contents 0.1 mth.121.04.02 ...faculty.essex.edu/~bannon/m121.r/notes/mth.121.review.03.pdfron bannon...

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Calculus IMTH-121

Essex County CollegeDivision of Mathematics

AssessmentsCourse Overview

Contents

0.1 mth.121.04.02 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 mth.121.04.03 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120.3 mth.121.04.04 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170.4 mth.121.04.05 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250.5 mth.121.04.06 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290.6 mth.121.04.07 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370.7 mth.121.04.08 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420.8 mth.121.04.09 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450.9 mth.121.05.01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500.10 mth.121.05.02 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560.11 mth.121.05.03 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

The question presented here are representative of problems students will seeonline or from the textbook. Errors should be reported to

Ron Bannon [email protected].

This document may be shared with students and instructors of MTH 121 atEssex County College, Newark, NJ.

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0.1 mth.121.04.02

1. Consider the function (Figure 1, page 2).

April 2, 2011

S E C T I O N 4.2 Extreme Values 385

Exercises1. The following questions refer to Figure 15.

(a) How many critical points does f (x) have on [0, 8]?(b) What is the maximum value of f (x) on [0, 8]?(c) What are the local maximum values of f (x)?(d) Find a closed interval on which both the minimum and maximum values of f (x) occur at critical points.(e) Find an interval on which the minimum value occurs at an endpoint.

83 4 5 6 721

2

3

4

5

6

1

f (x)

x

y

FIGURE 15

solution(a) f (x) has three critical points on the interval [0, 8]: at x = 3, x = 5 and x = 7. Two of these, x = 3 and x = 5, arewhere the derivative is zero and one, x = 7, is where the derivative does not exist.(b) The maximum value of f (x) on [0, 8] is 6; the function takes this value at x = 0.(c) f (x) achieves a local maximum of 5 at x = 5.(d) Answers may vary. One example is the interval [4, 8]. Another is [2, 6].(e) Answers may vary. The easiest way to ensure this is to choose an interval on which the graph takes no local minimum.One example is [0, 2].

2. State whether f (x) = x!1 (Figure 16) has a minimum or maximum value on the following intervals:(a) (0, 2) (b) (1, 2) (c) [1, 2]

1 2 3x

y

FIGURE 16 Graph of f (x) = x!1.

solution f (x) has no local minima or maxima. Hence, f (x) only takes minimum and maximum values on an intervalif it takes them at the endpoints.(a) f (x) takes no minimum or maximum value on this interval, since the interval does not contain its endpoints.(b) f (x) takes no minimum or maximum value on this interval, since the interval does not contain its endpoints.(c) The function is decreasing on the whole interval [1, 2]. Hence, f (x) takes on its maximum value of 1 at x = 1 andf (x) takes on its minimum value of 1

2 at x = 2.

In Exercises 3–20, find all critical points of the function.

3. f (x) = x2 ! 2x + 4

solution Let f (x) = x2 ! 2x + 4. Then f "(x) = 2x ! 2 = 0 implies that x = 1 is the lone critical point of f .

4. f (x) = 7x ! 2

solution Let f (x) = 7x ! 2. Then f "(x) = 7, which is never zero, so f (x) has no critical points.

5. f (x) = x3 ! 92x2 ! 54x + 2

solution Let f (x) = x3 ! 92x2 ! 54x + 2. Then f "(x) = 3x2 ! 9x ! 54 = 3(x + 3)(x ! 6) = 0 implies that

x = !3 and x = 6 are the critical points of f .

6. f (t) = 8t3 ! t2

solution Let f (t) = 8t3 ! t2. Then f "(t) = 24t2 ! 2t = 2t (12t ! 1) = 0 implies that t = 0 and t = 112 are the

critical points of f .

Figure 1: Partial graph of f .

(a) How many critical points does f (x) have on [0, 8]?

Solution: 3; at x = 3, 5, 7.

(b) What is the maximum value of f (x) on [2, 6]?

Solution: 5; at x = 5.

(c) What is the local minimum value of f (x) on [1, 6]?

Solution: 3; at x = 3.

(d) Find a closed interval on which both the minimum and maximum values of f (x)occur at critical points.

Solution: [2, 6] is one simple example.

(e) Find an interval on which the minimum value occurs at an endpoint.

Solution: [0, 1] is one simple example.

