, there exists a number δ > 0, such that for all x,

0 < x - x0 < δ implies f(x) - L < ε.

C) if given any number ε > 0, there exists a number δ > 0, such that for all x,

0 < x - x0 < ε implies f(x) - L > δ.

D) if given a number ε > 0, there exists a number δ > 0, such that for all x,

0 < x - x0 < δ implies f(x) - L > ε.

100) Identify the incorrect statements about limits.

I. The number L is the limit of f(x) as x approaches x0 if f(x) gets closer to L as x approaches x0.

II. The number L is the limit of f(x) as x approaches x0 if, for any ε > 0, there corresponds a δ > 0 such that

f(x) - L < ε whenever 0 < x - x0 < δ.

III. The number L is the limit of f(x) as x approaches x0 if, given any ε > 0, there exists a value of x for which

f(x) - L < ε.

A) I and II B) II and III C) I and III D) I, II, and III

Use the graph to find a δ > 0 such that for all x, 0 < x - x0 < δ ⇒ f(x) - L < ε. Write proofs also as shown in this video

for all linear function cases: http://youtu.be/oY-I0BD1Xg8

101)

x

y

0 x

y

0

y = x + 3

4.2

4

3.8

0.8 1 1.2

NOT TO SCALE

f(x) = x + 3

x0 = 1

L = 4

ε = 0.2

A) 3 B) 0.4 C) 0.2 D) 0.1

25

102)

x

y

0 x

y

0

y = 5x - 1

9.2

9

8.8

2

1.96 2.04

NOT TO SCALE

f(x) = 5x - 1

x0 = 2

L = 9

ε = 0.2

http://youtu.be/oY-I0BD1Xg8A) 0.08 B) 0.4 C) 7 D) 0.04

103)

f(x) = -5x - 1

x0 = -1

L = 4

ε = 0.2

x

y

0 x

y

0

y = -5x - 1

4.2

4

3.8

-1

-1.04 -0.96

NOT TO SCALE

A) 0.04 B) -0.04 C) 7 D) 0.4

26

104)

f(x) = -x + 2

x0 = -2

L = 4

ε = 0.2

x

y

0 x

y

0

y = -x + 2

4.2

4

3.8

-2.2 -2 -1.8

NOT TO SCALE

A) 0.4 B) -0.2 C) 6 D) 0.2

105)

x

y

0 x

y

0

y = 4

3x + 2

3.5

3.3

3.1

0.8 1 1.1

NOT TO SCALE

f(x) = 4

3x + 2

x0 = 1

L = 3.3

ε = 0.2

A) 0.3 B) 2.3 C) -0.3 D) 0.1

27

106)

f(x) = - 3

2x + 1

x0 = -2

L = 4

ε = 0.2

x

y

0 x

y

0

y = - 3

2x + 1

4.2

4

3.8

-2.1 -2 -1.9

NOT TO SCALE

A) -0.2 B) 0.1 C) 6 D) 0.2

107)

x

y

0 x

y

0

y = x

1.66

1.41

1.16

1.3575 2 2.7675

NOT TO SCALE

f(x) = x

x0 = 2

L = 2

ε = 1

4

A) 1.41 B) 0.6425 C) -0.59 D) 0.7675

28

108)

x

y

0 x

y

0

y = x - 2

1.25

1

0.75

2.5625 3 3.5625

NOT TO SCALE

f(x) = x - 2

x0 = 3

L = 1

ε = 1

4

A) 2 B) 0.5625 C) 0.4375 D) 1

109)

x

y

0 x

y

0

y = 2x2

3

2

1

1

0.71 1.22

NOT TO SCALE

f(x) = 2x2

x0 = 1

L = 2

ε = 1

A) 0.22 B) 0.51 C) 1 D) 0.29

29

110)

x

y

0 x

y

0

y = x2 - 1

9

8

7

3

2.83 3.16

NOT TO SCALE

f(x) = x2 - 1

x0 = 3

L = 8

ε = 1

A) 0.17 B) 5 C) 0.16 D) 0.33

ESSAY

111) Watch and take notes on this video which shows what happens when you have the wrong limit andyou try to prove it correct vs. when you have the correct limit: http://youtu.be/5iFwvdkc3OU


A function f(x), a point x0, the limit of f(x) as x approaches x0, and a positive number ε is given. Find a number δ > 0 such

that for all x, 0 < x - x0 < δ ⇒ f(x) - L < ε. (see video for f(x) = 3x2 below for the method to use to do these problems)

112) f(x) = 6x + 7, L = 19, x0 = 2, and ε = 0.01

A) 0.001667 B) 0.005 C) 0.008333 D) 0.003333


that for all x, 0 < x - x0 < δ ⇒ f(x) - L < ε.

113) f(x) = 6x - 9, L = -3, x0 = 1, and ε = 0.01

A) 0.001667 B) 0.01 C) 0.003333 D) 0.000833

Find the limit L for the given function f, the point c, and the positive number ε. Then find a number δ > 0 such that, for all

x, 0 < |x - c|< δ ⇒ |f(x) - L| < ε.

114) f(x) = x2 -7x -18

x + 2, c = -2, ε = 0.02

A) L = 0; δ = 0.02 B) L = -11; δ = 0.02 C) L = -7; δ = 0.03 D) L = -18; δ = 0.03

A function f(x), a point c, the limit of f(x) as x approaches c, and a positive number ε is given. Find a number δ > 0 such

that for all x, 0 < x - c < δ ⇒ f(x) - L < ε.

115) f(x) = 3x2, L =12, c = 2, and ε = 0.1 http://youtu.be/Emw2prCwxEUA) δ = 0.00832 B) δ = 2.00832 C) δ = 1.99165 D) δ = 0.00835

116) f(x) = -8x + 6, L = -18, c = 3, and ε = 0.01

A) δ = 0.00125 B) δ = 0.0025 C) δ = 0.005 D) δ = -0.003333

30

117) f(x) = 9x - 10, L = 17, c = 3, and ε = 0.01

A) δ = 0.002222 B) δ = 0.000556 C) δ = 0.001111 D) δ = 0.003333


that for all x, 0 < x - x0 < δ ⇒ f(x) - L < ε.

118) f(x) = -2x - 6, L = -12, x0 = 3, and ε = 0.01

A) 0.005 B) 0.01 C) 0.0025 D) -0.003333

119) f(x) = 3x2, L =48, x0 = 4, and ε = 0.1

A) 3.99583 B) 4.00416 C) 0.00417 D) 0.00416

ESSAY

Prove the limit statement

120) limx→1

(5x - 4) = 1 Review this video seen earlier for how to write a proof of a limit statement:

http://www.youtube.com/watch?v=NAAhJ5ucYJg

121) limx→6

2x2 - 7x- 30

x - 6 = 17


Use your calculator to plot the function near the point x0 being approached. From your plot guess the value of the

limit.

122) limx→25

x - 5

x - 25 http://youtu.be/D6l-uv_Q56I

A)1

5B) 0 C)

1

10D) 5

123) limx→0

81 + x - 81 - x

x

A) 9 B) 0 C)1

9D)

1

18

124) limx→0

25 - x - 5

x

A) - 1

10B) 10 C)

1

10D) 5

125) limx→ -1

x2 - 1

x2 + 3 - 2

A)1

4B) 2 C) 1 D) 4

31

ESSAY

Solve the problem BY HAND with ALGEBRA

126) Evaluate lim

x → 5 x2 - 25

x - 5. http://youtu.be/beP0b18Fktg

Solve the problem.

127) Evaluate lim

x → 4

x - 4

x - 2. http://youtu.be/beP0b18Fktg (same as previous video)

128) Evaluate lim

x → 9

x - 3

x - 9.

129) Evaluate lim

x → 3

2x - 6

x2 - 4x + 3.

130) Evaluate lim

x → 3

x2 - x - 6

x - 3.

131) Find the trigonometric limit: lim

θ→0

θ2

cosθ. Use calculator methods

132) Find the trigonometric limit: lim

θ→0

tanθ

θ. Use calculator methods

Note: The following problems pertain to sums, differences, products, and quotients oflimits.

133) Go tohttp://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx#Extras_Limit_LimitProp Write down the nine properties of limits listed there. The proofs of each property are also shown. Know the

proofs for properties 7, 1, and 2 for a test.

http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx#Extras_Limit_LimitProp



134) Provide a short sentence that summarizes the general limit principle given by the formal notation

limx→a

[f(x) ± g(x)] = limx→a

f(x) ± limx→a

g(x) = L ± M, given that limx→a

f(x) = L and limx→a

g(x) = M.

A) The limit of a sum or a difference is the sum or the difference of the limits.

B) The limit of a sum or a difference is the sum or the difference of the functions.

C) The sum or the difference of two functions is continuous.

D) The sum or the difference of two functions is the sum of two limits.

32

135) The statement "the limit of a constant times a function is the constant times the limit" follows from a

combination of two fundamental limit principles. What are they?

A) The limit of a product is the product of the limits, and a constant is continuous.

B) The limit of a constant is the constant, and the limit of a product is the product of the limits.

C) The limit of a function is a constant times a limit, and the limit of a constant is the constant.

D) The limit of a product is the product of the limits, and the limit of a quotient is the quotient of the limits.

Give an appropriate answer.

136) Let limx → 2

f(x) = -5 and limx → 2

g(x) = -8. Find limx → 2

[f(x) - g(x)].

A) -5 B) -13 C) 3 D) 2

137) Let limx → -9

f(x) = -1 and limx → -9

g(x) = 10. Find limx → -9

[f(x) · g(x)].

A) -9 B) 9 C) 10 D) -10

138) Let limx → -6

f(x) = 9 and limx → -6

g(x) = 8. Find limx → -6

f(x)

g(x).

A)9

8B) 1 C) -6 D)

8

9

33

Answer KeyTestname: LIMITS AND CONTINUITY WORKSHEET

1)

2)

3) see answers in the book. For #2 discuss in class.

4) see the applet

5) B

6) B

7) D

8) A

9) C

10) A

11) A

12) B

13) B

14) D

15) B

16) C

17) D

18) B

19) D

20) A

21) D

22) B

23) C

24) A

25) C

26) D

27) B

28) A

29) C

30) D

31) A

32) B

33) D

34) B

35) C

36) D

37) A

38) A

39) B

40) A

41) B

42) A

43) B

44) C

45) D

46) B

47) A

48) D

49) C

50) D

34


51) D

52) A

53) C

54) B

55) B

56) D

57) B

58) A

59) C

60) A

61) B

62) C

63) A

64) A

65) D

66) A

67) C

68) B

69) B

70) A

71) B

72) B

73) C

74) A

75)

76)

77) A

78) B

79) C

80) B

81) A

82) D

83) B

84) D

85) C

86) B

87) A

88) A

89) A

90) B

91) Let f(x) = 6x3 + 5x2 + 4x + 7 and let y0 = 0. f(-2) = -29 and f(-1) = 2. Since f is continuous on [-2, -1] and since y0 = 0 is

between f(-2) and f(-1), by the Intermediate Value Theorem, there exists a c in the interval (-2 , -1) with the property

that f(c) = 0. Such a c is a solution to the equation 6x3 + 5x2 + 4x + 7 = 0.

92) Let f(x) = 2x4 + 10x3 - 6x - 6 and let y0 = 0. f(-5) = 24 and f(-4) = -110. Since f is continuous on [-5, -4] and since y0 =

0 is between f(-5) and f(-4), by the Intermediate Value Theorem, there exists a c in the interval (-5, -4) with the

property that f(c) = 0. Such a c is a solution to the equation 2x4 + 10x3 - 6x - 6 = 0.

93) C

94) C

95) A

35


96) B

97) B

98)

99) B

100) C

101) C

102) D

103) A

104) D

105) D

106) B

107) B

108) C

109) A

110) C

111)

112) A

113) A

114) B

115) A

116) A

117) C

118) A

119) D

120)

Let ε > 0 be given. Choose δ = ε/5. Then 0 < x - 1 < δ implies that

(5x - 4) - 1 = 5x - 5

= 5(x - 1)

= 5 x - 1 < 5δ = ε

Thus, 0 < x - 1 < δ implies that (5x - 4) - 1 < ε

121) Let ε > 0 be given. Choose δ = ε/2. Then 0 < x - 6 < δ implies that

2x2 - 7x- 30

x - 6 - 17 =

(x - 6)(2x + 5)

x - 6 - 17

= (2x + 5) - 17 for x ≠ 6

= 2x - 12

= 2(x - 6)

= 2 x - 6 < 2δ = ε

Thus, 0 < x - 6 < δ implies that 2x2 - 7x- 30

x - 6 - 17 < ε

122) C

123) C

124) A

125) D

126) 10

127) 4

128)1

6

129) 1

130) 5

36


131) 0

132) 1

133)

134) A

135) B

136) C

137) D

138) A

37

Derivatives Worksheet 1 - Understanding the Derivative

Show all work on your paper as described in class. Video links are included throughout for instruction on how to do

the various types of problems. Important: Work the problems to match everything that was shown in the videos. For

example: Suppose a video shows 3 ways to do a problem, (such as algebraically, graphically, and numerically), then

your work should show these 3 ways also. That is , each video is a model for the work I want to see on your paper.

ESSAY.

1) A car is traveling on a straight track at a constant velocity moving away from the starting line. At time t =

0 hours the car is at point A, 5 miles from the starting line, and at time t = 2 hours the car is at point B, 25

miles away from the starting line. http://youtu.be/LKn61xIo4lsa) Express the distance of the car from the starting line, d(t), as a function of time, t.

b) Compute the average velocity of the car on its journey from point A to point B using a difference

quotient.

c) Graph d(t) and illustrate the average velocity calculation with auxiliary lines on the graph.

2) A car is traveling on a straight track at a constant velocity moving away from the starting line. At time t =

0 hours the car is at point A, 10 miles away from the starting line, and at time t = 6 hours the car is at point

B, 100 miles away from the starting line.

a) Express the distance of the car from the starting line, d(t), as a function of time, t.


quotient.


3) A toy car is traveling on a straight track at a constant velocity moving away from the starting line. At time

t = 3 seconds the car is at point A, 7 ft. from the starting line, and at time t = 6 seconds the car is at point B,

13 ft. away from the starting line.

a) Express the distance of the car from the starting line, d(t), as a function of time, t.


quotient.


4) A person is walking on a straight track such that his/her position, d(t), in feet from the starting line as a

function of time, t seconds, is given by d(t) = t2 . http://youtu.be/1z2q_dT7oQ8a) Compute the average velocity of the person over the time interval from t = 1 sec. to t = 3 sec. using a

difference quotient.

b) Graph d(t) and illustrate the average velocity calculation with auxiliary lines on the graph.

c) What constant velocity could the person have walked over the same time interval to arrive at the same

position at the end of the time interval?

1

5) A person is walking on a straight track such that his/her position, d(t), in feet from the starting line as a

function of time, t seconds, is given by d(t) = -t2 + 4t .

a) Where is the person at t = 0 sec.?

b) Where is the person at t = 4 sec.?

c) Describe the journey of the person on the track and include information about the maximum distance

from the starting line the person reaches.

d) Compute the average velocity of the person over the time interval from t = 0 sec. to t = 4 sec. using a

difference quotient.

e) Graph d(t) and illustrate the average velocity calculation with auxiliary lines on the graph.

f) What constant velocity could the person have walked over the same time interval to arrive at the same

position at the end of the time interval? Describe the journey of the person in that case.

FROM THE BOOK

6) Do section 2.1 # 1


The function s = f(t) gives the position of a body moving on a coordinate line, with s in meters and t in seconds.

NOTE: "Displacement" means the change in position (end position - start position), not the total distance

traversed while going forward or back on the way from the start to the end.

7) s = 2t2 + 4t + 2, 0 ≤ t ≤ 2

Find the body's displacement and average velocity for the given time interval.

http://youtu.be/WX5WNbR1WFcA) 20 m, 10 m/sec B) 16 m, 16 m/sec C) 16 m, 8 m/sec D) 12 m, 12 m/sec

8) s = 4t - t2, 0 ≤ t ≤ 4


A) 32 m, -4 m/sec B) 32 m, 8 m/sec C) 0 m, 0 m/sec D) -32 m, -8 m/sec

9) s = - t3 + 7t2 - 7t, 0 ≤ t ≤ 7


A) 637 m, 91 m/sec B) -49 m, -14 m/sec C) 49 m, 7 m/sec D) -49 m, -7 m/sec

ESSAY.

10) A ball is thrown upward. The height of the ball above the ground is given in the table below:

t (sec) 0 1 2 3 4 5 6

h (ft.) 6 90 142 162 150 106 30

You want to estimate the exact (instantaneous) velocity of the ball at t = 2 seconds. Use a 1 second time

interval to compute the average velocity of the ball over the time interval from t = 2 to 3 seconds, and then

use that as the estimate of the instantaneous) velocity at t = 2 seconds. http://youtu.be/Gke1L4mVARI

2


t (sec) 0 1 2 3 4 5 6

h (ft.) 6 90 142 162 150 106 30

You want to estimate the exact (instantaneous) velocity of the ball at t = 4 seconds. Use a 1 second time

interval to compute the average velocity of the ball over the time interval from t = 4 to 5 seconds, and then

use that as the estimate of the instantaneous velocity at t = 4 seconds.


t (sec) 0 1 2 3 4 5 6

h (ft.) 6 90 142 162 150 106 30

You are tired of estimation, so this time you want to find the instantaneous velocity of the ball at t = 1

seconds. Try using a 0 second time interval to compute the average velocity of the ball over the time

interval from t = 4 to 4 seconds (not a misprint!), and then use that as the answer to the instantaneous

velocity at t = 4 seconds. What is the trouble? !!

13) Go to http://en.wikipedia.org/wiki/History_of_calculus , read, and write a brief paragraph

on what Isaac Newton and Gottried Leibniz had to do with Calculus, including dates. Also, from the

Newton section, include info about Newton's idea about "x = x + o". (After your study of derivatives in

this class you should go back and read this wikipedia page again - you'll be amazed that you can actually

understand it!!)

14) Walking Man: A man is walking on a straight path with a position function d(t) where d is his postion on

the path in feet and t is time in seconds. The following video shows this motion:

Go to: http://youtu.be/rWOv5Eb7ojw This video is called "Walking Man" . Hit Full Screen on the

video.

a) Graph d(t) on the interval of 0 to 5 seconds on your own paper. ( d(t) is given on the video.)

b) Play the video and observe the motion of the man and the graph below.

c) Describe how the position of the man changes over time.

d) Describe how the velocity and acceleration change over time.

e) What is the relationship between the graphic of the man walking on the path and the graph below?

f) Playback the walk and stop it at 4 seconds. Write down the position and velocity values showing.

g) Compute d(t) for t = 4 sec. using the given d(t) function. Verify that your calculated d(4) value is the

same as the position indicated by the video.

h) Mark the cooresponding point on your d(t) graph.

15) Refering to the preceding Walking Man problem, now you want to use the d(t) function to estimate the

instantaneous velocity of the man at t = 3 sec. and see if your calculations match up with the velocity value

indicated on the video.

a) Using the d(t) function, calculate the average velocity of the man on a very small time interval from t =

3 to t = 3.01 sec. (that's a .01 second time interval).

b) Using the d(t) function, find the average velocity of the man on a very small time interval from t = 3 to t

= 3.0001 sec. (that's a .0001 second time interval !).

c) What does the video show as the instantaneous velocity at t = 3 seconds? Are you getting there with

your average velocity calculations? How could you get even better results?

3

16) Refering to the Walking Man problem, now you want to use d(t) to estimate the instantaneous velocity of

the man at t = 2 sec. and see if your calculations match up with the velocity from the video . Run the video,

stop at t = 2 sec. and see what velocity the video shows .

a) Using the d(t) function, find the average velocity of the man on a very small time interval from t = 2 to t

= 2.001 sec. (that's a .001 second time interval).

b) Using the d(t) function, find the average velocity of the man on a very small time interval from t = 2 to t

= 2.00001 sec. (that's a .00001 second time interval !).

c) What does the video show as the instantaneous velocity at t = 2 seconds? Are you getting there with

your average velocity calculations? How could you get even better results?

17) Refering to the Walking Man problem, and using your knowledge of limits:

a) Use numerical methods (calculator) to find limh→0

d(3+h)-d(3)

h , that is , find the limit of the average

velocity from t = 3 to 3+h seconds as h goes to 0 sec. This will be the true instantaneous velocity at t = 3

seconds.

b) Run the video and stop it at 3 seconds and see if velocity value shown verifies with your answer .


Use algebraic methods to find limh→0

d(3+h)-d(3)

h , that is , find the limit of the average velocity from t = 3

to 3+h seconds as h goes to 0 sec. This will be the true instantaneous velocity at t = 3 seconds.

http://youtu.be/N0_kHV0kd0Q

Note: limh→0

f(3+h)-f(3)

h , for a function f(x) in general, is called the Derivative of f(x) at the point x = 3.

And the general Point Definition of the Derivative of f(x) at the point x = a is limh→0

f(a+h)-f(a)

h . It gives

the instantaneous rate of change of the function f(x) at the point x = a. If the function models distance as a

function of time, as in the Walking Man problem, then this derivative is the instantaneous velocity at time t =

a units.


Use algebraic methods to find limh→0

d(t+h)-d(t)

h , that is , find the limit of the average velocity from t to

t+h seconds as h goes to 0 sec. This will yield an expression that will give the instantaneous velocity at

ANY time t seconds we choose!! http://youtu.be/K_sYh2kjaZQNote: This answer is the expression for the instantaneous velocity function, v(t) .

4

20) Based on your answer to the previous problem, find the following velocity values and then run the

Walking Man video to see if your calculations check out.

a) v(0), (the velocity at t = 0 sec.)

b) v(1)

c) v(2.5)

d) v(4)

e) v(5)

20f) Take notes and know for a test: The general Limit Definition of the Derivative of f at x

is limh→0

f(x+h)-f(x)

h . This limit produces a function that gives the instantaneous rate of

change of the function f at any point x (any point for which the limit exists). If the functionmodels distance as a function of time, as in the Walking Man problem, then this derivative isthe instantaneous velocity function, v(t) and gives us the velocity of an object at any time twe choose.

NOTE: The more consise notation for the derivative of f at x, or the derivative of f(x), is f'(x)and is read "f prime of x". Thus we have:

f '(x) = the derivative of f(x) = rate of change of f at x = limh→0

f(x+h)-f(x)

h . See defintions in

the book sections 2.1 to 2.3 Know these defintions for a test!

Solve the problem.

21) Find f'(x) using the limit definition of the derivative given f(x) = x2 + x + 1 .

http://youtu.be/m1JtDxOpCmk

22) Find g'(z) using the limit definition of the derivative given g(z) = z2 - 5z + 7, .

23) Find f'(x) using the limit definition of the derivative given f(x) = 3x2 -4x + 1 .

24) Find f'(x) using the limit definition of the derivative given f(x) = x3 + 2x

25) Use the limit definition of the derivative to find f'(x) given f(x) = 3x + 2 .

26) Use the limit definition of the derivative to find g'(x) given g(x) = 4x3 - 1 .

27) A ball thrown vertically upward at time t = 0 (seconds) has height y(t) = 96t - 16t2 (ft) at time t.

a) Graph y(t)

b) Find the velocity function, v(t) , using the limit definition of the derivative.

c) What is the velocity of the ball at t = 2 sec.?

d) What is the velocity of the ball at t = 4 sec. ? What does the negative sign mean?

e) When is the velocity function equal to 0? What physical events are occuring at each of the times when

the velocity is 0?

http://youtu.be/nNvPsEqUDhI

5

28) A ball thrown vertically upward at time t = 0 (seconds) has height h(t) = 64t - 16t2 (ft) at time t. The

velocity function is v(t) = 64 - 32t

a) Graph h(t)

b) Use 2nd calc max to find the time, tmax , when the ball reaches its maximum height, and illustrate the

point ( tmax , hmax) on your graph.

c) Using the velocity function, v(t) , evaluate v( tmax) .

d) Do you think the ball indeed comes to a stop instantaneously before coming back down? Why?

29) Walking Man 2: A man is walking on a straight path with a position function d(t) where d is his postion

on the path in feet and t is time in seconds. The following video shows this motion:

Go to: http://youtu.be/dAJE_cbXQWg This video is called "Walking Man 2" (it has a different d(t)

function than Walking Man did . Hit Full Screen on the video.

a) Graph d(t) on the interval of 0 to 5 seconds on your own paper. ( d(t) is given on the video.)

b) Play the video and observe the motion of the man and the graph below it.

c) Describe how the position of the man changes over time.

d) Describe how the velocity and acceleration change over time, particularly, when are they positive and

when are they negative and why?

e) What is the relationship between the graphic of the man walking on the path and the graph below?

f) Playback the walk and stop it at 2 seconds. Write down the velocity value showing. What special is

happening at t = 2 seconds? at 4 seconds? Illustrate these points on your graph.

g) What special is happening at t = 3 seconds? Illustrate that point on your graph.

h) Mark the cooresponding point on your d(t) graph.

NOTE: We know the Derivative of a function f(x) is the (instantaneous) rate of change of f at x. There is

another interpretation: The Derivative of f(x) gives the slope of the tangent line to the graphof f(x) at the point x. The following problems help you understand this concept.

30) A ball thrown vertically upward at time t = 0 (seconds) has height f(t) = 96t - 16t2 (ft) at time t.

a) Go to http://youtu.be/BpZnYjZxWG4 . Hit Full Screen on the video. For best quality select 720p.

Watch the video several times and observe the graph and what the graph is showing.

b) Below is a picture of the video at the beginning, when the time increment is h = 2 sec. Use the

information on the picture to find the equation of the secant line shown and the equation of the tangent

line.

c) Watch the video live and pause the video when h = 1 sec. and find the equation of the secant line shown.

d) Watch the video live and pause the video when h = .2 sec. and find the equation of the secant line

shown.

e) Describe the relationship between the average rate of change of the function, the secant line, the

derivative of f(t), and the tangent line.

6

NOTE: The slope of the tangent line to the graph of a function f(x) at a point x0 is the

derivative of f(x) at x0 , f'(x0 ) .

7


Find the slope of the line tangent to the graph at the given point. Use the limit definition of the derivative, or

numerical methods (small h = .001), to find the slope.

31) y = 8x + 7, x = -2

A) m = -16 B) m = -8 C) m = 16 D) m = 8

32) y = x2 + 6x - 8, x = -2 http://youtu.be/k2e0CU8fcTwA) m = -4 B) m = 2 C) m = -6 D) m = -8

Find an equation for the tangent line to the curve at the given point. AND graph the curve and the tangent line on

your TI and copy to your paper. It better look exactly tangent or something is wrong!

33) y = x2 - 2, (-3, 7) Hint: To save yourself some time, use y' = 2x

A) y = -3x - 11 B) y = -6x - 11 C) y = -6x - 22 D) y = -6x - 20

34) h(x) = t3 - 16t + 3, (4, 3) Hint: To save yourself some time, use h'(x) = 3t2 - 16

A) y = 3 B) y = 35t - 125 C) y = 32t - 125 D) y = 32t + 3

ESSAY.

35) For the graph of f(x) below, estimate f'(x) at the given x values by using the grids and the fact that f'(x) =

slope of the tangent line at the point x. http://youtu.be/jj3hwpe85Z8

x-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11

y5

4

3

2

1

-1

-2

-3

-4

-5

x-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11

y5

4

3

2

1

-1

-2

-3

-4

-5

a) x = -2

b) x = -1

c) x = -0.4

d) x = 0

e) x = 1

f) x = 2

g) x = 3.5

h) x = 5

8


Estimate the slope of the curve at the indicated point.

36) There are no grids for these so you just pick the only value that is possible among the choices.

A) 1 B) Undefined C) 0 D) -1

37)

A) 1 B) 0 C) -1 D) Undefined

38)

A) Undefined B) 1 C) -1 D) 0

39)

A) 1 B) Undefined C) 0 D) -1

40)

A) Undefined B) 0 C) 1 D) -1

9

41)

A) -2 B)1

2C) 2 D) -

1

2

42)

A) - 1

2B) -2 C) 2 D)

1

2

43)

A)1

20B) -2 C) -

1

20D) 2

FROM THE BOOK

44) Do section 2.1 #3, 5, 7, 13, 23; section 2.2 #1, 3, 5, 7, 9, 11, 15, 19; For 25 and 27 estimate the derivative with

numerical methods (small h = .001); 39, 43, 45, 47


Solve the problem.

10

45) The graph of y = f(x) in the accompanying figure is made of line segments joined end to end. Graph the

derivative of f. http://youtu.be/Rb4-iTE3ays

x

y

(-5, 0)

(-3, 2)

(0, -1)

(3, 5) (6, 5)

x

y

(-5, 0)

(-3, 2)

(0, -1)

(3, 5) (6, 5)

x

y

x

y

A)

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

B)

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

C)

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

D)

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

11

46) Use the following information to graph the function f over the closed interval [-5, 6].

i) The graph of f is made of closed line segments joined end to end.

ii) The graph starts at the point (-5, 1).

iii) The derivative of f is the step function in the figure shown here.

So the step function shown is the derivative of f(x), you draw the graph of the original f(x)

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y

6

4

2

-2

-4

-6

x

y

x

y

A)

x

y

(-5, 1)

(-3, 5)

(0, 2)

(3, 5)

(6, -1)

x

y

(-5, 1)

(-3, 5)

(0, 2)

(3, 5)

(6, -1)

B)

x

y

(-5, 1)

(-3, 6)

(0, 2)

(3, 5)

(6, 0) x

y

(-5, 1)

(-3, 6)

(0, 2)

(3, 5)

(6, 0)

C)

x

y

(-5, 1)

(-3, 5)

(0, 1)

(3, 5)

(6, -1)

x

y

(-5, 1)

(-3, 5)

(0, 1)

(3, 5)

(6, -1)

D)

x

y

(-5, 1)

(-3, 5)

(0, 2)

(3, 6)

(6, 0) x

y

(-5, 1)

(-3, 5)

(0, 2)

(3, 6)

(6, 0)

12

47) Match the graph of the function to the graph of its derivative. http://youtu.be/TnYLrTMqG2Y

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

13

48) Match the graph of the function with the graph of its derivative.

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

14

49) Match the graph of the function to the graph of its derivative.

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

15

The graph of a function is given. Choose the answer that represents the graph of its derivative.

50)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

A)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

B)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

C)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

D)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

16

51)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

A)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

B)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

C)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

D)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

17

52)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

A)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

B)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

C)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

D)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

18

53)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

A)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

B)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

C)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

D)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

19

54)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

A)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

B)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

C)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

D)

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

x-15 -10 -5 5 10 15

y15

10

5

-5

-10

-15

FROM THE BOOK

55) Do section 2.3 #1-11 odd, 17 - 35 odd, 41 http://youtu.be/QDHwSUCYwO0

ESSAY.

56) Read and take notes on "An Alternate Notation for the Derivative" in section 2.4 in the book. It

explains that dy

dx is another notation for f '(x), the derivative.

FROM THE BOOK

57) Do section 2.4 #1-5 odd, 11, 15, 21

20

ESSAY.

58) Go to http://youtu.be/fE5EXbNxLEQ and take notes on the "2nd Derivative" video and

pay attention to the relationship between f" and the concavity of f and inflection points.

FROM THE BOOK

59) Read section 2.5 and take notes

60) Do section 2.5 #1-13 odd, 17- 21 odd, 27, 29. Video for 11: http://youtu.be/zZXW8T8EKksVideo for 21: http://youtu.be/V_GkVXyLXD4

21


Solve the problem.

61) Select an appropriate graph of a twice-differentiable function y = f(x) that passes through the points

(- 2,1) , - 6

3,5

9, (0,0),

6

3,5

9 and ( 2,1), and whose first two derivatives have the following sign patterns.

y′ : + - + -

- 2 0 2

y′′ :

- + -

- 6

3

6

3

http://youtu.be/bJWayQoZFZEA)

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

B)

x-3 -2 -1 1 2 3

y2

1.5

1

0.5

-0.5

-1

-1.5

-2

x-3 -2 -1 1 2 3

y2

1.5

1

0.5

-0.5

-1

-1.5

-2

C)

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

D)

x-3 -2 -1 1 2 3

y4

3

2

1

-1

-2

-3

-4

x-3 -2 -1 1 2 3

y4

3

2

1

-1

-2

-3

-4

ESSAY.

62) Sketch a continuous curve y = f(x) with the following properties:

f(2) = 3; f′′(x) > 0 for x > 4; and f′′(x) < 0 for x < 4 .


63) The graph below shows the first derivative of a function y = f(x). Select a possible graph of f that passes

through the point P. http://youtu.be/dIbjDzq3WWU

f′

22

x

y

P

x

y

P

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

23

64) The graph below shows the first derivative of a function y = f(x). Select a possible graph f that passes

through the point P.

f′

x

y

P

x

y

P

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

24

65) The graph below shows the first derivative of a function y = f(x). Select a possible graph f that passes

through the point P.

f′

x

y

P

x

y

P

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

25

66) Using the following properties of a twice-differentiable function y = f(x), select a possible graph of f.

x y Derivatives

x < 2 y′ > 0,y′′ < 0

-2 11 y′ = 0,y′′ < 0

-2 < x < 0 y′ < 0,y′′ < 0

0 -5 y′ < 0,y′′ = 0

0 < x < 2 y′ < 0,y′′ > 0

2 -21 y′ = 0,y′′ > 0

x > 2 y′ > 0,y′′ > 0

A)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

B)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

C)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

D)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

FROM THE BOOK

67) Read section 2.6 Differentiability and take notes. Know the proof of "A differentiablefunction is continuous" for a test.

68) Do section 2.6 #1-9 odd. Video for #5: http://youtu.be/csRj7wNmXzQ

26


Given the graph of f, find any values of x at which f ′ is not defined.

69)

A) x = 0 B) x = 1 C) x = -1 D) x = 2

70)

A) x = -3, 0, 3 B) x = -2, 2 C) x = -3, 3 D) x = -2, 0, 2

71)

A) x = 2 B) x = -2, 0, 2 C) x = 0 D) x = -2, 2

72)

A) x = 1 B) x = 2 C) x = 0 D) x = 0, 1, 2

73)

A) x = 2 B) x = 1, 2, 3

C) x = 1, 3 D) Defined for all values of x

27

74)

A) x = -1, 0, 1 B) x = 0

C) x = -1, 1 D) Defined for all values of x

75)

A) x = 0 B) x = -2, 2

C) x = -2, 0, 2 D) Defined for all values of x

76)

A) x = 3 B) x = 0, 3

C) x = 0 D) Defined for all values of x

NOTE: Now we get some rules that give us short-cuts to differentiation ! We learn how to find the

derivative of a constant, a constant multiple, sum and difference, powers, and putting it all together,

polynomials. (Other functions come later.) Using these short-cuts without understanding what the

derivative really is is just useless rote memorization - That's why we spent all this time on understanding the

derivative. So now that you understand derivatives you won't have to use the limit definition of the

derivative to find derivative functions from now on! Well, if you run into a function displayed as data (as is

frequently the case in the real world), then you can always fall back on your understanding to solve the

problem. This way you have every situation covered.

FROM THE BOOK

77) Go to http://youtu.be/QVNz5LVNKjA and take notes on the this video about the

derivative of a constant, a constant multiple, sum and difference, and powers. Know fora test. Then read section 3.1 and take notes from the book. AND know (for a test) theproofs of Theorem 3.1, 3.2, and the Power rule (also explained in the video). I want youto know why these short-cut rules work!!

28


Find the derivative using the short-cut rules.

78) y = 14 - 14x2 http://youtu.be/nl0NGmZHd5gA) 14 - 28x B) 14 - 14x C) -28x D) -28

Find the derivative.

79) y = 5 - 4x3

A) -12x B) -8x2 C) 5 - 12x2 D) -12x2

80) y = 2x4 - 8x3 + 9 http://youtu.be/3cbiT9J_-3IA) 8x3 - 24x2 - 7 B) 4x3 + 3x2 C) 8x3 - 24x2 D) 4x3 + 3x2 - 7

81) s = 3t2 + 4t + 1

A) 3t2 + 4 B) 6t2 + 4 C) 3t + 4 D) 6t + 4

82) y = 7x-2 + 14x3 + 10x

A) -14x-3 + 42x2 + 10 B) -14x-1 + 42x2

C) -14x-3 + 42x2 D) -14x-1 + 42x2 + 10

83) y = 5x2 + 12x + 4x-3

A) 5x + 4x-4 B) 10x + 12 + 12x-4 C) 10x + 12 - 12x-4 D) 10x - 12x-4

84) w = z-5 - 1

z http://youtu.be/3cbiT9J_-3I

A) -5z-6 - 1

z2B) z-6 +

1

z2C) 5z-6 -

1

z2D) -5z-6 +

1

z2

85) r = 4

s3 -

5

s

A)12

s4 -

5

s2B)

4

s4 -

5

s2C) -

12

s2 +

5

s2D) -

12

s4 +

5

s2

86) y = 1

7x2 +

1

5x

A) - 2

7x3 -

1

5x2B) -

1

7x3 +

1

5x2C) -

2

7x -

1

5x2D)

2

7x3 +

1

5x2

Find the second derivative. Graph the derivative and graph the 2nd derivative. Illustrate that the 2nd derivative is

giving you the slope values of the derivative function.

87) y = 6x2 + 3x - 4 http://youtu.be/zXkPGvHEJKQA) 12 B) 6 C) 0 D) 12x + 3

29

Find the second derivative.

88) y = 7x4 - 4x2 + 8

A) 28x2 - 8x B) 28x2 - 8 C) 84x2 - 8x D) 84x2 - 8

89) s = 2t3

3 + 2

A) 4t + 2 B) 2t2 C) 4t D) 2t

90) y = 5x2 + 11x + 5x-3

A) 10 + 60x-1 B) 10 - 60x-5 C) 10 + 60x-5 D) 10x + 11 - 15x-4

91) r = 3

s3 -

5

s

A)3

s5 -

5

s3B)

36

s5 -

10

s3C) -

9

s4 +

5

s2D)

36

s5 +

10

s3

FROM THE BOOK

92) Do section 3.1 #1-47 odd ; video of selected problems: http://youtu.be/cctPUbcBGNk


Solve the problem.

93) The position of a body moving on a coordinate line is given by s = t2 - 6t + 7, with s in meters and t in

seconds. When, if ever, during the interval 0 ≤ t ≤ 6 does the body change direction?

http://youtu.be/E0TsxC0htLwA) t = 6 sec B) t = 12 sec

C) t = 3 sec D) no change in direction

****Videos will have been added after this point to the online

version of these notes by the time you get here!****

94) At time t, the position of a body moving along the s-axis is s = t3 - 15t2 + 72t m. Find the body's

acceleration each time the velocity is zero.

A) a(6) = 6 m/sec2, a(4) = -6 m/sec2 B) a(12) = 72 m/sec2, a(8) = 12 m/sec2

C) a(6) = 0 m/sec2, a(4) = 0 m/sec2 D) a(6) = -6 m/sec2, a(4) = 6 m/sec2

95) At time t ≥ 0, the velocity of a body moving along the s-axis is v = t2 - 9t + 8. When is the body moving

backward?

A) 1 < t < 8 B) 0 ≤ t < 1 C) t > 8 D) 0 ≤ t < 8

30

96) At time t ≥ 0, the velocity of a body moving along the s-axis is v = t2 - 11t + 10. When is the body's

velocity increasing?

A) t < 5.5 B) t < 10 C) t > 5.5 D) t > 10

97) A ball dropped from the top of a building has a height of s = 256 - 16t2 meters after t seconds. How long

does it take the ball to reach the ground? What is the ball's velocity at the moment of impact?

A) 16 sec, -512 m/sec B) 4 sec, 128 m/sec C) 8 sec, -64 m/sec D) 4 sec, -128 m/sec

98) A rock is thrown vertically upward from the surface of an airless planet. It reaches a height of

s = 120t - 2t2 meters in t seconds. How high does the rock go? How long does it take the rock to reach its

highest point?

A) 3570 m, 30 sec B) 3600 m, 60 sec C) 1800 m, 30 sec D) 7080 m, 60 sec

99) The area A = πr2 of a circular oil spill changes with the radius. At what rate does the area change with

respect to the radius when r = 9 ft?

A) 18 ft2/ft B) 18π ft2/ft C) 81π ft2/ft D) 9π ft2/ft

100) The driver of a car traveling at 60 ft/sec suddenly applies the brakes. The position of the car is s = 60t - 3t2,

t seconds after the driver applies the brakes. How far does the car go before coming to a stop?

A) 600 ft B) 10 ft C) 1200 ft D) 300 ft

101) The size of a population of mice after t months is P = 100(1 + 0.2t + 0.02t2). Find the growth rate at t = 13

months.

A) 172 mice/month B) 72 mice/month C) 144 mice/month D) 36 mice/month

102) The number of gallons of water in a swimming pool t minutes after the pool has started to drain is

Q(t) = 50(20 - t)2. How fast is the water running out at the end of 11 minutes?

A) 900 gal/min B) 450 gal/min C) 2025 gal/min D) 4050 gal/min

FROM THE BOOK

103) Do section 3.1 # 55 (graph the function and the tangent line together - see that your tangent line is the

perfect fit to the curve), 57, 65, 67, 69, 71

NOTE: Now we learn how to find the derivative of exponential functions, f(x) = ax

104) Read section 3.2 "Derivatives of Exponential Functions and the Number e" ; "A formula

for the Dervative of ax ", and take notes.

105) Do section 3.2 # 1-25 odd, 39, 41, 43, 45

NOTE: Now we learn short-cut rules for finding the derivative of the product and quotient

of two functions.

31

106) Go to http://math.ucsd.edu/~wgarner/math20a/prodrule.htm and take notes on the

derivation of the Product Rule (know all the steps for a test!) Note: The way the rule looksin the book is a bit different but it is equivalent. We'll use the rule as stated on this site.

107) Go to http://math.ucsd.edu/~wgarner/math20a/quotrule.htm and take notes on the

derivation of the Quotient Rule (know all the steps for a test!) Note: The way the rulelooks in the book is a bit different but it is equivalent. We'll use the rule as stated on thissite.


Find y ′.

108) y = (2x - 3)(2x + 1)

A) 8x - 8 B) 8x - 2 C) 4x - 4 D) 8x - 4

109) y = (3x - 3)(5x3 - x2 + 1)

A) 15x3 + 18x2 - 54x + 3 B) 60x3 - 18x2 + 54x + 3

C) 45x3 + 54x2 - 18x + 3 D) 60x3 - 54x2 + 6x + 3

110) y = (x2 - 3x + 2)(2x3 - x2 + 5)

A) 10x4 - 28x3 + 21x2 + 6x - 15 B) 2x4 - 24x3 + 21x2 + 6x - 15

C) 10x4 - 24x3 + 21x2 + 6x - 15 D) 2x4 - 28x3 + 21x2 + 6x - 15

111) y = 1

x2 + 7 x2 -

1

x2 + 7

A)4

x5 + 14x B) -

4

x5 - 14x C) -

1

x5 + 14x D)

4

x3 + 14x

Find the derivative of the function.