2. Consider the graph (Figure 2, page 3) of f (x) = 2x−1 for this interval (1, 5). Circle oneof the following choices.

(a) a local minima

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-1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8 5.6 6.4 7.2 8

-0.8

0.8

1.6

2.4

3.2

Figure 2: Graph of f .

Solution: No.

(b) a local maxima

Solution: No.

(c) both local minima and maxima

Solution: No.

(d) neither a local minima or maxima

Solution: This is the only correct answer.

3. Find all critical points of the function.

f (x) = 2x2 + 7x+ 17

Solution:

f (x) = 2x2 + 7x+ 17

f ′ (x) = 4x+ 7

The critical point is (−7/4, f (−7/4))


f (x) = x3 − 15x2

2− 42x+ 12

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Solution:

f (x) = x3 − 15x2

2− 42x+ 12

f ′ (x) = 3x2 − 15x− 42

= 3 (x− 7) (x+ 2)

The critical points are (7, f (7)) and (−2, f (−2))


f (x) = x−2 − x−3

Solution:

f (x) = x−2 − x−3

f ′ (x) = −2x−3 + 3x−4

=3− 2x

x4

The critical point is (3/2, f (3/2))


f (z) =1

z − 8− 1

z

Solution:

f (z) =1

z − 8− 1

z

f ′ (z) =64− 16z

z2 (z − 8)2

The critical point is (4, f (4))


f (t) = t− 10√t+ 1

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Solution:

f (t) = t− 10√t+ 1

f ′ (t) =

√t+ 1− 5√t+ 1

The critical point are (−1, f (−1)) and (24, f (24))


f (x) = 12x+ |3x+ 1|

Solution: You may want to graph this one out to see that the derivative is undefinedat x = −1/3 (sharp corner); the slope to the left of x = −1/3 is 9, and the slope to theright of x = −1/3 is 15. You should note on the graph that there’s a sharp corner atx = −1/3. The critical point is (−1/3, f (−1/3))

9. Find all critical points of the function.1

g (θ) = sin2 (27θ)

Solution: Many students may need to review MTH120.

g (θ) = sin2 (27θ)

g′ (θ) = 2 sin (27θ) cos (27θ) · 27

= 27 sin (54θ)

The critical point are (π/54 · k, g (π/54 · k)) where k is an arbitrary integer.

10. Find all critical points of the function f (x) = x ln (4x).

Solution:

f (x) = x ln (4x)

f ′ (x) = 1 + ln (4x)

The critical point is (e−1/4, f (e−1/4))

1Use k for an arbitrary integer.

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f (x) = 9xe4x

Solution:

f (x) = 9xe4x

f ′ (x) = 9e4x (1 + 4x)

The critical point is (−1/4, f (−1/4))


f (x) = cos−1 (x) + 8x, −1 < x < 1.

Solution:

f (x) = cos−1 (x) + 8x, −1 < x < 1

f ′ (x) =8√

1− x2 − 1√1− x2

The critical point are(−√

63/8, f(−√

8/8))

and(√

63/8, f(√

63/8))

13. Find all critical points and the extreme values of the function f (x) = 6x3 − 27x2 + 36xon the interval [0, 3].

Solution:

f (x) = 6x3 − 27x2 + 36x

f ′ (x) = 18x2 − 54x+ 36

= 18 (x− 1) (x− 2)

f (0) = 0

f (1) = 15

f (2) = 12

f (3) = 27

The critical point are (1, f (1)) and (2, f (2)); the absolute maximum occurs at (3, f (3));the absolute minimum occurs at (0, f (0));

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14. Find the critical points off (x) = 4 sinx+ 4 cosx

and determine the extreme values on [0, π/2].

Solution:

f (x) = 4 sinx+ 4 cosx

f ′ (x) = 4 cosx− 4 sinx

f (0) = 4

f (π/4) = 4√

2

f (π/2) = 4

The critical point is (π/4, f (π/4)); the absolute maximum occurs at (π/4, f (π/4));the absolute minimum occurs at two points (0, f (0)) and (π/2, f (π/2)).

15. Plot f (x) = 5√x− x on an appropriate viewing rectangle and determine the maximum

value using the graph and find its exact value using calculus.

Solution:

f (x) = 5√x− x

f ′ (x) =5− 2

√x

2√x

The critical numbers are x = 0 and x = 25/4; the global maximum occurs at x = 25/4.Your graph (Figure 3, page 8) happy should look similar to mine.

16. Find the maximum and minimum values of the function y = 14x3 + 14x2 − 14x on theinterval [−2, 2].

Solution:

y = 14x3 + 14x2 − 14x

y′ = 42x2 + 28x− 14

= 14 (3x− 1) (x+ 1)

y (−2) = −28 minimum

y (−1) = 14

y (1/3) = −70

27y (2) = 140 maximum

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-5 0 5 10 15 20 25 30

-2.5

2.5

5

7.5

Figure 3: Graph of f on (−5, 30).

17. Find the maximum and minimum values of

f (x) =x2 − 7

x− 4

on the interval [5, 8].

Solution:

f (x) =x2 − 7

x− 4

f ′ (x) =(x− 7) (x− 1)

(x− 4)2

f (5) = 18 maximum

f (7) = 14 minimum

f (8) =57

4

18. Find the minimum and maximum values of the function f (x) = 6 sinx cosx+ 3 on theinterval [0, π/2].

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Solution:

Solution:

f (x) = 6 sin x cosx+ 3

f ′ (x) = 6 cos 2x

f (0) = 3 minimum

f (π/4) = 6 maximum

f (π/2) = 3 minimum

19. Find the maximum and minimum values of the function y = 5 sin3 θ − 5 cos2 θ on theinterval [0, 2π].

Solution: This trigonometry will probably be beyond most students’ abilities. How-ever, you may want to consider using a computer algebra system to help out. Especiallya graph (Figure 4, page 10).

y = 5 sin3 θ − 5 cos2 θ

y′ = 15 sin2 θ cos θ + 10 cos θ sin θ Not easy to analyse.

= 5 cos θ sin θ [3 sin θ + 2]

Using the graph as an aid it is clear that x = π/2 is where the maximum occurs; andthe minimum values occur at: x = 0, π, 3π/2, 2π.

20. Verify that Rolle’s Theorem applies to the given functions and intervals. Find all valuesof c satisfying the conclusion of the theorem.

(a) f (x) =x2

12x− 27on the interval [3, 9]

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0 0.8 1.6 2.4 3.2 4 4.8 5.6

-5

-2.5

2.5

5

Figure 4: Graph of f on [0, 2π).

Solution: The function is continuous on the interval.

f (x) =x2

12x− 27f (3) = 1

f (9) = 1

f ′ (x) =2x (2x− 9)

3 (4x− 9)2

f ′ (c) =2c (2c− 9)

3 (4c− 9)2

0 =2c (2c− 9)

3 (4c− 9)2

c =9

2

(b) f (x) =x2

12x− 27on the interval [−9, −3]

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Solution: The function is continuous on the interval.

f (x) =x2

12x− 27

f (−9) = −3

5

f (−3) = −1

7f (−9) 6= f (−3)

Rolle’s Theorem does not apply.

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0.2 mth.121.04.03

1. Find a point c satisfying the conclusion of the MVT for the given function and interval.

y =√x, [4, 25]

Solution: The function is continuous on the interval [4, 25].

y =√x

y′ =1

2√x

m =y (25)− y (4)

25− 4=

3

21=

1

7

y′ (c) =1

2√c

=1

7

c =49

4

2. Consider the following (Figure 5, page 12).

April 2, 2011

S E C T I O N 4.3 The Mean Value Theorem and Monotonicity 407

8. y = ex ! x, [!1, 1]solution Let f (x) = ex ! x, a = !1, b = 1. Then f "(x) = ex ! 1, and by the MVT, there exists a c # (!1, 1) suchthat

ec ! 1 = f "(c) = f (b) ! f (a)

b ! a= (e ! 1) ! (e!1 + 1)

1 ! (!1)= 1

2(e ! e!1) ! 1.

Solving for c yields

c = ln

!e ! e!1

2

"

$ 0.1614 # (!1, 1).

9. Let f (x) = x5 + x2. The secant line between x = 0 and x = 1 has slope 2 (check this), so by the MVT,f "(c) = 2 for some c # (0, 1). Plot f (x) and the secant line on the same axes. Then plot y = 2x + b for different valuesof b until the line becomes tangent to the graph of f . Zoom in on the point of tangency to estimate x-coordinate c of thepoint of tangency.

solution Let f (x) = x5 + x2. The slope of the secant line between x = 0 and x = 1 is

f (1) ! f (0)

1 ! 0= 2 ! 0

1= 2.