112) y = x2 - 3x + 2

x7 - 2

A) y ′ = -5x8 + 18x7 - 14x6 - 3x + 6

(x7 - 2)2B) y ′ =

-5x8 + 19x7 - 14x6 - 4x + 6

(x7 - 2)2

C) y ′ = -5x8 + 18x7 - 13x6 - 4x + 6

(x7 - 2)2D) y ′ =

-5x8 + 18x7 - 14x6 - 4x + 6

(x7 - 2)2

113) y = x3

x - 1

A) y ′ = -2x3 + 3x2

(x - 1)2B) y ′ =

2x3 + 3x2

(x - 1)2C) y ′ =

2x3 - 3x2

(x - 1)2D) y ′ =

-2x3 - 3x2

(x - 1)2

32

114) g(x) = x2 + 5

x2 + 6x

A) g ′(x) = 2x3 - 5x2 - 30x

x2(x + 6)2B) g ′(x) =

4x3 + 18x2 + 10x + 30

x2(x + 6)2

C) g ′(x) = 6x2 - 10x - 30

x2(x + 6)2D) g ′(x) =

x4 + 6x3 + 5x2 + 30x

x2(x + 6)2

115) y = x2 + 8x + 3

x

A) y ′ = 3x2 + 8x - 3

xB) y ′ =

3x2 + 8x - 3

2x3/2C) y ′ =

2x + 8

xD) y ′ =

2x + 8

2x3/2

116) y = x2 + 2x - 2

x2 - 2x + 2

A) y ′ = 4x2 + 8x

(x2 - 2x + 2)2B) y ′ =

4x2 - 8x

(x2 - 2x + 2)2C) y ′ =

-4x2 + 8x

(x2 - 2x + 2)2D) y ′ =

-4x2 - 8x

(x2 - 2x + 2)2

117) r = θ - 4

θ + 4

A) r ′ = 4

θ(θ + 4)2B) r ′ =

4

θ + 4

C) r ′ = 8

(θ + 4) θ2 - 16D) r ′ = -

4

θ(θ + 4)2

118) z = 3x2ex

A)dz

dx = 6xex + 3x2ex B)

dz

dx = 3xex + 6x2ex C)

dz

dx = 3xex + 3x2ex D)

dz

dx = 6xex - 3x2ex

FROM THE BOOK

119) Do section 3.3 #1-33 odd, 41, 45, 47

NOTE: Now we learn the Chain Rule for finding the derivative of the Composition of twofunctions. First a review of Composition.


Find the requested composition or operation.

120) f(x) = 3x + 13, g(x) = 4x - 1

Find (f ∘ g)(x).

A) 12x + 16 B) 12x + 51 C) 12x + 10 D) 12x + 12

33

121) f(x) = x + 7, g(x) = 8x - 11

Find (f ∘ g)(x).

A) 8 x + 7 - 11 B) 2 2x + 1 C) 8 x - 4 D) 2 2x - 1

122) f(x) = 4x2 + 2x + 8, g(x) = 2x - 5

Find (g ∘ f)(x).

A) 4x2 + 4x + 11 B) 8x2 + 4x + 21 C) 8x2 + 4x + 11 D) 4x2 + 2x + 3

123) f(x) = x - 3

7, g(x) = 7x + 3

Find (g ∘ f)(x).

A) x B) 7x + 18 C) x - 3

7D) x + 6

Perform the requested composition or operation.

124) Find (f ∘ g)(8) when f(x) = -4x - 5 and g(x) = -4x2 + 5x + 3.

A) -34 B) -5658 C) -49 D) 847

125) Find (g ∘ f)(4) when f(x) = 9x - 6 and g(x) = -8x2 - 2x + 9.

A) -1149 B) -7251 C) -291 D) -285

Use the graphs to evaluate the expression.

126) (g ∘ f)(-3)

y = f(x) y = g(x)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) 1.5 B) 0.5 C) -2.5 D) -1

34

127) (f ∘ g)(-2)

y = f(x) y = g(x)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) -2 B) -1.5 C) 0 D) 1

128) (f ∘ g)(0)

y = f(x) y = g(x)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, -1)

(2, 2)

(0, -2)

(-1, -1)

(-2, 2)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, -1)

(2, 2)

(0, -2)

(-1, -1)

(-2, 2)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, 2)

(2, 4)

(-1, -2)

(-2, -4)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, 2)

(2, 4)

(-1, -2)

(-2, -4)

A) -1 B) 1 C) -2 D) -3

35

129) (g ∘ f)(-1)

y = f(x) y = g(x)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, -1)

(2, 2)

(0, -2)

(-1, -1)

(-2, 2)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, -1)

(2, 2)

(0, -2)

(-1, -1)

(-2, 2)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, 2)

(2, 4)

(-1, -2)

(-2, -4)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(1, 2)

(2, 4)

(-1, -2)

(-2, -4)

A) -2 B) -3 C) -1 D) -4

130) g(f(4))

y = f(x) y = g(x)

x-5 5

y5

-5

x-5 5

y5

-5

x-5 5

y5

-5

x-5 5

y5

-5

A) -2 B) 8 C) 4 D) 6

Use the tables to evaluate the expression if possible.

131) Find (g ∘ f)(11).

x 7 9 11

f(x) 49 81 121

x 81 121 64 49 100

g(x) 67 107 50 35 86

A) 89 B) 38 C) 67 D) 107

132) Find (f ∘ g)(3).

x 11 7 3 5

f(x) 22 14 6 10

x 5 3 6 4

g(x) 9 5 11 7

A) 10 B) 3 C) 14 D) 5

36

133) Find (g ∘ f)(10).

x 10 13 11 22

f(x) 11 20 55 57

x 12 22 10 11

g(x) 23 19 22 21

A) 55 B) 21 C) 10 D) 19

134) Find (f ∘ f)(8).

x 8 11 9 7

f(x) 9 8 43 45

x 10 8 11 9

g(x) 19 15 21 17

A) 43 B) 8 C) 17 D) 21

135) Find (g ∘ g)(3).

x 3 6 4 8

f(x) 4 6 13 15

x 5 8 3 4

g(x) 9 5 8 7

A) 7 B) 13 C) 15 D) 5

ESSAY.

136) Go to http://math.ucsd.edu/~wgarner/math20a/chainrule.htm and take notes on the

derivation of the Chain Rule (know all the steps for a test!)

FROM THE BOOK

137) Read section 3.4 Chain Rule and take notes

138) Do section 3.4 #1-43 odd, 59, 79

NOTE: Now we'll learn about the derivatives of Trigonometric functions

37

ESSAY.

139) Consider the graph of f(x) = sin(x) below.

x-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11

y5

4

3

2

1

-1

-2

-3

-4

-5

x-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11

y5

4

3

2

1

-1

-2

-3

-4

-5

a) Make a table of estimated slope values of the graph of f(x) for x values incrementing by 0.5 on the

interval from x = 0 to 6.5

b) Graph the derivative of f(x) using your data from (a). What classic trig function does it look like?

140) Another way to estimate the graph of the derivative of f(x) = sin(x) (or any function) is to graph the

difference quotient with h = .001, or other small h value, in your TI as follows:

y1 = (f(x+.001)-f(x))/.001 . So in the case of f(x) = sin(x) that would be: y1= (sin(x+.001)-sin(x))/.001

Use this method to graph the derivative of f(x) = sin(x) and report which trig function results.

141) Here is a limit that is critical to the proof that the derivative of sin(x) is cos(x).

Consider limh→0

sin(h)

h

a) Find the limit by graphical methods, that is, graph y1= sin(x)/x (remember, your calculator must be in

radian mode, not degrees) on x: -1 to 1 ; y: 0 to 2 and copy to your paper. Use 2nd calc value and put in x

= .001 . What is your estimate for the limit using this "h" value of .001?

b) Evaluate using smaller h values. What do you conclude the limit is?

142) Here is another limit that is critical to the proof that the derivative of sin(x) is cos(x).

Consider limh→0

1-cos(h)

h

a) Find the limit by graphical methods, that is, graph y1= (1-cos(x))/x (remember, your calculator must

be in radian mode, not degrees) on x: -1 to 1 ; y: -1 to 1 and copy to your paper. Use 2nd calc value and

put in x = .001 . What is your estimate for the limit using this "h" value of .001?

b) Evaluate using smaller h values. What do you conclude the limit is?

143) Go to http://www-math.mit.edu/~djk/18_01/chapter05/proof02.html and copy down the proof that the

derivative of sin(x) is cos(x), and know the proof for a test.

38

FROM THE BOOK

144) Find the informal justification of d

dx(sin(x)) = cos(x) in section 3.5, write it down with an illustration and

know for a test.

145) Find the informal justification of d

dx(cos(x)) = -sin(x) in section 3.5, write it down and know for a test.

ESSAY.

146) Estimate the graph of the derivative of f(x) = cos(x) using the difference quotient with h = .001 in your TI

as follows: y1= (cos(x+.001)-cos(x))/.001

Use this method to graph the derivative of f(x) = cos(x) and report which trig function results.

FROM THE BOOK

147) Find the proof of d

dx(tan(x)) =

1

cos2 x in section 3.5, write it down and know for a test.

ESSAY.

148) Estimate the graph of the derivative of f(x) = tan(x) using the difference quotient with h = .001 in your TI

as follows: y1= (tan(x+.001)-tan(x))/.001 Then put in y2 = 1/cos(x)^2 . What do you notice?

FROM THE BOOK

149) Do section 3.5 #3 - 41 odd, 45, 47, 49

ESSAY.

Solve the problem.

150) A rocket is launched vertically and is tracked by a radar station located on the ground 6 mi from the

launch pad. Suppose that the elevation angle θ of the line of sight to the rocket is increasing at 5° per

second when θ = 60°. What is the velocity of the rocket at this instant?

151) Write an equation of the line that is tangent to the curve y = x sin x at the point P with x-coordinate π

2 .

FROM THE BOOK

152) In section 3.6, copy down and know the proofs of

a) d

dx(x

1

2 ) = 1

2x

-1

2 b) d

dx(ln(x)) =

1

x c)

d

dx(ax) = ln(a)ax

d) d

dx(arctan(x)) =

1

1+x2 e)

d

dx(arcsin(x)) =

1

1-x2

39

153) Do section 3.6 #1-33 odd, 43

154) Read section 3.7 Implicit Functions and take notes

ESSAY.

155) Go to http://www.flashandmath.com/mathlets/calc/implicit/implicit.html Enter in the

implicit equation x2 + y2 = 4 into the applet (you would type in x^2+y^2=4 according to the info on the

site on how to enter syntax).

a) Copy the graph to your paper.

b) Use implicit differentiation to find dy

dx

c) Use dy

dx to find the slope of the curve at the point ( 2 , 2 )

d) Find the equation of the tangent line to the curve at ( 2 , 2 ) and graph the line on your graph.


Use implicit differentiation to find dy/dx.

156) 2xy - y2 = 1

A)y

x - yB)

y

y - xC)

x

y - xD)

x

x - y

157) x3 + 3x2y + y3 = 8

A) - x2 + 3xy

x2 + y2B) -

x2 + 2xy

x2 + y2C)

x2 + 2xy

x2 + y2D)

x2 + 3xy

x2 + y2

158) y x + 1 = 4

A)y

2(x + 1)B) -

y

2(x + 1)C) -

2y

x + 1D)

2y

x + 1

159) xy + x + y = x2y2

A)2xy2 + y + 1

-2x2y - x - 1B)

2xy2 - y

2x2y + xC)

2xy2 + y

2x2y - xD)

2xy2 - y - 1

-2x2y + x + 1

160) cos xy + x7 = y7

A)7x6 - x sin xy

7y6B)

7x6 + y sin xy

7y6 - x sin xyC)

7x6 - y sin xy

7y6 + x sin xyD)

7x6 + x sin xy

7y6

40

At the given point, find the slope of the curve or the line that is tangent to the curve, as requested. Then Go tohttp://www.flashandmath.com/mathlets/calc/implicit/implicit.html , enter in the implicit equation, copy

the graph to your paper and illustrate the slope or tangent line on your graph. (For a multiplication you have to type

for example: 9*x instead of 9x)

161) y6 + x3 = y2 + 9x, slope at (0, 1)

A) - 3

2B)

3

2C)

9

8D)

9

4

162) x6y6 = 64, slope at (2, 1)

A) -32 B) - 1

2C) 2 D) -

1

4

163) 5x2y - π cos y = 6π, slope at (1, π)

A) - π

2B) π C) 0 D) -2π

164) 6x2y - π cos y = 7π, tangent at (1, π)

A) y = - π

2x +

3π

2B) y = -2πx + π C) y = πx D) y = -2πx + 3π

165) y4 + x3 = y2 + 12x, tangent at (0, 1)

A) y = 6x + 1 B) y = - 3x - 1 C) y = - 2x D) y = 3x + 1

FROM THE BOOK

166) Do section 3.7 #1-17 odd, 19-25 ALL (for 19-25 go tohttp://www.flashandmath.com/mathlets/calc/implicit/implicit.html enter in the implicit

equation, copy the graph to your paper and illustrate the slope or tangent line requested on your

graph.)

Note: Local Linearity - Tangent Line Approximation. When you zoom in on a point

on the graph of f(x) where f(x) is differentiable, the graph appears very much straight like a line the closer

you zoom in - Local Linearity. This is because the requirement of the derivative existing at the point

prohibits that jagged edge look from occuring there, in other words, the graph is smooth at the point. Any

time you zoom in on something smoothly curved it starts looking more straight. Thus, using a tangent line at

the point to approximate values of the function near the point will be pretty good. That's what these next

problem focus on.

ESSAY.

Find the linearization L(x) (the tangent line) of f(x) at x = a AND evaluate the difference between L and f at a point

nearby. That is, find L(a + 0.1) - f(a + 0.1) to see how far off the linear approximation is from the actual function

value if we move 0.1 away to the right. Graph f(x) and L(x) to verify if your difference result makes sense.

167) f(x) = 4x2 - 4x + 4, a = -5

168) f(x) = 8x + 36, a = 0

41

169) f(x) = sin x, a = 0

170) f(x) = tan x, a = π

FROM THE BOOK

171) Do section 3.9 #1 - 7 odd, 11, 13, 19, 25

Note: Four Important Theorems: Mean Value Theorem; Rolle's Theorem; IncreasingFunction Theorem; Constant Function Theorem

172) In Section 3.10: Read and take notes about The Mean Value Theorem

ESSAY.

173) Go to http://math.ucsd.edu/~wgarner/math20a/mvt.htm and take notes on the proofs of the

Mean Value Theorem and Rolle's Theorem. Know the proofs for a test.


Find the value or values of c that satisfy the equation f(b) - f(a)

b - a = f′(c) in the conclusion of the Mean Value Theorem

for the function and interval.

174) f(x) = x2 + 3x + 4, [-3, 2]

A) -3, 2 B) 0, - 1

2C) -

1

2D) -

1

2,

1

2

175) f(x) = x + 112

x, [7, 16]

A) -4 7, 4 7 B) 4 7 C) 7, 16 D) 0, 4 7

176) f(x) = ln (x - 3), [4, 6] Round to the nearest thousandth.

A) ±4.820 B) 4.820 C) 5.820 D) 5.885

Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval.

177) f(x) = x1/3, -4,2

A) Yes B) No

178) g(x) = x3/4, 0,5

A) Yes B) No

179) s(t) = t(3 - t), -1,5

A) Yes B) No

42

ESSAY.

Answer the question.

180) A trucker handed in a ticket at a toll booth showing that in 2 hours he had covered 230 miles on a toll road

with speed limit 65 mph. The trucker was cited for speeding. Why?

181) A marathoner ran the 26.2 mile New York City Marathon in 2.7 hrs. Did the runner ever exceed a speed of

9 miles per hour?


182) The function f(x) = -5x 0 ≤ x < 1

0 x = 1is zero at x = 0 and x = 1 and differentiable on (0, 1), but its derivative on

(0,1) is never zero. Does this example contradict Rolle's Theorem?

183) Decide if the statement is true or false. If false, explain.

The points (-1, -1) and (1, 1) lie on the graph of f(x) = 1

x. Therefore, the Mean Value Theorem says that

there exists some value x = c on (-1, 1) for which f′(x) = 1 - (-1)

1 - (-1) = 1.

FROM THE BOOK

184) In Section 3.10 read and take notes on The Increasing Function Theorem and The Constant Function

Theorem. Know the proofs for a test.


Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing.

185) f′(x) = (3 - x)(4 - x)

A) Decreasing on (-∞, 3); increasing on (4, ∞)

B) Decreasing on (3, 4); increasing on (-∞, 3) ∪ (4, ∞)

C) Decreasing on (-∞, -3) ∪ (-4, ∞); increasing on (-3, -4)

D) Decreasing on (-∞, 3) ∪ (4, ∞); increasing on (3, 4)

186) f′(x) = x1/3(x - 6)

A) Decreasing on (0, 6); increasing on (6, ∞)

B) Decreasing on (-∞, 0) ∪ (6, ∞); increasing on (0, 6)

C) Decreasing on (0, 6); increasing on (-∞, 0) ∪ (6, ∞)

D) Increasing on (0, ∞)

187) f′(x) = (x - 1) e-x

A) Increasing on (-∞, -1); decreasing on (-1, ∞) B) Decreasing on (-∞, 1); increasing on (1, ∞)

C) Decreasing on (-∞, -1); increasing on (-1, ∞) D) Increasing on (-∞, 1); decreasing on (1, ∞)

188) f′(x) = (x + 5)2 e-x

A) Decreasing on (-∞, -5); increasing on (-5, ∞)

B) Never decreasing; increasing on (-∞, -5) ∪ (-5, ∞)

C) Never decreasing; increasing on (-∞, 5) ∪ (5, ∞)

D) Never increasing; decreasing on (-∞, -5) ∪ (-5, ∞)

43

Answer KeyTestname: DERIVATIVES WORKSHEET 1 - UNDERSTANDING THE DERIVATIVE

1) d(t) = 10t+5, 10 mi/hr

2) d(t) = 15t+10, 15 mi/hr

3) d(t) = 2t+1, 2 ft/sec

4) 4 ft/sec, The person could walk at a 4 ft/sec velocity to arrive at the same end point at the same time.

5) a) at the start line, b) back at the start line, c) goes forward until the 2 sec. mark where he/she is 4 ft. from the

start, then turns around and goes back to the start line, d) 0 ft/sec. f) he/she could have walked 0 ft/sec and

ended up at the same place, that is, just stay at the start line and not move.


7) C

8) C

9) D

10) 20 ft/sec (positive means the ball is going up)

11) -44 ft/sec (negative means the ball is going up) This shows the difference between velocity and speed. The ball

is going at a speed of 44 ft/sec in the downward direction. The velcity of the ball is -44 ft/sec)

12) (90-90)/(1-1) = 0/0 is undefined. That's the trouble , so we can quit and go home cuz there's no way to do it ...

OR IS THERE?!!

13) Paragraph essay

14) discuss in class

15) a) 6.01 ft/sec, b) 6.0001 ft/sec, c) discuss in class

16) a) 4.001 ft/sec, b) 4.00001 ft/sec, c) discuss in class

17) 6 ft/sec

18) 6 m/sec

19) 2t

20) a) 0, b) 2 c) 5 d) 8 e) 10

21) 2x + 1

22) a) 2z - 5

23) 6x-4

24) 3x2 + 2

25) f'(x) = 3

26) 12x2

27) b) v(t) = 96 - 32t, c) 32 ft/sec. , d) -32 ft/sec, the ball is going down.

28) b) tmax = 2 sec. , and hmax = 64 ft. , c) v(2) = 0 ft/sec , d) Yes, it does, because ... you answer why in class!


30) b) secant line: y = 32t + 48 , tangent line: y = 64t + 16 , c) y = 48t + 32 , d) y = 60.8t + 19.2 , e) check answer in class.

31) D

32) B

33) B

34) C

35) The following were obtained by plugging into the actual f '(x) function, so yours will be approximate: a) 5.6 b) 1.8

c) .096 d) -.8 e) -2.2 f) -2.4 g) -.45 h) 4.2

36) C

37) A

38) C

39) D

40) A

41) B

42) A

44


43) D


45) B

46) A

47) B

48) A

49) D

50) D

51) C

52) B

53) B

54) B


56)


58)

59) section 2.5


61) B

62) Answers will vary. A general shape is indicated below:

63) A

64) C

65) B

66) D

67)


69) A

70) B

71) C

72) C

73) D

45


74) B

75) C

76) B

77)

78) C

79) D

80) C

81) D

82) A

83) C

84) D

85) D

86) A

87) A

88) D

89) C

90) C

91) B


93) C

94) A

95) A

96) C

97) D

98) C

99) B

100) D

101) B

102) A


104) Read and take notes


106) http://math.ucsd.edu/~wgarner/math20a/prodrule.htm

107) http://math.ucsd.edu/~wgarner/math20a/quotrule.htm108) D

109) D

110) A

111) A

112) D

113) C

114) C

115) B

116) C

117) A

118) A

46


119) see answers in the book. You can also get the answer to any derivative problem by going to

http://www.wolframalpha.com/ Type in: derivative of (type in your function) and click the"=" button.

120) C

121) D

122) C

123) A

124) D

125) B

126) B

127) C

128) C

129) A

130) C

131) D

132) A

133) B

134) A

135) D

136) http://math.ucsd.edu/~wgarner/math20a/chainrule.htm137) Read and take notes

138) see answers in the book. You can also see answers by going to http://www.wolframalpha.com/ Typein: derivative of (type in your function) and click the "=" button.

139) a) show table b) y = cos(x) .