A plot of f (x), the secant line between x = 0 and x = 1, and the line y = 2x ! 0.764 is shown below at the left. Theline y = 2x ! 0.764 appears to be tangent to the graph of y = f (x). Zooming in on the point of tangency (see below atthe right), it appears that the x-coordinate of the point of tangency is approximately 0.62.

y = x5 + x2

y = 2x ! .764

x1

2

4

y

x

y

0.3

0.6

0.5

0.4

0.56 0.6 0.640.52

10. Plot the derivative of f (x) = 3x5 ! 5x3. Describe its sign changes and use this to determine the local extremevalues of f (x). Then graph f (x) to confirm your conclusions.

solution Let f (x) = 3x5 ! 5x3. Then f "(x) = 15x4 ! 15x2 = 15x2(x2 ! 1). The graph of f "(x) is shown belowat the left. Because f "(x) changes from positive to negative at x = !1, f (x) changes from increasing to decreasing andtherefore has a local maximum at x = !1. At x = 1, f "(x) changes from negative to positive, so f (x) changes fromdecreasing to increasing and therefore has a local minimum. Though f "(x) = 0 at x = 0, f "(x) does not change sign atx = 0, so f (x) has neither a local maximum nor a local minimum at x = 0. The graph of f (x), shown below at the right,confirms each of these conclusions.

x

10

5

!10

!5

y

x1 2!2 !1

20

40

60

80

y

1

2!2 !1

11. Determine the intervals on which f "(x) is positive and negative, assuming that Figure 13 is the graph of f (x).

x654321

y

FIGURE 13

solution The derivative of f is positive on the intervals (0, 1) and (3, 5) where f is increasing; it is negative on theintervals (1, 3) and (5, 6) where f is decreasing.

Figure 5: Graph of f on (0, 6).

(a) Determine the interval(s) on which f ′ (x) is positive and negative, assuming thatthe figure (Figure 5, page 12) is the graph of f (x) on (0, 6).

Solution: f ′ (x) > 0 on (0, 1) ∪ (3, 5); f ′ (x) < 0 on (1, 3) ∪ (5, 6).

(b) Determine the interval(s) on which f (x) is increasing or decreasing, assuming thatthe figure (Figure 5, page 12) is the graph of f ′ (x) on (0, 6).

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Solution: f (x) is increasing on on (0, 1) ∪ (3, 5); f (x) is decreasing on (1, 3) ∪(5, 6).

3. Find the critical points of the function and use the first derivative test to determinewhether the critical point is a local minimum or maximum (or neither).

y = −x2 + 2x+ 5

Determine the intervals on which the function is increasing or decreasing.

Solution:

y = −x2 + 2x+ 5

y′ = −2x+ 2

The critical point is (1, 6); simple sign analysis on the first derivative indicates thatthis point is a maximum.


f (x) =3

4x4 + 2x3 − 3

2x2 − 6x


Solution:

f (x) =3

4x4 + 2x3 − 3

2x2 − 6x

f ′ (x) = 3x3 + 6x2 − 3x− 6

= 3 (x+ 2) (x+ 1) (x− 1)

The critical values are x = −2, −1, 1; simple sign analysis on the first derivativeindicates that a maximum occurs at x = −1, and minimums at x = −2, 1.


f (x) = x1/3(x2 + 15x− 100

)

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Solution:

f (x) = x1/3(x2 + 15x− 100

)= x7/3 + 15x4/3 − 100x1/3

f ′ (x) =7x2 + 60x− 100

3x2/3

=(7x− 10) (x+ 10)

3x2/3

The critical values are x = −10, 0, 10/7; simple sign analysis on the first derivativeindicates that a maximum occurs at x = 0, and minimums at x = −10, 10/7.


f (x) = − 5

x2 + 4


Solution:

f (x) = − 5

x2 + 4

f ′ (x) =10x

(x2 + 4)2

The critical point is (0, −5/4); simple sign analysis on the first derivative indicates thatthis point is a minimum.


y = 5θ + 5 sin θ + 5 cos θ,

on (0, 2π).

Solution:

y = 5θ + 5 sin θ + 5 cos θ

y′ = 5 + 5 cos θ − 5 sin θ

The critical values are θ = π/2, π; simple sign analysis on the first derivative indicatesthat a maximum occurs at x = π, and the minimum at x = π/2.