140) y = cos(x)

141) a) .99999983 b) 1

142) a) .0005 b) 0

143) http://www-math.mit.edu/~djk/18_01/chapter05/proof02.html

144) see section 3.5


146) y = -sin(x)

147) see section 3.5 Derivative of the Tangent Function

148) y = 1

cos2 x is the resulting graph, the estimated graph of the derivative of tan(x) and the graph of

1

cos2 x

coincide.

149) see answers in the book. You can also see more info at WolframAlpha: http://www.wolframalpha.com/Type in: derivative of (type in your function) and click the "=" button.

150)2π

3 mi/s ≈ 7540 mi/h

151) y = x


153) see answers in the book. Also, you can check answers using the WolframAlpha site

154) read book and take notes

155) b) y' = -x

y c) slope = -1 d) y = -x + 2 2

156) B

47


157) B

158) B

159) D

160) C

161) D

162) B

163) D

164) D

165) A

166) see answers in the book. Also you can check the graphs of the equations by entering in the equations into the

WolframAlpha site. (Type in like: graph y^6+x^3=y^2+9x ). You can even get the derivative answers by

typing in like: derivative y^6+x^3=y^2+9x

167) L(x) = -44x - 96 , L(a + 0.1) - f(a + 0.1) = -.04

168) L(x) = 2

3x + 6 , L(a + 0.1) - f(a + 0.1) = .00036631

169) L(x) = x , L(a + 0.1) - f(a + 0.1) = .00016658

170) L(x) = x - π , L(a + 0.1) - f(a + 0.1) = -.00033467


172) Section 3.10 in the book

173) http://math.ucsd.edu/~wgarner/math20a/mvt.htm174) C

175) B

176) B

177) B

178) A

179) B

180) As the trucker's average speed was 115 mph, the Mean Value Theorem implies that the trucker must have been

going that speed at least once during the trip.

181) Yes, the Mean Value Theorem implies that the runner attained a speed of 9.7 mph, which was her average speed

throughout the marathon.

182) This example does not contradict Rolle's Theorem because the function f is not continuous on the closed interval

[0, 1]. In particular, f is not continuous at the right end point x = 1.

183) False. The function has a non-removable discontinuity at x = 0. The mean value theorem does not apply.

184) section 3.10

185) B

186) C

187) B

188) B

48

Derivatives Worksheet 2 - Using the Derivative





More videos will have been added to the online version of this worksheet by the time you get here!

FROM THE BOOK

1) In Section 4.1 Read and take notes on: Local Maxima and Minima; How to detect a local maximum or

minimum (critical point and critical value); Local Extrema and Critical Points; The First-Derivative Test

for Local Maxima and Minima; The Second-Derivative Test for Local Maxima and Minima; Concavity and

Inflection Points

ESSAY.

2) Go to http://youtu.be/KqbsIJ3v4Y4 . "Increasing, Decreasing, and Tangent line slopes.avi" Use full

screen. Take notes on this video and answer the following:

a) What is this video showing? What does the blue segment represent?

b) (Fill in on your paper) When the blue line goes up from left to right then the sign of the slope is ________,

and the function is _____________. When the blue line goes down from left to right then the sign of the slope

is ________, and the function is _____________.

c) Stop the video at a = -1 . What is happening there?

3) Go to http://youtu.be/CWznFtlzbs4 "Derivative Increasing, Decreasing, and Concavity.avi" Use full

screen. Take notes on this video and answer the following:

a) What is this video showing?

b) Stop the video at a = 0. What is happening here?

4) Go to http://www.youtube.com/watch?v=aJuJOB6NTuc "Increasing/Decreasing , Local

Maximums/Minimums". Take notes on this video and answer the following:

a) Is the point A a local minimum? What is the derivative of the function there?

b) What kind of point is point F?

c) f '(6) is what value?

d) f '(7) is what value?

e) Exlain why point c is both a local min and a local max.

5) Go to http://www.youtube.com/watch?v=-W4d0qFzyQY "Finding Intervals of

Increase/Decrease Local Max/Mins" Take notes on this video and answer the following:

a) If you want to find all the local extrema (local maximums or minimums), what do you do with the first

derivative, and how do you tell if you have found a local maximum or a local minimum?

1

6) Go tohttp://www.youtube.com/user/EducatorVids?v=Z7QWpBU1ePU&feature=pyv&ad=8624604428

&kw=second%20derivatives%20test "First Derivative Test, Second Derivative Test" Take notes on this

video and answer the following:

a) What is the "first derivative test" and what is it used for. What is the "2nd derivative test" and what is it used

for?

7) Go to http://www.youtube.com/watch?v=c1N8zyVhWxM "Concavity, Inflection Points and

Second Derivatives" and take notes on this video.

8) Go to http://www.youtube.com/watch?v=wRBCvDy2jEY "Second Derivative Test" and take

notes on this video.

9) Go to http://www.youtube.com/watch?v=QtXCIxB6kW8 "Finding Local

Maximums/Minimums - Second Derivative Test" and take notes on this video.

NOTE: Now we do problems using the concepts and techniques in the videos. It's always a good idea to

verify answers also by going to http://www.wolframalpha.com and entering in like: local extrema of

(x-5)e^(-x) , or like: inflection points of (x-3)^2(x+2) You get a graph and the answers.


Using the derivative of f(x) given below, determine the critical points of f(x).

10) f′(x) = (x + 10)(x + 9)

A) -10, -9 B) -19 C) 0, -10, -9 D) 9, 10

11) f′(x) = (x - 3)2(x + 2)

A) -3, 0, 2 B) -3, -2, 3 C) -3, 2 D) -2, 3

12) f′(x) = (x - 5) e-x

A) 6 B) -6 C) -5 D) 5

Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing.

13) f′(x) = (3 - x)(4 - x)

A) Decreasing on (-∞, -3) ∪ (-4, ∞); increasing on (-3, -4)

B) Decreasing on (-∞, 3); increasing on (4, ∞)



14) f′(x) = x1/3(x - 6)

A) Decreasing on (0, 6); increasing on (6, ∞)

B) Increasing on (0, ∞)



2

15) f′(x) = (x - 1) e-x

A) Decreasing on (-∞, 1); increasing on (1, ∞) B) Increasing on (-∞, 1); decreasing on (1, ∞)

C) Decreasing on (-∞, -1); increasing on (-1, ∞) D) Increasing on (-∞, -1); decreasing on (-1, ∞)

16) f′(x) = (x + 5)2 e-x

A) Decreasing on (-∞, -5); increasing on (-5, ∞)

B) Never decreasing; increasing on (-∞, -5) ∪ (-5, ∞)

C) Never increasing; decreasing on (-∞, -5) ∪ (-5, ∞)

D) Never decreasing; increasing on (-∞, 5) ∪ (5, ∞)

Use the maximum/minimum finder on a graphing calculator (that's 2nd calc max or 2nd calc min on the TI) to

determine the approximate location of all local extrema.

17) f(x) = 0.1x3 -15x2 - 23x - 14

A) Approximate local maximum at -100.761; approximate local minimum at 0.761

B) Approximate local maximum at -0.761; approximate local minimum at 100.761

C) Approximate local minimum at -100.761; approximate local maximum at 0.761

D) Approximate local minimum at -0.761; approximate local maximum at 100.761

18) f(x) = 0.1x4 - x3- 15x2 + 59x + 6

A) Approximate local maximum at 1.652; approximate local minima at -6.73 and 12.445

B) Approximate local maximum at 1.646; approximate local minima at -6.693 and 12.611

C) Approximate local maximum at 1.815; approximate local minima at -6.837 and 12.498

D) Approximate local maximum at 1.735; approximate local minima at -6.777 and 12.542

19) f(x) = x4 - 3x3- 21x2 + 74x - 71



C) Approximate local maximum at 1.54; approximate local minima at -3.006 and 3.817


20) f(x) = x4 - 4x3- 53x2 - 86x + 62



C) Approximate local maximum at -0.944; approximate local minima at -3.192 and 7.136


21) f(x) = 0.1x5 + 5x4 - 8x3- 15x2 - 6x + 77

A) Approximate local maxima at -41.132 and -0.273; approximate local minima at -0.547 and 1.952

B) Approximate local maxima at -41.126 and -0.329; approximate local minima at -0.543 and 1.896

C) Approximate local maxima at -41.043 and -0.177; approximate local minima at -0.587 and 1.992

D) Approximate local maxima at -41.191 and -0.214; approximate local minima at -0.55 and 2.011

3

Find the open intervals on which the function is increasing and decreasing. Identify the function's local and

absolute extreme values, if any, saying where they occur.

22)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

A) increasing on (-2, 2); decreasing on (-6, -2) and (2, 6);

absolute maximum at (2, 4); absolute minimum at (-2, -4)

B) increasing on (-2, 2); decreasing on (-6, -2) and (2, 6);

no absolute maximum; no absolute minimum

C) increasing on (-2, 2); decreasing on (0, 6);


D) increasing on (-2, 2); decreasing on (-6, 0);


23)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) increasing on (-3, 0); decreasing on (-5, -3) and (2, 5)

absolute maximum at (-5, 0); local maximum at (0, -1) and (2, -1);

absolute minimum at (5, -4)

B) increasing on (-3, 0); decreasing on [-5, -3) and (2, 5]

absolute maximum at (-5, 0); absolute minimum at (5, -4)

C) increasing on (-3, 0); decreasing on (-5, -3) and (2, 5)

absolute maximum at (-5, 0); local minimum at (-3, -4) and (5, -4)

D) increasing on (-3, 1); decreasing on (-5, -3) and (0, 5)

absolute maximum at (-5, 0); no absolute minimum

4

24)

x-5 -4 -3 -2 -1 1 2 3 4 5

y8

7

6

5

4

3

2

1

-1x-5 -4 -3 -2 -1 1 2 3 4 5

y8

7

6

5

4

3

2

1

-1

A) increasing on (-2, 0) and (2, 4); decreasing on (0, 2);

absolute maximum at (4, 6); absolute minimum at (-2, 0) and (2, 0)

B) increasing on (-2, 0) and (2, 4); decreasing on (0, 2);

absolute maximum at (4, 6); local maximum at (0, 2); absolute minimum at (-2, 0) and (2, 0)

C) increasing on (-2, 0) and (2, 4); decreasing on (0, 2);

absolute maximum at (4, 6) and(0,2); absolute minimum at (-2, 0) and (2, 0)

D) increasing on (2, 4); decreasing on (0, 2);

absolute maximum at (4, 6); local maximum at (0, 2); absolute minimum at (-2, 0) and (2, 0)

Find the largest open interval where the function is changing as requested.

25) Increasing y = 7x - 5

A) (-5, ∞) B) (-∞, 7) C) (-5, 7) D) (-∞, ∞)

26) Increasing f(x) = 1

4x2 -

1

2x

A) (-1, 1) B) (-∞, -1) C) (1, ∞) D) (-∞, ∞)

27) Increasing y = (x2 - 9)2

A) (-∞, 0) B) (3, ∞) C) (-3, 0) D) (-3, 3)

28) Increasing f(x) = x2 - 2x + 1

A) (-∞, 0) B) (0, ∞) C) (-∞, 1) D) (1, ∞)

29) Increasing f(x) = 1

x2 + 1

A) (1, ∞) B) (0, ∞) C) (-∞, 1) D) (-∞, 0)

30) Decreasing f(x) = 4 - x

A) (-∞, 4) B) (-4, ∞) C) (4, ∞) D) (-∞, -4)

31) Decreasing f(x) = ∣x - 8∣

A) (-8, ∞) B) (8, ∞) C) (-∞, -8) D) (-∞, 8)

5

32) Decreasing f(x) = x3 - 4x

A)2 3

3, ∞ B) -

2 3

3,

2 3

3C) -∞, -

2 3

3D) -∞, ∞

Identify the function's local and absolute (global) extreme values, if any, saying where they occur. They mean to do

it by hand using f'(x) tp find the critical points and then use the first and/or 2nd derivative test to classify the point

(to decide whether it is a max or a min or neither).

33) f(x) = -x3- 9x2 - 24x + 2

A) local maximum at x = 4; local minimum at x = 2

B) local maximum at x = -4; local minimum at x = -2

C) local maximum at x = -2; local minimum at x = -4

D) local maximum at x = 2; local minimum at x = 4

Identify the function's local and absolute extreme values, if any, saying where they occur.

34) f(r) = (r - 9) 3

A) local minimum: x = 0; local maximum: x = 9 B) local minimum: x = 0

C) no local extrema D) local minimum: x = 9

35) h(x) = x - 2

x2 + 3x + 6

A) local minimum at x = -2; no local maxima

B) no local extrema

C) local minimum at x = -5; local maximum at x = 6

D) local minimum at x = -2; local maximum at x = 6

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

A) absolute maximum: 1 at x = -1

B) absolute minimum: 1 at x = -1

C) no local extrema

D) relative minimum: 1 at x = -1; relative maximum: -1 at x = 1

Identify the function's extreme values in the given domain, and say where they are assumed. Tell which of the

extreme values, if any, are absolute.

37) f(x) = (x + 7)2, -∞ < x ≤ 0

A) local maximum: 49 at x = 0;

local and absolute absolute minimum: 0 at x = -7

B) no local extrema; no absolute extrema

C) local and absolute minimum: 0 at x = -7

D) local and absolute maximum: 49 at x = 0;

local and absolute minimum: 0 at x = -7

38) f(x) = x2 - 8x, -∞ < x ≤ 8

A) local minimum: 0 at x = 8; local and absolute maximum:-16 at x = 4

B) local minimum: -16 at x = 4; local and absolute maximum: 0 at x = 8

C) local and absolute minimum: -16 at x = 4; local and absolute maximum: 0 at x = 8

D) local and absolute minimum: -16 at x = 4; local maximum: 0 at x = 8

6

39) g(t) = t3

3 -

9

2t2 + 8t, 0 ≤ t < ∞

A) local minimum: 23

6 at x =1; local and absolute maximum: -

160

3 at x = 8

B) local and absolute minimum: - 160

3 at x = 8; local maximum:

23

6 at x =1

C) local minimum: 0 at x = 0; local and absolute minimum: - 160

3 at x = 8; local maximum:

23

6 at x =1

D) local minimum: 23

6 at x =1; local maximum: 0 at x = 0; absolute maximum: -

160

3 at x = 8

40) h(x) = x3 + 2x2 + 5x + 4, -∞ < x ≤ 0

A) local and absolute maximum: 4 at x = 0;

local and absolute minimum: 3 at x = -1

B) local and absolute minimum: 3 at x = -1;

C) local and absolute maximum: 4 at x = 0; D) no local extrema; no absolute extrema

41) f(x) = 4 - x2, -2 ≤ x < 2

A) local and absolute minimum: 0 at x = -2;

local and absolute maximum: 2 at x = 0

B) local and absolute minimum: 0 at x = -2 and x = 2;

local and absolute maximum: 2 at x = 0

C) no local extrema; no absolute extrema

D) local and absolute maximum: 0 at x = -2;

local and absolute minimum: 2 at x = 0

Find the extrema of the function on the given interval, and say where they occur.

42) sin 4x, 0 ≤ x ≤ π

2

A) local maxima: 1 at x = π

8 and 0 at x =

π

2;

local minimum: -1 at x = 3π

8

B) local maxima: 1 at x = π

4 and 0 at x =

π

2;

local minima: 0 at x = 0 and -1 at x = 3π

8

C) local maxima: 1 at x = π

8 and 0 at x =

π

2;

local minima: 0 at x = 0 and -1 at x = 3π

8

D) local maxima: 1 at x = π

8 and 0 at x =

π

4;

local minimum: 0 at x = 0

7

43) sin x + cos x, 0 ≤ x ≤ 2π

A) local maxima: 1 at x = 0 and - 2 at x = 7π

4;

local minima: 1 at x = 2π and 2 at x = π

4

B) local maxima: 1 at x = 0 and - 2 at x = 5π

4;

local minima: 1 at x = 2π and 2 at x = π

4

C) local maxima: 1 at x = 2π and 2 at x = π

4;

local minima: 1 at x = 0 and - 2 at x = 5π

4

D) local maxima: 1 at x = 2π and 2 at x = π

4;

local minima: 1 at x = 0 and - 2 at x = 7π

4


44) Find the absolute maximum and minimum values of f(x) = 2x - ex on [0, 1].

A) Maximum = ln 4 - 2 at x = ln 2, minimum = 1 at x = 0

B) Maximum = ln 2 - 2 at x = ln 2, minimum = -1 at x = 0

C) Maximum = 0 at x = ln 2, minimum = 1 at x = 0

D) Maximum = ln 4 - 2 at x = ln 2, minimum = -1 at x = 0

Use the graph of the function f(x) to locate the local extrema and identify the intervals where the function is concave

up and concave down.

45)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) Local minimum at x = 1; local maximum at x = -1; concave up on (0, ∞); concave down on (-∞, 0)

B) Local minimum at x = 1; local maximum at x = -1; concave down on (-∞, ∞)

C) Local minimum at x = 1; local maximum at x = -1; concave up on (-∞, ∞)

D) Local minimum at x = 1; local maximum at x = -1; concave down on (0, ∞); concave up on (-∞, 0)

8

46)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) Local maximum at x = 1; local minimum at x =-1; concave up on (0, ∞); concave down on (-∞, 0)

B) Local minimum at x = 1; local maximum at x =-1; concave up on (0, ∞); concave down on (-∞, 0)

C) Local minimum at x = 1; local maximum at x =-1; concave down on (0, ∞); concave up on (-∞, 0)

D) Local maximum at x = 1; local minimum at x =-1; concave up on (-∞, ∞)

47)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) Local minimum at x = 3; local maximum at x = -3 ; concave down on (0, ∞); concave up on (-∞, 0)

B) Local maximum at x = 3; local minimum at x = -3 ; concave up on (0, -3) and (3, ∞); concave down on

(-3, 3)

C) Local minimum at x = 3; local maximum at x = -3 ; concave up on (0, -3) and (3, ∞); concave down on

(-3, 3)

D) Local minimum at x = 3; local maximum at x = -3 ; concave up on (0, ∞); concave down on (-∞, 0)

9

48)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) Local minimum at x = 3 ; local maximum at x = -3 ; concave up on (-∞, -3) and (3, ∞); concave down

on (-3, 3)

B) Local minimum at x = 3 ; local maximum at x = -3 ; concave up on (0, ∞); concave down on (-∞, 0)

C) Local minimum at x = 3 ; local maximum at x = -3 ; concave down on (0, ∞); concave up on (-∞, 0)

D) Local minimum at x = 3 ; local maximum at x = -3 ; concave down on (-∞, -3) and (3, ∞); concave up

on (-3, 3)

49)

x-10 10

y

10

-10

x-10 10

y

10

-10

A) Local minimum at x = 0; local maximum at x = 2; concave down on (0, ∞); concave up on (-∞, 0)

B) Local minimum at x = 0; local maximum at x = 2; concave up on (0, ∞); concave down on (-∞, 0)

C) Local minimum at x = 2; local maximum at x = 0; concave up on (0, ∞); concave down on (-∞, 0)

D) Local minimum at x = 2; local maximum at x = 0; concave down on (0, ∞); concave up on (-∞, 0)

Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points.

10

50) y = 8x

x2 + 16

x

y

x

y

A) absolute maximum: 0, 1

2

no inflection point

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

B) local minimum: -4, - 1

2

local maximum: 4, 1

2

inflection point: (0,0)

x-4 -2 2 4

y3

2

1

-1

-2

-3

x-4 -2 2 4

y3

2

1

-1

-2

-3

C) local minimum: (4, -1)

local maximum: (-4, 1)

inflection point: (0, 0)

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

D) local minimum: (-4, -1)

local maximum: (4, 1)

inflection points: (0, 0), (-4 3, -2 3),

(4 3, 2 3)

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

11

51) y = 2x3 - 9x2 + 12x

x

y

x

y

A) local minimum: (2, 4)


inflection point: 3

2,

9

2

x-8 -4 4 8

y

24

12

-12

-24

x-8 -4 4 8

y

24

12

-12

-24

B) local minimum: (0, 0)



x-8 -4 4 8

y

432

216

-216

-432

x-8 -4 4 8

y

432

216

-216

-432

C) no extrema


x-8 -4 4 8

y

72

36

-36

-72

x-8 -4 4 8

y

72

36

-36

-72

D) local minimum: (1, 10)

no inflection point

x-8 -4 4 8

y

24

12

-12

-24

x-8 -4 4 8

y

24

12

-12

-24

12

52) y = x1/3(x2 - 175)

x

y

x

y

A) local minimum: 5, -1503

5

local maximum: -5, 1503

5


x-20 -10 10 20

y400

300

200

100

-100

-200

-300

-400

x-20 -10 10 20

y400

300

200

100

-100

-200

-300

-400


no inflection points

x-20 -10 10 20

y400

300

200

100

-100

-200

-300

-400

x-20 -10 10 20

y400

300

200

100

-100

-200

-300

-400

C) local minimum: ± 75, - 75

2


inflection points: ±5, - 25

3

x-20 -10 10 20

y100

75

50

25

-25

-50

-75

-100

x-20 -10 10 20

y100

75

50

25

-25

-50

-75

-100

D) no extrema


x-20 -10 10 20

y600

450

300

150

-150

-300

-450

-600

x-20 -10 10 20

y600

450

300

150

-150

-300

-450

-600

13

53) y = x2

x2 + 2

x

y

x

y

A) local minimum: (0, 0)


x-3 -2 -1 1 2 3

y0.375

0.25

0.125

-0.125

-0.25

-0.375

x-3 -2 -1 1 2 3

y0.375

0.25

0.125

-0.125

-0.25

-0.375


inflection points: - 6

3,

1

4,

6

3,

1

4

x-3 -2 -1 1 2 3

y0.75

0.5

0.25

-0.25

-0.5

-0.75

x-3 -2 -1 1 2 3

y0.75

0.5

0.25

-0.25

-0.5

-0.75

C) local minimum: 0, - 1

2


x-3 -2 -1 1 2 3

y0.75

0.5

0.25

-0.25

-0.5

-0.75

x-3 -2 -1 1 2 3

y0.75

0.5

0.25

-0.25

-0.5

-0.75

D) local minimum: 0, 1

2


x-3 -2 -1 1 2 3

y1.5

1

0.5

-0.5

-1

-1.5

x-3 -2 -1 1 2 3

y1.5

1

0.5

-0.5

-1

-1.5

14

54) y = x + cos 2x, 0 ≤ x ≤ π

x1 2 3

y

4

3

2

1

-1

x1 2 3

y

4

3

2

1

-1

A) local minimum: 5π

12,

5π - 6 3

12

local maximum: π

12,

π + 6 3

12

inflection points: π

4,

π

4,

3π

4,

3π

4

x1 2 3

y

4

3

2

1

-1

x1 2 3

y

4

3

2

1

-1

B) no local extrema

inflection point: π

2,

π

2

x1 2 3

y

4

3

2

1

-1

x1 2 3

y

4

3

2

1

-1

15

C) local minimum: π

4, -1

local maximum: 3π

4, 3

inflection point: π

2, 1

x1 2 3

y

4

3

2

1

-1

x1 2 3

y

4

3

2

1

-1

D) local minimum: (1.444, -0.246)

local maximum: (0.126, 1.031)

inflection points: (0.785, 0.393), (2.356, 1.178)

x1 2 3

y

4

3

2

1

-1

x1 2 3

y

4

3

2

1

-1

55) y = x 17 - x2

x

y

x

y

16

A) local maximum: 34

3,

2 · 173/2 · 3

9

no inflection point.

x-16 -12 -8 -4 4 8 12 16

y32

24

16

8

-8

-16

-24

-32

x-16 -12 -8 -4 4 8 12 16

y32

24

16

8

-8

-16

-24

-32

B) local minimum: - 34

2, -

17

2

local maximum: 34

2,

17

2


x-4 -3 -2 -1 1 2 3 4

y8

6

4

2

-2

-4

-6

-8

x-4 -3 -2 -1 1 2 3 4

y8

6

4

2

-2

-4

-6

-8

C) local maximum: (0, 17)

no inflection points.

x-4 -3 -2 -1 1 2 3 4

y8

6

4

2

-2

-4

-6

-8

x-4 -3 -2 -1 1 2 3 4

y8

6

4

2

-2

-4

-6

-8

D) local minimum: - 51

3, -

2 · 173/2 · 3

9

local maximum: 51

3,

2 · 173/2 · 3

9


x-4 -3 -2 -1 1 2 3 4

y32

24

16

8

-8

-16

-24

-32

x-4 -3 -2 -1 1 2 3 4

y32

24

16

8

-8

-16

-24

-32

Sketch the graph and show all local extrema and inflection points.

17

56) y = -x4 + 2x2 - 7

x-10 -5 5 10

y

10

5

-5

-10

x-10 -5 5 10

y

10

5

-5

-10

A) Absolute maxima: (-1, -6), (1, -6)

Local minimum: (0, -7)

Inflection points: -1

3,

2

3,

1

3,

2

3

x-10 -5 5 10

y

10

5

-5

-10

x-10 -5 5 10

y

10

5

-5

-10

B) Absolute minima: (-1, 6), (1, 6)

Local maximum: (0, 7)

Inflection point: -1

3,

58

9,

1

3,

58

9

x-10 -5 5 10

y

10

5

-5

-10

x-10 -5 5 10

y

10

5

-5

-10

C) Absolute maxima: (-1, -6), (1, -6)

Inflection points: -1

3,

2

3,

1

3,

2

3

x-10 -5 5 10

y

10

5

-5

-10

x-10 -5 5 10

y

10

5

-5

-10

D) Absolute maxima: (-1, -6), (1, -6)

Local minimum: (0, -7)

No inflection points

x-10 -5 5 10

y

10

5

-5

-10

x-10 -5 5 10

y

10

5

-5

-10

18

57) y = x - sin x, 0 ≤ x ≤ 2π

xπ2

π 3π2

2π

y

6

4

2

xπ2

π 3π2

2π

y

6

4

2

A) Local minimum: (0, 0)

Local maximum: (2π, 2π)


xπ2

π 3π2

2π

y

6

4

2

xπ2

π 3π2

2π

y

6

4

2

B) Local minimum: (0, 0)


Inflection point: (π, π)

xπ2

π 3π2

2π

y

6

4

2

xπ2

π 3π2

2π

y

6

4

2

C) Local minimum: (0, 0)



xπ2

π 3π2

2π

y

6

4

2

xπ2

π 3π2

2π

y

6

4

2

D) Local minimum: (0, 0)


Inflection point: (π, π)

xπ2

π 3π2

2π

y

6

4

2

xπ2

π 3π2

2π

y

6

4

2

19

58) y = ln (8 - x2)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A) Local minimum (0, ln 8)

No inflection point

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

B) No extrema

No inflection point

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

C) Local maximum (0, ln 8)

No inflection point

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

D) Local minimum (0, -ln 8)

No inflection point

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

x-5 -4 -3 -2 -1 1 2 3 4 5

y6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

20

59) y = ex - 3e-x - 4x

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

A) Local minimum (2, -1)

No inflection point

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

B) Local maximum (0, -2)

Local minimum (ln 3, 2 - 4 ln 3)

No inflection point

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

C) Local minimum 1

2 ln 3, - 2 ln 3

No inflection point

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

D) Local maximum (0, -2)

Local minimum (ln 3, 2 - 4 ln 3)

Inflection point 1

2 ln 3, - 2 ln 3

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

For the given expression y′, find y'' and sketch the general shape of the graph of y = f(x).

21

60) y' = x2

3 - 5

x

y

x

y

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

22

61) y′ = x2(2 - x)

x

y

x

y

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

23

62) y' = sin x, 0 ≤ x ≤ 2π

x

y

x

y

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

24

63) y′ = x-2/3(x - 6)

x

y

x

y

A)

x

y

x

y

B)

x

y

x

y

C)

x

y

x

y

D)

x

y

x

y

Solve the problem.

25

64) The graphs below show the first and second derivatives of a function y = f(x). Select a possible graph of f

that passes through the point P.

f′ f′′

x

y

P

x

y

P

x

y

P

x

y

P

A)

x

y

x

y

[NOTE: Graph vertical scales

may vary from graph to graph.]

B)

x

y

x

y



C)

x

y

x

y



D)

x

y

x

y



26

65) The graphs below show the first and second derivatives of a function y = f(x). Select a possible graph f that

passes through the point P.

f′ f′′

x

y

P

x

y

P

x

y

P

x

y

P

A)

x

y

x

y



B)

x

y

x

y



C)

x

y

x

y



D)

x

y

x

y



27



f′ f′′

x

y

P

x

y

P

x

y

P

x

y

P

A)

x

y

x

y



B)

x

y

x

y



C)

x

y

x

y



D)

x

y

x

y



28



f′ f′′

x

y

P

x

y

P

x

y

P

x

y

P

A)

x

y

x

y



B)

x

y

x

y



C)

x

y

x

y



D)

x

y

x

y



29



f′ f′′

x

y

P

x

y

P

x

y

P

x

y

P

A)

x

y

x

y



B)

x

y

x

y



C)

x

y

x

y



D)

x

y

x

y



Graph the rational function.

30

69) y = x - 4

x2 - 7x + 12

x-8 -4 4 8

y

8

4

-4

-8

x-8 -4 4 8

y

8

4

-4

-8

A)

x-8 -4 4 8

y

8

4

-4

-8

x-8 -4 4 8

y

8

4

-4

-8

B)

x-8 -4 4 8

y

8

4

-4

-8

x-8 -4 4 8

y

8

4

-4

-8

C)

x-8 -4 4 8

y

8

4

-4

-8

x-8 -4 4 8

y

8

4

-4

-8

D)

x-8 -4 4 8

y

8

4

-4

-8

x-8 -4 4 8

y

8

4

-4

-8

31

70) y = x2 + x - 56

x2 - x - 42

x-10 -8 -6 -4 -2 2 4 6 8 10

y10

8

6

4

2

-2

-4

-6

-8

-10

x-10 -8 -6 -4 -2 2 4 6 8 10

y10

8

6

4

2

-2

-4

-6

-8

-10

A)

x-10 -8 -6 -4 -2 2 4 6 8 10

y10

8

6

4

2

-2

-4

-6

-8

-10

x-10 -8 -6 -4 -2 2 4 6 8 10

y10

8

6

4

2

-2

-4

-6

-8

-10

B)

x-10 -8 -6 -4 -2 2 4 6 8 10

y10

8

6

4

2

-2

-4

-6

-8

-10

x-10 -8 -6 -4 -2 2 4 6 8 10

y10

8

6

4

2

-2

-4

-6

-8

-10

C)

x-10 -8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-10 -8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

D)

x-10 -8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-10 -8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

32

71) y = x2

x2 + 2

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

A)

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

B)

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

C)

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

D)

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

x-3 -2 -1 1 2 3

y

1.5

1

0.5

-0.5

33

72) y = x - 1

x2 - 1

-6 -4 -2 2 4 6

4

2

-2

-4

-6 -4 -2 2 4 6

4

2

-2

-4

A)

-6 -4 -2 2 4 6

6

4

2

-2

-4

-6

-6 -4 -2 2 4 6

6

4

2

-2

-4

-6

B)

-6 -4 -2 2 4 6

4

2

-2

-4

-6 -4 -2 2 4 6

4

2

-2

-4

C)

-6 -4 -2 2 4 6

6

4

2

-2

-4

-6

-6 -4 -2 2 4 6

6

4

2

-2

-4

-6

D)

-6 -4 -2 2 4 6

4

2

-2

-4

-6 -4 -2 2 4 6

4

2

-2

-4

34

73) y = 18x

x2 + 9

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

A)

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

B)

x-4 -2 2 4

y3

2

1

-1

-2

-3

x-4 -2 2 4

y3

2

1

-1

-2

-3

C)

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

D)

x-4 -2 2 4

y6

4

2

-2

-4

-6

x-4 -2 2 4

y6

4

2

-2

-4

-6

35

74) y = 4

x2

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

A)

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

B)

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

C)

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

D)

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

x-8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

36

75) y = x

x2 - 25

x-10 10

y

5

-5

x-10 10

y

5

-5

A)

x-10 10

y

5

-5

x-10 10

y

5

-5

B)

x-10 10

y

10

-10

x-10 10

y

10

-10

C)

x-10 10

y

5

-5

x-10 10

y

5

-5

D)

x-10 10

y

5

-5

x-10 10

y

5

-5

37

Solve the problem.

76) Using the following properties of a twice-differentiable function y = f(x), select a possible graph of f.

x y Derivatives

x < 2 y′ > 0,y′′ < 0

-2 11 y′ = 0,y′′ < 0

-2 < x < 0 y′ < 0,y′′ < 0

0 -5 y′ < 0,y′′ = 0

0 < x < 2 y′ < 0,y′′ > 0

2 -21 y′ = 0,y′′ > 0

x > 2 y′ > 0,y′′ > 0

A)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

B)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

C)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

D)

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

x-4 -3 -2 -1 1 2 3 4

y24

16

8

-8

-16

-24

38

77) Select an appropriate graph of a twice-differentiable function y = f(x) that passes through the points

(- 2, 1), - 6

3,

5

9, (0, 0),

6

3,

5

9 and ( 2, 1), and whose first two derivatives have the following sign

patterns.

y′ : + - + -

- 2 0 2

y′′ :

+ - +

- 6

3

6

3

A)

x-3 -2 -1 1 2 3

y2

1.5

1

0.5

-0.5

-1

-1.5

-2

x-3 -2 -1 1 2 3

y2

1.5

1

0.5

-0.5

-1

-1.5

-2

B)

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

C)

x-3 -2 -1 1 2 3

y4

3

2

1

-1

-2

-3

-4

x-3 -2 -1 1 2 3

y4

3

2

1

-1

-2

-3

-4

D)

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

x-3 -2 -1 1 2 3

y16

12

8

4

-4

-8

-12

-16

ESSAY.


78) If f(x) is a differentiable function and f′ (c) = 0 at an interior point c of f's domain, and if f′′(x) > 0 for all x in

the domain, must f have a local minimum at x = c? Explain.

79) Sketch a smooth curve through the origin with the following properties:

f′(x) > 0 for x < 0; f′ (x) < 0 for x > 0; f′′(x) approaches 0 as x approaches -∞; and

f′′(x) approaches 0 as x approaches ∞.

39


Identify the function's local and absolute extreme values, if any, saying where they occur.

80) f(x) = x3- 3x2 + 3x - 1

A) local maximum at x = 1

B) local minimum at x = 1

C) local maximum at x = 1; local minimum at x = -1

D) no local extrema

81) f(x) = x3 + 6.5x2 + 12x + 3

A) local maximum at x = -3; local minimum at x = - 4

3

B) local maximum at x = 4

3; local minimum at x = 3

C) local maximum at x = 4; local minimum at x = 1

D) local maximum at x = - 1; local minimum at x = -4

ESSAY.


82) The accompanying figure shows a portion of the graph of a function that is twice-differentiable at all x

except at x = p. At each of the labeled points, classify y′ and y′′ as positive, negative, or zero.

40

83) The graph below shows the position s = f(t) of a body moving back and forth on a coordinate line.

(a) When is the body moving away from the origin? Toward the origin? At approximately what times is

the (b) velocity equal to zero? (c) Acceleration equal to zero? (d) When is the acceleration positive?

Negative?

84) For x > 0, sketch a curve y = f(x) that has f(1) = 0 and f′(x) = - 1

x. Can anything be said about the concavity

of such a curve? Give reasons for your answer.

85) Sketch a continuous curve y = f(x) with the following properties:

f(2) = 3; f′′(x) > 0 for x > 4; and f′′(x) < 0 for x < 4 .


Find the largest open interval where the function is changing as requested.

86) Decreasing f(x) = - x + 3

A) (-∞, 3) B) (-3, ∞) C) (-∞, -3) D) (3, ∞)


87) Suppose the derivative of the function y = f(x) is y' = (x - 2)2(x + 7). At what points, if any, does the graph

of f have a local minimum or local maximum?

A) local maximum at x = -7 B) local minimum at x = -7

C) local minimum at x = 2 D) no local minimum or local maximum

88) Suppose that the second derivative of the function y = f(x) is y'' = (x - 4)(x + 8). For what x-values does the

graph of f have an inflection point?

A) 4, -8 B) -4, 8 C) 4, 8 D) -4, -8

FROM THE BOOK

89) Do section 4.1 # 1-21 odd, 28-31 ALL, 35, 37, 41, 45

Curve Sketching

Solve the problem.

90) Find the intervals on which the function f(x) = 4x5 - 5x4 is increasing and decreasing. Sketch the graph of

y = f(x), and identify any local maxima and minima. Any global extrema should also be identified.

41

91) Find the interval on which the function f(x) = 5x2 + 10x - 7 is increasing and decreasing. Sketch the graph

of y = f(x), and identify any local maxima and minima. Any global extrema should also be identified.

92) Find the interval on which the function f(x) = (x - 2)2(x + 3)2 is increasing and decreasing. Sketch the graph


93) Find the intervals on which the function f(x) = x2/3(10 - x) is increasing and decreasing. Sketch the graph


94) Find the intervals on which the function f(x) = 3x5 - 20x3 is increasing and decreasing. Sketch the graph of


95) Find the intervals on which the function f(x) = 3

x is increasing and decreasing. Sketch the graph of of


96) Find the intervals on which the function f(x) = x3 - 27x is increasing and decreasing. Sketch the graph of


Optimization

97) Find the maximum possible area of a rectangle with perimeter 180 meters.

98) A rectangular corral is to be constructed with an internal divider parallel to two opposite sides, and 1200

meters of fencing will be used to make the corral (including the divider). What is the maximum possible

area that such a corral can have?

99) What is the greatest amount by which a number in the interval [0, 1] can exceed its cube?

100) A farmer has 600 yards of fencing with which to build a rectangular corral. He will use a long straight wall

to form one side of the corral (to save fencing). What is the maximum possible area that can be enclosed in

this way?

101) Two nonnegative numbers have sum 1. What is the maximum possible value of the sum of their squares?

102) Find the shape of the right circular cylinder of maximal surface area (including the top and the bottom)

inscribed in a sphere of radius R.

103) Two cubes have total volume 250 cm3. What is the maximum possible surface area they can have? The

minimum?

104) A rectangle has each diagonal of length 5. What is the maximum possible perimeter of such a rectangle?

105) The product of two positive integers is 1600. What is the minimum possible value of their sum?

42

106) A rectangle has its base on the x-axis and its upper two vertices on the graph of y = 12 - x2. What is the

maximum possible area of such a rectangle?

107) The sum of two nonnegative numbers x and y is 8. Find the maximum possible value of the expression

x2 + y3.

108) A mass of clay of volume 432 in.3 is formed into two cubes. What is the minimum possible total surface

area of the two cubes? What is the maximum?

109) A rancher has 1700 meters of fencing with which to build two widely separated corrals; one is to be

square, and the other is to be twice as long as it is wide. What is the maximum possible total area that the

rancher can thereby enclose?

110) Find the maximum possible volume of a right circular cylinder inscribed in a sphere of radius R.

111) A farmer has 480 meters of fencing. He wishes to enclose a rectangular plot of land and to divide the plot

into three equal rectangles with two parallel lengths of fence down the middle. What dimensions will

maximize the enclosed area? Be sure to verify that you have found the maximum enclosed area.

112) The sum of two nonnegative numbers is 10. Find the minimum possible value of the sum of their cubes.

113) A rectangle has a line of fixed length L reaching from one vertex to the midpoint of one of the far sides.

What is the maximum possible area of such a rectangle?

114) Write an equation of the straight line through the point (7, 1) that cuts off the least area from the first

quadrant. Be sure to verify that your area is minimal.

115) A circle is dropped into the graph of the parabola y = x2. How small can the radius of the circle be and yet

allow the circle to touch the parabola at two different points?

116) Find the point on the graph of y = x that is closest to the point (3, 0). Be sure to verify that it is indeed

closest.

117) The sum of two nonnegative numbers is 48. What is the smallest possible value of the sum of their

squares?

118) A rectangular box is to have a base three times as long as it is wide. The total surface area of the box is to

be 20 ft2. Find the maximum possible volume of such a box.

119) A right circular cone has a slant height of 10 ft. Find the maximum possible volume of such a cone. Verify

that your answer is maximal.

120) A rectangular poster is to contain 8250 cm2 of print in the shape of a smaller rectangle; the margins at top

and bottom must each be 22 cm, and those at the sides must each be 15 cm. What are the dimensions of

such a poster having the least possible total area?

43

121) A wastebasket is to have as its base an equilateral triangle, its sides are to be vertical, and its volume is to

be 8 ft3. What is the minimum possible surface area of such a wastebasket?

122) A railroad will operate a special a special excursion train if at least 200 people subscribe. The fare will be

$8 per person if 200 people subscribe, but will decrease 1 cent for each additional person who subscribes.

What number of passengers will bring the railroad maximum revenue?

123) An aquarium has a square base made of slate costing 8¢/in.2 and four glass sides costing 3¢/in.2. The

volume of the aquarium is to be 36,000 in.3. Find the dimensions of the least expensive such aquarium.

124) A poster is to contain 96 in.2 of print, and each copy must have 3-in. margins at top and bottom and 2-in.

margins on each side. What are the dimensions of such a poster having the least possible area?

125) A wall 8 m high stands 4 m away from a building. What is the length of the shortest ladder that will lean

over the wall and touch the building? Use as independent variable the angle that the ladder makes with

the ground.

126) A rectangular sheet of thin metal is 5 m wide and 8 m long. Four small equal squares are cut from its

corners and the projections of the resulting cross-shaped piece of metal are bent upward and welded to

make an open-topped box with a rectangular base. What is the maximum possible volume of such a box?

127) You plan to build a playing field in the shape of a rectangle with a semicircular region at each end, so that

races can be held around the perimeter of the field. If you want the total perimeter of the field to be 1000

m, what dimensions should your field have to maximize the area of the rectangular portion of the field?

44

128) You are planning to close off a corner of the first quadrant with a line segment 15 units long running from

(x, 0) to (0, y). Show that the area of the triangle enclosed by the segment is largest when x = y.


129) A private shipping company will accept a box for domestic shipment only if the sum of its length and girth

(distance around) does not exceed 90 in. What dimensions will give a box with a square end the largest

possible volume?

A) 15 in. × 15 in. × 30 in. B) 15 in. × 15 in. × 75 in.

C) 15 in. × 30 in. × 30 in. D) 30 in. × 30 in. × 30 in.

45

130) A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas

the semicircle is of tinted glass that transmits only one-fifth as much light per unit area as clear glass does.

The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the

thickness of the frame.

A)width

height =

20

10 + πB)

width

height =

5

10 + 4πC)

width

height =

20

10 + 4πD)

width

height =

20

5 + 4π

131) A trough is to be made with an end of the dimensions shown. The length of the trough is to be 24 feet long.

Only the angle θ can be varied. What value of θ will maximize the trough's volume?

θ θ

A) 54° B) 6° C) 30° D) 32°

46

132) A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as

shown in part (a) of the figure. What values of x and y give the largest volume?

A) x = 14 cm; y = 4 cm B) x = 12 cm; y = 6 cm

C) x = 11 cm; y = 7 cm D) x = 13 cm; y = 5 cm

133) The 9 ft wall shown here stands 30 feet from the building. Find the length of the shortest straight beam

that will reach to the side of the building from the ground outside the wall.

9' wall

30'

A) 53.3 ft B) 51.3 ft C) 39 ft D) 52.3 ft

47

134) The strength S of a rectangular wooden beam is proportional to its width times the square of its depth.

Find the dimensions of the strongest beam that can be cut from a 12-in.-diameter cylindrical log. (Round

answers to the nearest tenth.)

12"

A) w = 5.9 in.; d = 10.8 in. B) w = 7.9 in.; d = 10.8 in.

C) w = 6.9 in.; d = 9.8 in. D) w = 7.9 in.; d = 8.8 in.

135) The stiffness of a rectangular beam is proportional to its width times the cube of its depth. Find the

dimensions of the stiffest beam than can be cut from a 14-in.-diameter cylindrical log. (Round answers to

the nearest tenth.)

14"

A) w = 8.0 in.; d = 13.1 in. B) w = 6.0 in.; d = 13.1 in.

C) w = 8.0 in.; d = 11.1 in. D) w = 7.0 in.; d = 12.1 in.

136) A small frictionless cart, attached to the wall by a spring, is pulled 10 cm back from its rest position and

released at time t = 0 to roll back and forth for 4 sec. Its position at time t is

s = 1 - 10 cos πt. What is the cart's maximum speed? When is the cart moving that fast? What is the

magnitude of of the acceleration then?

A) 10π ≈ 31.42 cm/sec; t = 0.5 sec, 1.5 sec, 2.5 sec, 3.5 sec; acceleration is 0 cm/sec2

B) π ≈ 3.14 cm/sec; t = 0.5 sec, 1.5 sec, 2.5 sec, 3.5 sec; acceleration is 0 cm/sec2

C) 10π ≈ 31.42 cm/sec; t = 0.5 sec, 2.5 sec; acceleration is 1 cm/sec2

D) 10π ≈ 31.42 cm/sec; t = 0 sec, 1 sec, 2 sec, 3 sec; acceleration is 0 cm/sec2

137) At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical

miles per hour; a nautical mile is 2000 yards) and continued to do so all day. Ship B was sailing east at

7 knots and continued to do so all day. The visibility was 5 nautical miles. Did the ships ever sight each

other?

A) Yes. They were within 4 nautical miles of each other.

B) No. The closest they ever got to each other was 7.0 nautical miles.

C) No. The closest they ever got to each other was 6.0 nautical miles.

D) Yes. They were within 3 nautical miles of each other.

48

138) If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city,

where p(x) = 106 - x

28. How many candy bars must be sold to maximize revenue?

A) 2968 thousand candy bars B) 2968 candy bars

C) 1484 candy bars D) 1484 thousand candy bars

139) Find the number of units that must be produced and sold in order to yield the maximum profit, given the

following equations for revenue and cost:

R(x) = 60x - 0.5x2

C(x) = 9x + 4.

A) 52 units B) 69 units C) 55 units D) 51 units

140) Suppose c(x) = x3 - 18x2 + 10,000x is the cost of manufacturing x items. Find a production level that will

minimize the average cost of making x items.

A) 9 items B) 11 items C) 8 items D) 10 items

141) Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 3.

A) h = 3 2, w = 3 2

2, B) h = 3 2, w = 2 C) h =

3 2

2, w = 3 2 D) h = 2, w = 3 2

ESSAY.

142) A storage building is to be shaped like a box with a square base (its floor). The floor costs $3/m2, the walls

cost $7/m2, and the flat roof costs $5/m2. The volume of the building is to be 12,544 m3. What is the shape

of the least expensive such building?

143) You need a cardboard container in the shape of a right circular cylinder of volume 54π in.2. What radius r

and height h would minimize its total surface area (including top and bottom)?

144) What is the maximum possible volume of a right circular cylinder with total surface area 600π in.2

(including the top and the bottom)?

145) Find the minimum possible value of the sum of a real number and its square.

146) What point on the parabola y = x2 is closest to the point (3, 0)?

147) A rectangular box with square base and no top is to have a volume of exactly 1000 cm3. What is the

minimum possible surface area of such a box?

148) Find the coordinates of the point or points on the curve 2y2 = 5x + 5 which is (are) closest to the origin (0,

0).

149) The sum of the squares of two nonnegative real numbers x and y is 18. What is the minimum possible

value of x + y?

49

FROM THE BOOK

150) Do section 4.2 # 3 - 11 odd, 13-21 odd, 25, 29, 33

151) Do section 4.4 #1-9 odd, 17-37 odd, 47, 49

Rates and Related Rates


Solve the problem.

152) A company knows that the unit cost C and the unit revenue R from the production and sale of x units are

related by C = R2

134,000 + 8437. Find the rate of change of unit revenue when the unit cost is changing by

$11/unit and the unit revenue is $4000.

A) $184.25/unit B) $843.70/unit C) $513.98/unit D) $220.00/unit

153) Water is falling on a surface, wetting a circular area that is expanding at a rate of 6 mm2/s. How fast is the

radius of the wetted area expanding when the radius is 170 mm? (Round your answer to four decimal

places.)

A) 0.0353 mm/s B) 178.0234 mm/s C) 0.0112 mm/s D) 0.0056 mm/s

154) A wheel with radius 3 m rolls at 15 rad/s. How fast is a point on the rim of the wheel rising when the point

is π/3 radians above the horizontal (and rising)? (Round your answer to one decimal place.)

A) 90.0 m/s B) 45.0 m/s C) 22.5 m/s D) 11.3 m/s

155) Assume that the profit generated by a product is given by P(x) = 3 x, where x is the number of units sold.

If the profit keeps changing at a rate of $600 per month, then how fast are the sales changing when the

number of units sold is 1900? (Round your answer to the nearest dollar per month.)

A) $156,920/month B) $9/month C) $17,436/month D) $8718/month

156) A piece of land is shaped like a right triangle. Two people start at the right angle of the triangle at the same

time, and walk at the same speed along different legs of the triangle. If the area formed by the positions of

the two people and their starting point (the right angle) is changing at 2 m2/s, then how fast are the people

moving when they are 3 m from the right angle? (Round your answer to two decimal places.)

A) 1.33 m/s B) 0.67 m/s C) 0.33 m/s D) 4.48 m/s

Solve the problem. Round your answer, if appropriate.

157) Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1634p(1/2), where p is the

pressure (in psi). If the water pressure is changing at a rate of 0.209 psi/sec, find the acceleration (dv/dt) of

the water when p = 34.0 psi.

A) 140 ft/sec2 B) 996 ft/sec2 C) 29.3 ft/sec2 D) 47.6 ft/sec2

158) One airplane is approaching an airport from the north at 149 km/hr. A second airplane approaches from

the east at 246 km/hr. Find the rate at which the distance between the planes changes when the

southbound plane is 27 km away from the airport and the westbound plane is 17 km from the airport.

A) -257 km/hr B) -514 km/hr C) -128 km/hr D) -385 km/hr

50

159) Water is being drained from a container which has the shape of an inverted right circular cone. The

container has a radius of 6.00 inches at the top and a height of 10.0 inches. At the instant when the water in

the container is 9.00 inches deep, the surface level is falling at a rate of 1.2 in./sec. Find the rate at which

water is being drained from the container.

A) 110 in.3/s B) 159 in.3/s C) 129 in.3s D) 105 in.3/s

160) A man 6 ft tall walks at a rate of 5 ft/sec away from a lamppost that is 18 ft high. At what rate is the length

of his shadow changing when he is 45 ft away from the lamppost? (Do not round your answer)

A)5

4 ft/sec B)

75

2 ft/sec C)

5

2 ft/sec D)

5

8 ft/sec

161) The volume of a sphere is increasing at a rate of 6 cm3/sec. Find the rate of change of its surface area when

its volume is 32π

3 cm3. (Do not round your answer.)

A) 4 cm2/sec B) 12π cm2/sec C)8

3 cm2/sec D) 6 cm2/sec

162) The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base

increases at a rate of 3 cm2/sec. Find the rate at which the height of the box is decreasing when each side of

the base is 18 cm long. (Do not round your answer.)

A)125

81 cm/sec B)

125

486 cm/sec C)

125

8748 cm/sec D)

1

108 cm/sec

163) The radius of a right circular cylinder is increasing at the rate of 4 in./sec, while the height is decreasing at

the rate of 9 in./sec. At what rate is the volume of the cylinder changing when the radius is 6 in. and the

height is 12 in.?

A) -36π in.3/sec B) 252π in.3/sec C) -36 in.3/sec D) -42 in.3/sec

FROM THE BOOK

164) Do section 4.6 #1-9 odd, 15, 17, 21 - 31 odd, 35, 37, 43, 45

51

Answer KeyTestname: DERIVATIVES WORKSHEET 2 - USING THE DERIVATIVE

1) see section 4.1

2) Watch the video - pause it in various places and see what is happening.

3) Watch the video

4) Watch the video.

5) Watch the video

6) Watch the video

7) Watch the video

8) Watch the video

9) Watch the video

10) A

11) D

12) A

13) C

14) C

15) A

16) B

17) B

18) D

19) D

20) C

21) A

22) A

23) B

24) B

25) D

26) C

27) B

28) D

29) D

30) A

31) D

32) B

33) C

34) C

35) D

36) B

37) A

38) D

39) C

40) C

41) A

42) C

43) C

44) D

45) A

46) B

47) D

48) B

52


49) C

50) D

51) A

52) A

53) B

54) A

55) B

56) A

57) B

58) C

59) D

60) D

61) B

62) A

63) A

64) D

65) C

66) D

67) D

68) A

69) C

70) B

71) B

72) B

73) D

74) B

75) A

76) C

77) A

78) Yes. The point x = c is either a local maximum, a local minimum, or an inflection point. But, since f ′′(x) > 0 for all

x in the domain, there are no inflection points and the curve is everywhere concave up and thus cannot have a

local maximum. Hence, there is a local minimum at x = c.


80) D

81) A

53


82) a: both y′ and y′′ are undefined.

b: y′ =0 and y′′ > 0

c: y′ > 0 and y′′ = 0

d: y′ = 0 and y′′ = 0

e: y′ > 0 and y′′ = 0

f: y′ = 0 and y′′ < 0

g: y′ < 0 and y′′ = 0

83) (a). Moving towards to origin on (1, 2) and (5.7, 7); moving away from the origin on (0, 1), (2, 5.7), and (7, 10).

(b). Velocity is zero at the extrema. These occur at t ≈ 1 sec and t ≈ 5.7 sec.

(c). Acceleration is zero at the inflection points. These occur at t ≈2.3 sec, t ≈ 4 sec, t ≈ 5.1 sec, t ≈ 7 sec, and t ≈ 8.5

sec.

(d). Acceleration is positive where f(t) is concave up and negative where it is concave down. Acceleration is

positive on (2.3, 4), (5.1, 7), and (8.5, 10). Acceleration is negative on (0, 2.3), (4, 5.1), and (7, 8.5).

84)

Since f′′(x) = 1

x2 > 0 for all x > 0, then the function is everywhere concave up.


54


86) B

87) B

88) A

89) see answers in the book. You can also verify answers by graphing the function on your TI and do 2nd calc max

or min. Or you can go to WolframAlpha.com and enter in like: local extrema of x^3-x , or you can enter in like:

inflection points of x^3-x , or like: absolute maximum of y=-x^2+3 Try it!

90) local maximum at (0, 0)

local minimum at (1, -1)

increasing on (-∞, 0) ∪ (1, ∞)

decreasing on (0, 1)

x-4 -2 2 4

y

10

5

-5

-10

x-4 -2 2 4

y

10

5

-5

-10

91) global minimum at (-1, -12)

increasing on (-1, ∞)

decreasing on (-∞, -1)

x-4 -2 2 4

y

10

5

-5

-10

x-4 -2 2 4

y

10

5

-5

-10

55


92) global minima at (-3, 0) and (2, 0)

local maximum at (- 1

2,

625

16)

increasing on (-3, - 1

2) ∪ (2, ∞)

decreasing on (-∞, -3) ∪ (- 1

2, 2)

x-4 -2 2 4

y40

30

20

10

-10

x-4 -2 2 4

y40

30

20

10

-10

93) local minimum at (0, 0) (and a cusp)

local maximum at (4, 123

2)

increasing on (0, 4)

decreasing on (-∞, 0) ∪ (4, ∞)

x-10 -8 -6 -4 -2 2 4 6 8

y2018161412108642

-2-4-6-8

-10-12-14-16-18-20

x-10 -8 -6 -4 -2 2 4 6 8

y2018161412108642

-2-4-6-8

-10-12-14-16-18-20

56


94) local minimum at (2, -64)

local maximum at (-2, 64)

increasing on (-∞, -2) ∪ (2, ∞)

decreasing on (-2, 2) (but with a horizontal tangent at (0, 0))

x-4 -2 2 4

y100908070605040302010

-10-20-30-40-50-60-70-80-90

-100

x-4 -2 2 4

y100908070605040302010

-10-20-30-40-50-60-70-80-90

-100

95) no extrema

decreasing on (-∞, 0) ∪ (0, ∞)

x-10 -8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

x-10 -8 -6 -4 -2 2 4 6 8

y10

8

6

4

2

-2

-4

-6

-8

-10

57


96) local maximum at (-3, 54)

local minimum at (3, -54)

increasing on (-∞, -3) ∪ (3, ∞)

decreasing on (-3, 3)

x-10 -5 5 10

y200

100

-100

-200

x-10 -5 5 10

y200

100

-100

-200

97) A = 2025 m2

98) 60,000 m2

99)2 3

9

100) 45,000 yd2

101) 1

102) r = 1

2 +

1

2 5R, h = 2R

5 - 5

10

1/2

(Must satisfy r2 + 1

2h

2 = R2)

103) min = 150 3

4 cm2, max = 300 cm2

104) max = 10 2

105) S = 80

106) 32

107) max = 512

108) max = 432 in.2, min = 216 3

4 in.2

109) 180,625 m2 (using all the fence for the square corral)

110)4 3

9πR3

111) A = 7,200, width = 60, length = 120

112) S = 250

113) L2

114) x + 7y = 14

58


115) r > 1

2

116)5

2,

5

2

117) S = 1,152

118)5 10

3 ft3

119)2000π

9 3

120) 105 cm wide × 154 cm high

121) 12 3 ft2

122) 500

123) 30 in. × 30 in. × 40 in.

124) 12 in. × 18 in.

125) 4 1 + 22/3 3/2 m

126) 18 m3

127) The rectangular region should be 500

π m × 250 m, with the radius of each semicircular region being

250

π m.

128) If x , y represent the legs of the triangle, then x2 + y2 = 152.

Solving for y, y = 225 - x2

A(x) = xy = x 225 - x2

A'(x) = - x2

2 225 - x2 +

225 - x2

2

Solving A'(x) = 0, x = ± 15 2

2

Substitute and solve for y: (15 2

2)2

+ y2 = 225 ; y = 15 2

2 ∴ x = y.

129) A

130) C

131) C

132) B

133) D

134) C

135) D

136) A

137) C

138) D

139) D

140) A

141) C

142) 28 m × 28 m × 16 m

143) r = 3 in., h = 6 in.

144) 2000π in.3

59


145) - 1

4

146) (1, 1)

147) 3003

4 cm2

148) (-1, 0)

149) 18



152) A

153) D

154) C

155) C

156) B

157) C

158) A

159) A

160) C

161) D

162) C

163) B

164) See answers in the book

60

Integration Worksheet 1 - Understanding the Definite Integral







Estimate the value of the quantity.

1) The table shows the velocity of a remote controlled race car moving along a dirt path for 8 seconds.

Estimate the distance traveled by the car using 8 sub-intervals of length 1 with left-end point values.

Time

(sec)

Velocity

(in./sec)

0

1

2

3

4

5

6

7

8

0

10

16

12

22

25

27

12

5

A) 114 in. B) 124 in. C) 248 in. D) 129 in.

2) The table shows the velocity of a remote controlled race car moving along a dirt path for 8 seconds.

Estimate the distance traveled by the car using 8 sub-intervals of length 1 with right-end point values.

Time

(sec)

Velocity

(in./sec)

0

1

2

3

4

5

6

7

8

0

8

23

30

29

27

30

25

4

A) 176 in. B) 166 in. C) 172 in. D) 182 in.

1

3) Joe wants to find out how far it is across the lake. His boat has a speedometer but no odometer. The table

shows the boats velocity at 10 second intervals. Estimate the distance across the lake using right-end

point values.

Time

(sec)

Velocity

(ft/sec)

0

10

20

30

40

50

60

70

80

90

100

0

12

30

53

50

55

52

55

45

15

0

A) 3670 ft B) 367 ft C) 3770 ft D) 5500 ft

4) A piece of tissue paper is picked up in gusty wind. The table shows the velocity of the paper at 2 second

intervals. Estimate the distance the paper travelled using left-endpoints.

Time

(sec)

Velocity

(ft/sec)

0

2

4

6

8

10

12

14

16

0

8

12

6

21

26

16

10

2

A) 182 ft B) 202 ft C) 101 ft D) 179 ft

5) The velocity of a projectile fired straight into the air is given every half second. Use right endpoints to

estimate the distance the projectile travelled in four seconds.

Time

(sec)

Velocity

(m/sec)

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

139

134.1

129.2

124.3

119.4

114.5

109.6

104.7

99.8

A) 935.6 m B) 974.8 m C) 487.4 m D) 467.8 m

2

In the following problems, f(x) represents the velocity of an object in ft/sec at time x seconds moving on a straight

track. Estimate the distance the object has traveled during the given time interval using the given number of

rectangles and using left or right sums as instructed.

6) f(x) = x2 between x = 0 and x = 4 using a left sum with two rectangles of equal width.

A) 8 B) 38.75 C) 20 D) 40

7) f(x) = x2 between x = 0 and x = 1 using a right sum with two rectangles of equal width.

A) .625 B) .75 C) .125 D) .3145

8) f(x) = 1

x between x = 1 and x = 9 using a right sum with two rectangles of equal width.

A)56

45B)

24

5C)

8

15D)

56

5

9) f(x) = x2 between x = 2 and x = 6 using a left sum with four rectangles of equal width.

A) 86 B) 54 C) 69 D) 62

10) f(x) = 1

x between x = 1 and x = 6 using an left sum with two rectangles of equal width.

A)95

14B)

15

28C)

45

14D)

95

84

11) f(x) = x2 between x = 1 and x = 5 using a right sum with four rectangles of equal width.

A) 69 B) 41 C) 54 D) 30

3

Graph the function f(x) over the given interval. Partition the interval into 4 sub-intervals of equal length. Then add

to your sketch the rectangles associated with the left sums or right sums as directed. NOTE: These sums are called

Riemann sums, f(x) Δx∑ , where x is the left endpoint of a sub interval for a "left Riemann sum" and x is the right

endpoint for a "right Riemann sum". And Δx = right endpoint of interval - left endpoint of interval

number of sub intervals =

b - a

n

12) f(x) = 2x + 4, [0, 2], left-hand endpoint

x0.5 1 1.5 2

y8

7

6

5

4

3

2

1

x0.5 1 1.5 2

y8

7

6

5

4

3

2

1

A)

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

B)

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

C)

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

D)

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

x0.5 1 1.5 2

y

8

7

6

5

4

3

2

1

4

13) f(x) = -2x - 1, [0, 2], left-hand endpoint

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

A)

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

B)

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

C)

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

D)

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

x0.5 1 1.5 2

y

-1

-2

-3

-4

-5

-6

-7

-8

14) f(x) = x2 - 2, [0, 8], left-hand endpoint

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

5

A)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

B)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

C)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

D)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

15) f(x) = x2 - 2, [0, 8], right-hand endpoint

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

6

A)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

B)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

C)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

D)

x2 4 6 8

y646056524844403632282420161284

x2 4 6 8

y646056524844403632282420161284

16) f(x) = -3x2, [0, 4], left-hand endpoint

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

7

A)

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

B)

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

C)

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

D)

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

x1 2 3

y4

-4-8

-12-16-20-24-28-32-36-40-44-48-52-56-60

8

17) f(x) = cos x + 2, [0, 2π], left-hand endpoint

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

A)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

B)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

C)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

D)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

9

18) f(x) = cos x + 3, [0, 2π], right-hand endpoint

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

A)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

B)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

C)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

D)

xπ2

π 3π2

2π

y

5

4

3

2

1

xπ2

π 3π2

2π

y

5

4

3

2

1

Using the "Approximating Distance Traveled Using Tables and Graphs" applets

ESSAY.

19) Go to http://www.mathopenref.com/calcinttable.html Click "open in resizeable window" to enlarge the applet. Choose "#5 You Try" from the drop down menu.

Put in v(t) = 2t and select velocity to use: left Put in intervals = 4 Then press Enter.

a) Write down the table of t and v(t) values generated. (Note: t is in sec. and v(t) is in ft/sec.) Calculate the

the estimated total distance traveled using the left velocity values for each time sub interval.

b) What does the applet indicate as the total distance traveled? Did it match your answer?

10


Put in v(t) = 2t and select velocity to use: left Put in intervals = 8 Then press Enter.


the estimated total distance traveled using the left velocity values for each time sub interval.


c) Put in intervals = 16 . What does the applet indicate as the total distance traveled?

d) Put in intervals = 64 . What does the applet indicate as the total distance traveled?

e) Put in intervals = 128 . What does the applet indicate as the total distance traveled?

f) Put in intervals = 1000 . What does the applet indicate as the total distance traveled?

g) Do the total distance values seem to be converging to a value as the number of intervals gets larger?

What value?


Put in v(t) = 2t and select velocity to use: right Put in intervals = 8 Then press Enter.


the estimated total distance (ft) traveled using the right velocity values for each time sub interval.







What value?

h) In the previous problem the distance values increased toward a limit value, in this problem the values

decreased toward a limit value, why?


a) Put in various functions, starting and ending t values, number of intervals and experiment until you

really know what is going on with the applet. Show one such experimentation and comment on it.

b) Put in v(t) = sin(t) , start = 0 and end = 2pi , select velocity to use: right , intervals = 100 . What does the

applet show as the total distance traveled? (hint: E-17 means 10-17 a very small number!)

c) How can an object travel and end up having zero distance traveled? Explain what is really going on

here - what does the applet actually do in its calculations?

11

23) Go to http://www.mathopenref.com/calcintgraph.html This is a graphical applet. Click "open in resizeable window" to enlarge the applet. Choose "#5 You Try"

from the drop down menu.

Put in v(t) = t^2 and select velocity to use: left Put in intervals = 4 Then press Enter.

a) Copy the graph generated. (Note: t is in sec. and v(t) is in ft/sec.)

b) What does the applet indicate as the total distance traveled? (it says "pink area =")






What value?

24) Go to http://www.mathopenref.com/calcintgraph.html This is a graphical applet. Click "open in resizeable window" to enlarge the applet. Choose "#5 You Try"

from the drop down menu.

Put in v(t) = t^2 and select velocity to use: right Put in intervals = 4 Then press Enter.

a) Copy the graph generated. (Note: t is in sec. and v(t) is in ft/sec.)

b) What does the applet indicate as the total distance traveled? (it says "pink area =")






What value?

h) In the previous problem the distance values increased toward a limit value, in this problem the values

decreased toward a limit value, why? Does the function being increasing or decreasing on the interval

have anything to do with whether the left estimates are over or under?

i) Use the applet to experiment with increasing functions and decreasing functions until you know enough

to state a rule concerning the relationship between these 6 concepts: v(t) increasing, v(t) decreasing, Left

estimate, Right estimate, over estimate, under estimate.

FROM THE BOOK

25) Read and take notes on section 5.1

26) Do section 5.1 #1 - 23 odd

The Definite Integral


Evaluate the sum.

27)7

k = 1

k∑

A) 7 B) 14 C) 56 D) 28

12

28)14

k = 1

k3∑

A) 11,025 B) 1015 C) 2744 D) 3150

29)25

k = 4

6∑

A) 146 B) 132 C) 150 D) 126

30)6

k = 1

k2 - 8∑

A) 28 B) 83 C) 43 D) 91

FROM THE BOOK

31) Read section 5.2 and take notes on the Definite Integral, The Definite Integral as an Area, and More

general Riemann Sums

ESSAY.

32) a) Go to the internet and find out info on Bernhard Riemann and write a brief paragraph about him and hisRiemann sums.

b) Go to http://mathworld.wolfram.com/RiemannSum.html to see a Riemann sum applet(similar idea to the applet used above but has a different layout). You can use this applet in the some ofthe problems below.

33) There is a great Riemann sum program for the TI-84 . Go tohttp://www.calcblog.com/riemann-sum-program-ti83-ti84/ There you see a nice tutorial

about Riemann sums then you scroll down to the Using the RIEMANN Program section. It shows how to

download the program into your TI-84 from your computer, and various links show you how to get the

program into your computer in the first place. (I did it and it worked!) Then it shows how to use the

program. You can use this program in the some of the problems below. You must have this program or a simplerversion of it in your TI-84.

FROM THE BOOK

34) Do section 5.2 #1, 3, 7, 9; For [11-21 odd, 27] you either use your TI Riemann program or use the wolfram

Riemann sum program mentioned above; #23, 25, 29, 31, 33

13


Graph the integrand and use areas to evaluate the integral. Then run a Riemann program or applet with n = 100 sub

intervals to verify your answer. (n = 100 is a large enough amount of intervals to give a pretty close estimate for

these problems below because the widths of the intervals are not too large. (Only the problem with 16 - x2 may

have an answer that is significantly far off from the Riemann estimate with n=100 sub intervals - learn why in class)

35)7

-1

4 dx∫

A) 32 B) 8 C) 24 D) 16

36)10

6

x dx∫

A) 8 B) 16 C) 64 D) 32

37)4

0

8x dx∫

A) 128 B) 8 C) 32 D) 64

38)5

-5

(2x + 10) dx∫

A) 100 B) 20 C) 50 D) 200

39)9

-2

x dx∫

A)85

2B) 85 C) 11 D)

77

2

40)4

-4

16 - x2 dx∫

A) 4π B) 16π C) 16 D) 8π

The Fundamental Theorem of Calculus

14

FROM THE BOOK

41) Read section 5.3 and takes notes on The Fundamental Theorem of Calculus, and on the Definite Integral as

an Average. AND know the proof steps of The Fundamental Theorem of Calculus for a test . The proof

steps follow after the statement of the theorem.

The F.T.O.C says that if you can find an antiderivative F(t) for your function f(t), then instead of

computing area estimates or limits of sums to find b

a

f(t)∫ dt you can just subtract F(b) - F(a) . A very

simple subtraction! And this theorem ties the derivative ( F '(t) = f(t) ) to sums ( f(t) △t∑ ) in a very slick

way!

ESSAY.

42) Go to http://calculusapplets.com/fundtheorem.html a) Read and take notes on the first paragraph.

b) In the "Try the following" section, do and answer #3 Hit "launch the presentation" to get a large

re-sizeable view.

c) For the #3 Parabola example, we have f(x) = x2 and its derivative f '(x) = 2x . On the interval [1,3] as

shown, compute the change in f(x) , that is, compute f(3) - f(1).

d) Now, using your Riemann sum program or an applet, find 3

1

2x dx∫ using n = 100 sub intervals. Do

you get the same (or nearly the same) answer as in (c)?

e) What is the f(b) - f(a) total change value the applet shows?

f) What is the 3

1

2x dx∫ area value the applet shows? Are the answers to e and f equal?

g) Do your own function example f(x) and make up your own interval and verify that f(b) - f(a) equals

b

a

f '(x) dx∫

43) The rate at which the world's oil is being consumed is r(t) = 32e.05t billion barrels per year where t = years

after 2004.

a) Set up a definite integral that will give the total amount of barrels of oil consumed from 2004 to 2012

(assuming the rate functions remains accurate over that time)

b) Find(estimate) the value of the integral by using technology methods with n = 100 intervals

c) Now you want to answer the problem by using the Fundamental Theorem of Calculus but you need an

antiderivative (a function whose derivative is 32e^(.05t) ) , call it f(t), so you can find f(b) - f(a). Go to

http://www.wolframalpha.com and put in: antiderivative 32e^(.05t) and an antiderivative shows up.

15

44) A spill of radioactive iodine occurs and the rate of decay is given by r(t) = 2.4e-0.004t millirems/hour

where t = 0 hours is the time of the spill.

a) An acceptable rate of radiation is 0.6 millirems/hour. How many hours will it take to reach that rate?

b) Set up a definite integral that will give the total amount of radiation (in millirems) that will be emitted

from t = 0 hours to the time of the acceptable rate level?

c) Find(estimate) the value of the integral by using technology methods with n = 500 intervals


antiderivate (a function whose derivative is 2.4e-0.004t ) , call it f(t), so you can then find f(b) - f(a). Go to

http://www.wolframalpha.com and put in: antiderivative 2.4e^(-.004t) and an antiderivative shows

up.

45) A runner is jogging on a straight track and is speeding up and slowing down with velocity function v(t) =

cos(t) + 5 miles/hr where t is the hours after he/she started.

a) Set up a definite integral that will give the total distance he/she ran from the time period of t = 1 hour to

t = 2 hours.

b) Find(estimate) the value of the integral by using technology methods with n = 100 intervals


antiderivate (a function whose derivative is cos(t)+5 , call it f(t), so you can find f(b) - f(a). Go to

http://www.wolframalpha.com and put in: antiderivative cos(t)+5 and an antiderivative shows up.

FROM THE BOOK

46) Do section 5.3 #15, 17

ESSAY.

47) Go to http://calculusapplets.com/aveval.html and click "hide answer". Put in f(x) = x^2 and put in

ymax = 5 and hit enter.

You want to find the average value of f(x) on a = 0 to b = 2. Think of the function as a velocity of a man

walking in ft/sec. Think of the average value of the function as the constant velocity he could have walked

to cover the same ground as if he walked according to the function.

a) Click on the black square at the origin and move it up to what you think the average value would be.

(Another way to think of it is: Move the black square up to a point where the area under the resulting

rectangle equals the yellow area under the f(x) curve.) What is your average value guess?

b) Click "show answer" and see what the applet says. Write that down.

c) Compute the average value of f(x) = x2 on x = 0 to 2 by using the Average Value of f formula in section

5.3 in the book: Average value of f on a to b = 1

b-a

b

a

f(x) dx∫ . Use F(x) = x3

3 as an antiderivative of f(x)

= x2 . Did the answer match with the applet's answer for the average value?


Find the average value of the function over the given interval. If you can think of an antiderivative for the function

then use it, if not, go to http://www.wolframalpha.com and put in: definite integral and then put in the function

to integrate and the limits of integration given.

48) f(x) = 4x on [5, 7]

A) 12 B) 48 C) 24 D) 96

16

49) f(x) = 4 - x on [0, 4]

A) 32 B) 2 C) 8 D) 4

50) f(x) = x on [-6, 6]

A) 6 B)3

2C) 36 D) 3

51) y = x2 - 2x + 4; [0, 2]

A) 3 B)22

3C) 4 D)

10

3

52) y = 6 - x2; [-5, 2]

A) - 1

3B) -

13

3C) 3 D)

81

7

FROM THE BOOK

53) Do section 5.3 #9, 11, 19, 27, 31, 35, 37, 39, 41 (for 19, 27, 39, 41 use http://www.wolframalpha.com to

evaluate the definite integral involved, or, use your Riemann program with n = 100 sub-intervals))

54) Read and take notes on section 5.4 Properties of Limits of Integration, Properties of Sums and Constant

Multiples of the Integrand, Area Between Curves, Using the Fundamental Theorem to Compute Integrals


Use your technology methods to estimate the area under the graph of the given function on the stated interval as

instructed. If you can think of an antiderivative, F(x), then use that and the FTOC.

55) f(x) = x2 between x = 0 and x = 2 using a lower sum with two rectangles of equal width.

A) 4.5 B) 1 C) 5 D) 2.5

56) f(x) = x2 between x = 0 and x = 1 using an upper sum with two rectangles of equal width.

A) .125 B) .75 C) .3145 D) .625

57) f(x) = 1

x between x = 1 and x = 8 using a lower sum with two rectangles of equal width.

A)2359

20736B) -

1

3C) -

1

6D)

2359

6912

58) f(x) = 1

x between x = 2 and x = 5 using a upper sum with two rectangles of equal width.

A)117

980B) -

29

70C)

39

980D) -

29

35

59) f(x) = x2 between x = 4 and x = 8 using an upper sum with four rectangles of equal width.

A) 174 B) 149 C) 126 D) 165

17

Solve the problem.

60) Suppose that 2

1

f(x) dx∫ = -2. Find 2

1

6f(u) du∫ and 2

1

- f(u) du∫ .

A) 4; -2 B) 6; 2 C) -12; - 1

2D) -12; 2

61) Suppose that -1

-4

g(t) dt∫ = -12. Find -1

-4

g(x)

-12 dx∫ and

-4

-1

- g(t) dt∫ .

A) 1; 12 B) 1; -12 C) 0; -12 D) -1; 12

62) Suppose that f and g are continuous and that 8

4

f(x) dx∫ = -5 and 8

4

g(x) dx∫ = 9.

Find 8

4

4f(x) + g(x) dx∫ .

A) 13 B) -11 C) 16 D) 31


2

f(x) dx∫ = -2 and 6

2

g(x) dx∫ = 7.

Find 6

2

f(x) - 2g(x) dx∫ .

A) -18 B) 12 C) -16 D) -9


2

f(x) dx∫ = -5 and 6

2

g(x) dx∫ = 7.

Find 2

6

g(x) - f(x) dx∫ .

A) 12 B) -2 C) -12 D) 2

65) Suppose that h is continuous and that 5

-4

h(x) dx∫ = 7 and 6

5

h(x) dx = -9.∫ Find 6

-4

h(t) dt∫ and

-4

6

h(t) dt.∫

A) -2; 2 B) 2; -2 C) 16; -16 D) -16; 16

66) Suppose that f is continuous and that 4

-4

f(z) dz = 0∫ and 5

-4

f(z) dz = 4.∫ Find -5

4

2f(x) dx∫ .

A) -2 B) 8 C) -8 D) -4

18

Evaluate the integral. Think of antiderivative or use Wolframalpha to get one, then use the FTOC.

67)13

1

x dx∫

A) 13 - 1 B) 12 C) - 6 D) 6

68) 2π

3π/2

θ dθ∫

A)9π2

8B)

7π2

8C)

π2

2D)

π2

8

69)1/8

0

t2 dt∫

A)1

1536B) -

1

1536C) -

1

8D) 1536

70)

313

0

x2 dx∫

A) 169 B)3

13

3C)

13 13

3D)

13

3

71)2π

π

θ2 dθ∫

A)π3

3B)

27π3

24C)

7π3

3D)

π3

24

72)12

8

7 dx∫

A) 84 B) 28 C) -76 D) 0

73)15

2

z- 15 dz∫

A) - 19

2 15 B) -

19

2 + 2 15 C) -

15

2 15 D) - 15

FROM THE BOOK

74) Do section 5.4 #3-15 odd (for 5-15 odd, set up the integral, then evaluate using technology methods such

as the calculator program with n= 100 sub-intervals, or http://www.wolframalpha.com ,

19-25 odd, 37, 49; Review for Chapter 5 #5-13 odd, 17, 19, 21, 33, 47

19

75) Read and take notes on section 6.1 Visualizing antiderivatives using slopes, Computing values of an

Antiderivative using the FTOC.

76) Do section 6.1 #1-8 ALL, 15, 19, 21, 22, 23

Now we learn how to find Antiderivatives by hand (analytically, or,

algebraically)

77) Read and take notes on section 6.2 (all of it!)

78) Do section 6.2 #1 - 77 odd, 81


Evaluate the integral.

79)16

0

2 x∫ dx

A) 128 B) 192 C) 16 D)256

3

80)6

-2

6x5∫ dx

A) 1280 B) 279,552 C) -46,592 D) 46,592

81)4

1

t2 + 1

t dt∫

A)92

5B)

77

5C)

72

5D) 32

82)π/2

0

18 sin x dx∫

A) 0 B) 1 C) 18 D) -18

Find the total area of the region between the curve and the x-axis.

83) y = 2x + 7; 1 ≤ x ≤ 5

A) 52 B) 18 C) 26 D) 9

84) y = 2x - x2; 0 ≤ x ≤ 2

A)2

3B)

4

3C)

7

3D)

5

3

20

85) y = 3

x3; 1 ≤ x ≤ 3

A)1

3B)

4

3C)

1

2D) 3

86) y = -x2 + 9; 0 ≤ x ≤ 5

A)98

3B)

5

9C)

10

9D)

10

3

87) y = 1

x; 1 ≤ x ≤ 4

A) 2 B) 4 C)1

4D)

1

2

Find the average value of the function over the given interval.

88) f(x) = 10x on [1, 3]

A) 40 B) 10 C) 20 D) 80

89) f(x) = 4 - x on [0, 4]

A) 2 B) 32 C) 4 D) 8

90) y = x2 - 2x + 6; [0, 2]

A)16

3B) 5 C) 6 D)

34

3

91) y = 3 - x2; [-2, 2]

A) 3 B)5

3C) -

1

3D) 0

Construction of Antiderivatives for functions we can't use any techniques to obtain.

(There are still more techniques of integration to learn but still there will be functions forwhich we just cannot obtain a closed form antiderivative - we use the construction method

to find values of the antiderivative function in that case.)

FROM THE BOOK

92) Read and take notes on section 6.4 The Second Fundamental Theorem of Calculus. Know the proof for a

test of the statement F(x) = x

a

f(t)dt∫ is an antiderivative of f(x)

21

ESSAY.

93) a) Evaluate x

a

2t dt∫ using the FTOC and call the result F(x). ("a" is any constant)

b) Take the derivative of F(x) you found in part (a) and show you get 2x and thus demonstrating that

x

a

2t dt∫ is an antiderivative for 2x.

94) a) Make a table of values of F(x) = x

0

2t dt∫ for x = 0, 1, 2, 3, 4 by using your Riemann program with

n=100 sub-intervals (use left sums) to evaluate the integral at each x value, one at a time.

b) Graph your table of values. Does the graph of F(x) look like it's derivative function would be 2x ?

95) a) Make a table of values of F(x) = x

0

e-t2dt∫ for x = 0, .2, .4, .6, .8, 1 by using your Riemann program

with n=100 sub-intervals (use left sums) to evaluate the integral at each x value.

b) Graph your table of values. Does the graph of F(x) look like it's derivative function would be e-x2 ?

FROM THE BOOK

96) Do section 6.4 #5-19 odd, 25-35 odd

97) Read and take notes on section 6.5 Equations of Motion. Know (for a test) the derivation steps of how we

start with the acceleration of an object due to gravity, -g, and end up knowing the position function of the

object, s(t).

98) Do section 6.5 #3 (use g = 32 ft

sec2 ), 7, 8

ESSAY.

99) On Mars the acceleration due to gravity is 12 ft

sec2 . (On Earth gravity is much stronger at 32

ft

sec2 . ) In

the movie, John Carter, it shows Carter leaping about 100 feet up into the air on Mars. John Carter is an

Earth man who has been transported to Mars so his leg muscles have been built to handle Earth's gravity

while Mars gravity is a lot less. On Earth, Michael Jordan (a famous basketball player) had a vertical jump

velocity of 16 ft/sec. Suppose John Carter could triple that initial jump velocity due to being on Mars, so

his initial velocity would be v0 = 48 ft/sec .

a) How high could he jump on Mars?

b) How long could he stay in the air before he hit the ground?

c) The movie shows Carter jumping about 100 ft. high. Is that about right by the Calculus?

d) What would his speed be when he hit the ground?

22

100) referring to the previous problem, suppose the planet John Carter transported to had an acceleration due

to gravity of 2 ft

sec2 and his initial jump velocity was still v0 = 48 ft/sec .

a) How high could he jump under these conditions?

b) How long could he stay in the air before he hit the ground?

c) What would his speed be when he hit the ground?

More advanced Integration Methods to find antiderivatives by hand (algebraically) -

Integration by Substitution

FROM THE BOOK

101) Read and take notes on section 7.1 but where they use the letter "w" for the substitution, most people

(including me) use the letter "u" , so use "u" . Know the proof steps of why substitution works for a test.


Evaluate the integral using the given substitution.

102) x cos (4x2)∫ dx, u = 4x2

A) sin(4x2) + C B)x2

2 sin (4x2) + C C)

1

8 sin (4x2) + C D)

1

u sin (u) + C

103) 2 - sin t

2

2cos

t

2∫ dt, u = 2 - sin t

2

A)1

32 - sin

t

2

3sin

t

2 + C B)

2

32 - cos

t

2

3 + C

C) 2 2 - sin t

2

3 + C D) -

2

32 - sin

t

2

3 + C

104) x4(x5 - 4)5∫ dx , u = x5 - 4

A)1

6(x5 - 4)6 + C B)

1

30x30 - 4 + C C)

1

30(x5 - 4)6 + C D)

1

20(x5 - 4)4 + C

105)8s3 ds

8 - s4∫ , u = 8 - s4

A) -4s3 8 - s4 + C B)-2

2 8 - s4 + C C)

4s4

8 - s4D) -4 8 - s4 + C

106)dx

7x + 2∫ , u = 7x + 2

A)7

2

1

7x + 2 + C B) 2 7x + 2 + C C)

2

77x + 2 + C D)

1

7(7x + 2)3/2 + C

23

107) 18(6x - 8)-6∫ dx , u = 6x - 8

A) - 6

5(6x - 8)-5 + C B) -

3

5(6x - 8)-5 + C C) -

3

7(6x - 8)-7 + C D) (6x - 8)-5 + C

FROM THE BOOK

108) Do section 7.1 #3-39 odd, 47-75 odd, 91-97 odd, 101, 103, 109


Evaluate the integral.

109)x dx

(7x2 + 3)5∫

A) - 7

3(7x2 + 3)-4 + C B) -

1

56(7x2 + 3)-4 + C

C) - 7

3(7x2 + 3)-6 + C D) -

1

14(7x2 + 3)-6 + C

110)sin t

(3 + cos t)4 dt∫

A)3

(3 + cos t)3 + C B)

1

(3 + cos t)3 + C C)

1

5(3 + cos t)5 + C D)

1

3(3 + cos t)3 + C

111)ln x7

x dx∫

A)1

ln x7 + C B)

1

2(ln x7)2 + C C)

1

14(ln x7)2 + C D)

1

7(ln x7)2 + C

112)dx

x 9x2 - 6∫

A)6

6 sin-1

1

26 x + C B)

6

6 sec-1

1

26 x + C

C)1

3 sec-1 3x - 6 + C D)

1

3 sec-1 3 x + C

113)1

t2 sin

3

t + 5∫ dt

A)1

3 cos

3

t + 5 + C B) -cos

3

t + 5 + C C) -

1

3 cos

3

t + 5 + C D) 3 cos

3

t + 5 + C

24

114) sin (8x - 4) dx∫A)

1

8 cos (8x - 4) + C B) 8 cos (8x - 4) + C

C) -cos (8x - 4) + C D) - 1

8 cos (8x - 4) + C

115)dx

xln x3∫

A) ln x3+ C B)1

3ln ln x3 + C C)

1

3ln x3 + C D) ln ln x3 + C

116) x3∫ x4 + 2 dx

A)2

3x4 + 2 3/2 + C B) -

1

2x4 + 2 -1/2 + C

C)1

6x4 + 2 3/2 + C D)

8

3x4 + 2 3/2 + C

117) x5(x6 - 6)4∫ dx

A) x6 - 6 5 + C B)x6 - 6 3

18 + C C)

x6 - 6 5

6 + C D)

x6 - 6 5

30 + C

ESSAY.

Solve the problem.

118) Evaluate x2∫ 5x3 + 1 dx.

119) Evaluate t2

16t3 + 5∫ dt.

120) Evaluate ( x + 4)3

3 x∫ dx.


Use the substitution formula to evaluate the integral.

121)1

0

x + 16∫ dx

A) 17 17 - 64 B)34

317 -

128

3C)

51

217 -

51

2D)

34

317

25

122)0

-1

2t

2 + t2 3 dt∫

A) - 5

72B)

5

72C) -

5

18D) -

5

36

123)0

-1

3t

4 + t2 4 dt∫

A) - 61

8000B)

61

16000C) -

183

8000D) -

61

16000

124)1

0

4 r dr

4 + 2r2∫

A) - 2 6 + 4 B) 6 - 2 C) 2 6 - 4 D)6

2 - 1

125)4

1

4 - x

x dx∫

A) - 5

2B) 5 C) 10 D)

5

2

126)π

0

(1 + cos 5t) 2 sin 5t dt∫

A)1

15B)

8

3C)

8

15D)

1

5

127)π/2

0

cos x

(4 + 2 sin x)3 dx∫

A) - 5

288B)

5

288C)

5

576D) -

15

64

128)π/8

0

(1 + etan 2x)∫ sec2 2x dx

A) e B) - e

2C) 2e D)

e

2

129)3π/2

π

sin θ dθ

2 + cos θ∫

A) -ln 2 B) ln 3 C) 0 D) -ln 3

26

130)3π/4

0

tan x

3∫ dx

A)3 ln 2

2B)

3 2

2C)

-3 ln 2

2D)

-3 2

2

131)ln 3/4

0

4 e4x dx

1 + e8x∫

A)π

12B) -

π

6C) -

π

12D)

π

6

Find the area enclosed by the given curves.

132) Find the area of the region between the curve y = 6x/(1 + x2) and the interval -2 ≤ x ≤ 2 of the x-axis.

A) 6 ln 5 B) 6 e5 C) ln 5 D) 0

133) Find the area of the region between the curve y = 53-x and the interval 0 ≤ x ≤ 2 on the x-axis.

A) 120 ln 5 B)120

ln 5C)

125

ln 5D) 125

27

Answer KeyTestname: INTEGRATION WORKSHEET 1 - UNDERSTANDING THE DEFINITE INTEGRAL

1) B

2) A

3) A

4) B

5) D

6) A

7) A

8) A

9) B

10) C

11) C

12) C

13) D

14) C

15) D

16) C

17) B

18) D

19) The applet shows the answer

20) The applet shows the answers

21) The applet shows the answers. For (h) discuss in class.

22) see applet

23) The applet shows the answers

24) The applet shows the answers. For (i) discuss in class.

25)


27) D

28) A

29) B

30) C

31) section 5.2

32)

33)

34) see answers in the book. See your Riemann program or the Riemann applet

35) A

36) D

37) D

38) A

39) A

40) D

41)

42) see applet

43) a) 8

0

32e.05t∫ dt billion barrels b) 314.1 billion barrels (using a left sum, you could use a right sum or average

the two) c) 314.8 billion barrels

28


44) a) 346.6 hours b) 346.6

0

2.4e-0.004t∫ dt millirems c) 450.6 millirems (using a left sum, you could use a right

sum or average the two) d) 450.016 millirems

45) a) 2

1

(cos(t)+5)∫ dt miles b) 5.07 miles (using a left sum, you could use a right sum or average the two) c) 5.068

miles


47) a) your guess b) 1.3333333 c) 4

3 yes it matched exactly.

48) C

49) B

50) D

51) D

52) A


54)

55) B

56) D

57) C

58) B

59) A

60) D

61) B

62) B

63) C

64) C

65) A

66) C

67) D

68) B

69) A

70) D

71) C

72) B

73) B


75)

76) See answers in the book, for even numbers discuss in class.

77)


79) D

80) D

81) C

82) C

83) A

84) B

29


85) B

86) D

87) A

88) C

89) A

90) A

91) B

92)

93) discuss in class.

94) discuss in class.



97)

98) see answers in the book, for #8 discuss in class

99) a) s = 96 ft. max height b) 8 seconds in the air. c) Yes, the movie was in line with the Calculus!! d) -48 ft/sec ,

that's 48 ft/sec downward = 32.7 miles/hour

100) a) s = 576 ft. max height b) 48 seconds in the air c) -48 ft/sec , that's 48 ft/sec downward = 32.7 miles/hour

101)

102) C

103) D

104) C

105) D

106) C

107) B

108) see answers in the book.

109) B

110) D

111) C

112) B

113) A

114) D

115) B

116) C

117) D

118)2

45(5x3 + 1)3/2 + C

119)16t3 + 5

24 + C

120)( x + 4)4

6 + C

121) B

122) A

123) D

124) C

125) B

126) C

127) C

30


128) D

129) A

130) A

131) A

132) A

133) B

31

Integration Worksheet 2 - Using the Definite Integral






FROM THE BOOK

1) Read and take notes on section 8.1 Areas and Volumes

2) Do section 8.1 #1-8 ALL For # 1-6 find the area first using Calculus then by simple known geometric area

formulas - see if the answers match. Realize that you can always evaluate a definite integral using your

Riemann program or the wolfram site to verify answers. For #4,5 set up the integral then use

http://www.wolframalpha.com to get the antiderivative function.


Find the area of the shaded region.

3) f(x) = x3 + x2 - 6x

x-4 -2 2 4

y30

20

10

-10

-20

-30(-4, -24)

(0, 0)

(3, 18)

x-4 -2 2 4

y30

20

10

-10

-20

-30(-4, -24)

(0, 0)

(3, 18)

g(x) = 6x

A)343

12B)

81

12C)

937

12D)

768

12

1

4) f(x) = -x3 + x2 + 16x

g(x) = 4x

x-4 -2 2 4 6

y

30

20

10

-10

-20

-30

(-3, -12)

(0, 0)

(4, 16)

x-4 -2 2 4 6

y

30

20

10

-10

-20

-30

(-3, -12)

(0, 0)

(4, 16)

A)1153

12B)

343

12C)

937

12D) -

343

12

5) y = x2 - 4x + 3

x-3 -2 -1 1 2 3 4 5

y8

6

4

2

-2

-4

-6

-8

x-3 -2 -1 1 2 3 4 5

y8

6

4

2

-2

-4

-6

-8

y = x - 1

A)41

6B) 3 C)

9

2D)

25

6

2

6)

x1 2

y

1

-1

-2

x1 2

y

1

-1

-2

y = x2 - 2x

y = -x4

A) 2 B)76

15C)

7

15D)

22

15

7) y = 2x2 + x - 6 y = x2 - 4

x-3 -2 -1 1 2 3

y5

4

3

2

1

-1

-2

-3

-4

-5

-6

-7

(2, 4)

x-3 -2 -1 1 2 3

y5

4

3

2

1

-1

-2

-3

-4

-5

-6

-7

(2, 4)

A)11

6B)

9

2C)

19

3D)

8

3

3

8)

x2 4 6 8 10

y6

4

2

-2

-4

-6

x2 4 6 8 10

y6

4

2

-2

-4

-6

y = x - 4

y = 2x

A)32

3B)

64

3C) 32 D)

128

3

9) y = x4 - 32

x-4 -3 -2 -1 1 2 3 4

y5

-5

-10

-15

-20

-25

-30

-35

-40

x-4 -3 -2 -1 1 2 3 4

y5

-5

-10

-15

-20

-25

-30

-35

-40

y = -x4

A)256

5B)

516

5C)

512

5D)

2816

5

4

10)

x1 2 3

y3

2

1

-1

-2

-3

x1 2 3

y3

2

1

-1

-2

-3

y = 2

y = 2 sin(πx)

A) 8 B)4

πC) 4 +

4

πD) 4

11) y = sec2 x

xπ4

π2

y

3

2

1

xπ4

π2

y

3

2

1y = cos x

A)2

2B) 2 - 2 C) 1 -

2

2D) 1 + 2

Find the area enclosed by the given curves.

12) y = 2x - x2, y = 2x - 4

A)34

3B)

31

3C)

37

3D)

32

3

13) y = x, y = x2

A)1

3B)

1

2C)

1

6D)

1

12

14) y = x3, y = 4x

A) 8 B) 16 C) 2 D) 4

5

15) y = 1

2 x2, y = -x2 + 6

A) 32 B) 4 C) 8 D) 16

16) y = - 4sin x, y = sin 2x, 0 ≤ x ≤ π

A) 4 B)1

2C) 16 D) 8

17) Find the area of the region in the first quadrant bounded by the line y = 8x, the line x = 1, the curve y = 1

x,

and the x-axis.

A)5

4B)

3

2C) 6 D)

3

4

18) Find the area of the region in the first quadrant bounded on the left by the y-axis, below by the line y = 1

3x,

above left by y = x + 4, and above right by y = - x2 + 10.

A)73

6B)

39

2C) 15 D)

39

4

19) Find the area between the curves y = ln x and y = ln 2x from x = 1 to x = 5.

A) ln 2 B) ln 32 C) ln 16 - 8 D) ln 16

20) Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = e2x,

below by the curve y = ex, and on the right by the line x = ln 4.

A) 4 ln 4 B) 4 C)9

2D)

27

2

21) Find the area of the region between the curve y = 2x/(1 + x2) and the interval -3 ≤ x ≤ 3 of the x-axis.

A) 2 e10 B) ln 10 C) 2 ln 10 D) 0

FROM THE BOOK

22) Do section 8.1 #9-14 ALL, 23-28 ALL

23) Read and take notes on section 8.2 Volumes of Revolution, Volumes of regions with a known

cross-section, Arc Length (not parametric)

24) Do section 8.2 #1-15 odd, 21-41 odd, 45, 47, 49

6


Find the volume of the described solid.

25) The solid lies between planes perpendicular to the x-axis at x = 0 and x = 6. The cross sections

perpendicular to the x-axis between these planes are squares whose bases run from the parabola y = - 3 x

to the parabola y = 3 x.

A) 630 B) 648 C) 108 D) 324

26) The solid lies between planes perpendicular to the x-axis at x = -2 and x = 2. The cross sections

perpendicular to the x-axis between these planes are squares whose bases run from the semicircle

y = - 4 - x2 to the semicircle y = 4 - x2.

A)64

3B)

32

3C)

128

3D)

16

3

27) The base of the solid is the disk x2 + y2 ≤ 25. The cross sections by planes perpendicular to the y-axis

between y = - 5 and y = 5 are isosceles right triangles with one leg in the disk.

A)1000

3B)

250

3C)

1250

3D)

2000

3

28) The solid lies between planes perpendicular to the x-axis at x = - 2 and x = 2. The cross sections

perpendicular to the x-axis are semicircles whose diameters run from y = - 4 - x2 to y = 4 - x2.

A)16

3π B)

64

3π C)

32

3π D)

8

3π

29) The solid lies between planes perpendicular to the x-axis at x = - 4 and x = 4. The cross sections

perpendicular to the x-axis are circular disks whose diameters run from the parabola y = x2 to the

parabola y = 32 - x2.

A)256

3π B)

8192

5π C)

16384

15π D)

8192

15π

30) The base of a solid is the region between the curve y = 3cos x and the x-axis from x = 0 to x = π/2. The

cross sections perpendicular to the x-axis are squares with bases running from the x-axis to the curve.

A) 2π B)9

2π C)

3

2π D)

9

4π

7

Find the volume of the solid generated by revolving the shaded region about the given axis.

31) About the x-axis

x1 2 3

y10987654321

x1 2 3

y10987654321

y = - 2x + 4

A) 12π B)224

3π C)

32

3π D)

64

3π


x1 2 3 4 5

y20

16

12

8

4

x1 2 3 4 5

y20

16

12

8

4 y = 9 - x2

A)1053

5π B)

648

5π C)

3159

5π D) 18π

33) About the y-axis

x1 2 3 4 5 6 7 8

y8

7

6

5

4

3

2

1

x1 2 3 4 5 6 7 8

y8

7

6

5

4

3

2

1

x = 6y/7

A) 98π B) 21π C) 168π D) 84π

8


x1 2 3 4 5 6

y6

5

4

3

2

1

x1 2 3 4 5 6

y6

5

4

3

2

1

y = 5x

A)25

3π B) 25π C) 50π D) 625π


x1 2 3 4 5 6

y6

5

4

3

2

1

x1 2 3 4 5 6

y6

5

4

3

2

1

x = y2

3

A)27

5π B)

108

5π C)

45

2π D) 18π


x1 2 3 4

y2018161412108642

x1 2 3 4

y2018161412108642

y = 4 - x2

A)64

15π B)

8

3π C)

224

15π D)

256

15π

9


xπ2

y4

2

xπ2

y4

2

y = 3 sin x

A)9

2π2 - 9π B)

9

2π2 C)

9

2π2 - 3π D)

9

2π2 + 9π

Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the

x-axis.

38) y = x, y = 0, x = 2, x = 6

A) 16π B) 20π C)208

3π D)

4

3π

39) y = x2, y = 0, x = 0, x = 6

A)7776

5π B) 72π C) 324π D) 1944π

40) y = 2x + 3, y = 0, x = 0, x = 1

A) 2π B) π C) 4π D)3π

2

41) y = 1

x, y = 0, x = 1, x = 8

A)7

16π B) πln 8 C)

3

8π D)

7

8π

42) y = 2x, y = 2, x = 0

A) 1π B)8

3π C)

4

3π D) 6π

43) y = - 5x + 10, y = 5x, x = 0

A) 50π B) 150π C) 25π D) 10π

44) y = 3x, y = 3, x = 0

A) 9π B) 18π C)27

4π D)

27

2π

10

45) y = x2, y = 16, x = 0

A)6144

5π B)

4096

5π C)

1024

5π D)

128

3π

46) y = 9

x, y = - x + 10

A)512

3π B) 72π C)

1000

3π D) 128π

Find the volume of the solid generated by revolving the region about the given line.

47) The region bounded above by the line y = 16, below by the curve y = 16 - x2, and on the right by the line

x = 4, about the line y = 16

A)7168

15π B)

64

3π C)

8192

15π D)

1024

5π

48) The region in the second quadrant bounded above by the curve y = 4 - x2, below by the x-axis, and on the

right by the y-axis, about the line x = 1

A)256

15π B)

32

3π C)

56

3π D) 8π

Solve the problem.

49) The disk (x - 6)2 + y2 ≤ 4 is revolved about the y-axis to generate a torus. Find its volume. (Hint:

2

-2

4 - y2 dy∫ = 2π, since it is the area of a semicircle of radius 2.)

A) 24π2 B) 48π2 C) 12π2 D) 24π

50) The hemispherical bowl of radius 5 contains water to a depth 1. Find the volume of water in the bowl.

A)7

3π B)

14

3π C)

139

3π D) 88π

51) A water tank is formed by revolving the curve y = 5x4 about the y-axis. Find the volume of water in the

tank as a function of the water depth, y.

A) V(y) = 3π

2 5y3/2 B) V(y) =

2π

3 5y3/2 C) V(y) =

π

2 5y1/2 D) V(y) =

π

9y9

52) A right circular cylinder is obtained by revolving the region enclosed by the line x = r, the x-axis, and the

line y = h, about the y-axis. Find the volume of the cylinder.

A) πrh B) πrh2 C) 2πr2h D) πr2h

53) Find the volume that remains after a hole of radius 1 is bored through the center of a solid sphere of radius

2.

A)5

3π B)

8

3π C)

10

3π D)

32

3π

11

Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated

axis.


x1 2 3 4 5

y7

6

5

4

3

2

1

x1 2 3 4 5

y7

6

5

4

3

2

1

y = 2x - x2

A) 2π B) 4π C)8

3π D)

4

3π


x1 2 3 4 5

y

4

3

2

1

x1 2 3 4 5

y

4

3

2

1

y = 1 - x2

A)2

3π B)

3

2π C)

1

3π D) 1π


x1.8

y

4

3

2

1

x1.8

y

4

3

2

1

y =3sin(x2)

A) 12π B) 6π C) 9π D) 3π

12

Use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves

and lines about the y-axis.

57) y = 7x, y = - x

7, x = 1

A)33

7π B)

100

21π C)

50

21π D) 50π

58) y = x2, y = 4 + 3x, for x ≥ 0

A) 32π B) 192π C) 96π D) 64π

59) y = 2

x, y = 0, x = 1, x = 4

A)56

3π B)

64

3π C) 24π D)

28

3π

60) y = 4e-x2, y = 0, x = 0, x = 1

A) 8 1 - 1

e π B) 4(e - 1) π C) 4 1 -

1

e π D) 8 1 +

1

e π

61) y = 6ex2, y = 0, x = 0, x = 1

A) 6(e - 1) π B) 12e π C) 3(e - 1) π D) 12(e - 1) π


and lines about the x-axis.

62) x = 3 y, x = - 3y, y = 1

A)22

5π B) 6π C) 8π D)

11

5π

63) x = 7y - y2, x = 0

A)2401

3π B)

2401

6π C)

2401

12π D)

343

6π

64) y = x, y = 0, y = x - 6

A)225

2π B)

63

2π C) 27π D)

63

4π


about the given lines.

65) y = 4 - x2, y = 4, x = 2; revolve about the line y = 4

A)256

15π B)

224

15π C)

8

3π D)

32

5π

66) y = 2x, y = 0, x = 2; revolve about the line x = -3

A)52

3π B)

104

3π C) -

40

3π D)

52

3

13

Find the length of the curve.

67) y = 2x3/2 from x = 0 to x = 5

4

A)9

4B)

335

3C)

335

72D)

335

108

68) y = 1

6x3 +

1

2x from x = 1 to x = 5

A)316

15B)

79

5C)

632

15D)

127

6

Set up an integral for the length of the curve.

69) y = x4, 0 ≤ x ≤ 1

A)1

0

1 + 16x8 dx∫ B)1

0

1 + 4x6 dx∫ C)1

0

1 + 4x3 dx∫ D)1

0

1 + 16x6 dx∫

FROM THE BOOK

70) Read and take notes on section 8.5 Work (not force and pressure)

71) Do section 8.5 # 1, 5-17 odd, and Examples in the section #1-6 ALL


Solve the problem.

72) How much work is required to move an object from x = 0 to x = 3 (measured in meters) in the presence of a

constant force of 7 N acting along the x-axis?

A) 32 J B) ∞ C) 21 J D) 2 J

73) A 20-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the

system and that the chain has a density of 5 kg/m. How much work is required to wind the entire chain

onto the cylinder using the winch?

A) 1000g J B) 100g J C) 2000g J D) 1950g J

74) A swimming pool has the shape of a box with a base that measures 20 m by 19 m and a depth of 2 m. How

much work is required to pump the water out of the pool when it is full?

A) 28,302,400 J B) 14,896,000 J C) 760,000 J D) 7,448,000 J

14

75) A cylindrical water tank has height 10 m and radius 3 m (see figure). If the tank is full of water, how much

work is required to pump the water to the level of the top of the tank and out of the tank? Express the

answer in terms of π.

10 m

3 m

A) 490,000 J B) 44,100,000π J C) 4,410,000π J D) 1,455,300π J

76) A spherical water tank with an inner radius of 4 m has its lowest point 3 m above the ground. It is filled by

a pipe that feeds the tank at its lowest point (see figure). Neglecting the volume of the inflow pipe, how

much work is required to fill the tank if it is initially empty? Express the answer in terms of π.

4 m

3 m

A)1,254,400

3π J B)

627,200

3π J C)

5,017,600

3π J D)

2,508,800

3π J

77) A water trough has a semicircular cross section with a radius of 0.5 m and a length of 5 m (see figure).

How much work is required to pump water out of the trough when it is full? Round to two decimal places

when appropriate.

5 m

0.5 m

A) 2041.67 J B) 4083.33 J C) 60,331.25π J D) 24,500 J

15

78) A glass has circular cross sections that taper (linearly) from a radius of 8 cm at the top of the glass to a

radius of 7 cm at the bottom. The glass is 14 cm high and full of lemonade. How much work is required to

drink all the lemonade through a straw if your mouth is 6 cm above the top of the glass? Assume the

density of lemonade equals the density of water. Round to two decimal places when appropriate.

A) 3.08 J B) 4.7 J C) 0.98 J D) 53.88 J

THE END !!!

16

Answer KeyTestname: INTEGRATION WORKSHEET 2 - USING THE DEFINITE INTEGRAL

1)

2) see answers in the book. For the evens, discuss in class.

3) C

4) C

5) C

6) C

7) C

8) B

9) C

10) D

11) C

12) D

13) C

14) A

15) D

16) D

17) A

18) A

19) D

20) C

21) C

22) see answers in the book. For the evens, discuss in class.

23)

24)

25) B

26) C

27) A

28) A

29) C

30) D

31) C

32) B

33) D

34) B

35) B

36) C

37) A

38) C

39) A

40) C

41) D

42) B

43) A

44) D

45) B

46) A

47) D

48) C

17

Answer KeyTestname: INTEGRATION WORKSHEET 2 - USING THE DEFINITE INTEGRAL

49) B

50) B

51) B

52) D

53) C

54) C

55) A

56) B

57) B

58) D

59) A

60) C

61) A

62) A

63) B

64) B

65) D

66) B

67) D

68) A

69) D

70)

71) See answers in the book.

72) C

73) A

74) D

75) C

76) D

77) B

78) A

18

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