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Solution: The function is decreasing on (π/2, π); increasing on (0, π/2) or (π, 2π). Ihighly recommend using a graphic calculator to verify your analysis.


f (x) = 8x+ e−7x + 4

Solution:

f (x) = 8x+ e−7x + 4

f ′ (x) = 8− 7e−7x

The critical point occurs at x = − ln (8/7) /7; simple sign analysis on the first derivativeindicates that this point is a minimum.


Solution: Decreasing on (−∞, − ln (8/7) /7); increasing on (− ln (8/7) /7, ∞).

9. Find the critical points of the function and use the First Derivative Test to determinewhether the critical point is a local minimum or maximum (or neither).

f (x) = 4 arctanx− 2x+ 2

Solution:

f (x) = 4 arctanx− 2x+ 2

f ′ (x) =2 (1− x) (1 + x)

1 + x2

The critical values are x = ±1; ; simple sign analysis on the first derivative indicatesthat a maximum occurs at x = 1, and the minimum at x = −1.


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Solution: Decreasing on (−∞, −1) or (1, ∞); increasing on (−1, 1).

10. Find the largest value of C so that f (x) = 8− 9x+ cx2− x3 is decreasing on (−∞, ∞)whenever |c| < C.

Solution:

f (x) = 8− 9x+ cx2 − x3

f ′ (x) = −9 + 2cx− 3x2

This derivative is quadratic so we’ll look at the discriminant first.

0 = −9 + 2cx− 3x2

0 = 3x2 − 2cx+ 9

D = 2√c2 − 27

We want this D to be complex. That is, we do not want the derivative changing signs.

c2 − 27 < 0

Simple enough, the c must be on the interval(−3√

3, 3√

3); so the largest value of C

is 3√

3.

11. Which value of c satisfies the conclusion of the MVT for the function f (x) = 7x2+4x+6on the interval [6, 11]?

Solution: f (x) and its derivative is continuous on the given interval.

f (x) = 7x2 + 4x+ 6

m =f (11)− f (6)

11− 6= 123

f ′ (x) = 14x+ 4

f ′ (c) = 14c+ 4 = 123

c =17

2

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0.3 mth.121.04.04

1. For the graph (Figure 6, page 17) match the statement that represents company profitsas a function of time.

April 2, 2011

420 C H A P T E R 4 APPLICATIONS OF THE DERIVATIVE

2. Match each statement with a graph in Figure 14 that represents company profits as a function of time.(a) The outlook is great: The growth rate keeps increasing.(b) We’re losing money, but not as quickly as before.(c) We’re losing money, and it’s getting worse as time goes on.(d) We’re doing well, but our growth rate is leveling off.(e) Business had been cooling off, but now it’s picking up.(f) Business had been picking up, but now it’s cooling off.

(i) (ii) (iii) (iv) (v) (vi)

FIGURE 14

solution(a) (ii) An increasing growth rate implies an increasing f !, and so a graph that is concave up.(b) (iv) “Losing money” implies a downward curve. “Not as fast” implies that f ! is becoming less negative, so thatf !!(x) > 0.(c) (i) “Losing money” implies a downward curve. “Getting worse” implies that f ! is becoming more negative, so thecurve is concave down.(d) (iii) “We’re doing well” implies that f is increasing, but “the growth rate is leveling off” implies that f ! is decreasing,so that the graph is concave down.(e) (vi) “Cooling off” generally means increasing at a decreasing rate. The use of “had” implies that only the beginningof the graph is that way. The phrase “...now it’s picking up” implies that the end of the graph is concave up.(f) (v) “Business had been picking up” implies that the graph started out concave up. The phrase “…but now it’s coolingoff” implies that the graph ends up concave down.

In Exercises 3–18, determine the intervals on which the function is concave up or down and find the points of inflection.

3. y = x2 " 4x + 3

solution Let f (x) = x2 " 4x + 3. Then f !(x) = 2x " 4 and f !!(x) = 2 > 0 for all x. Therefore, f is concave upeverywhere, and there are no points of inflection.

4. y = t3 " 6t2 + 4

solution Let f (t) = t3 " 6t2 + 4. Then f !(t) = 3t2 " 12t and f !!(t) = 6t " 12 = 0 at t = 2. Now, f is concaveup on (2, #), since f !!(t) > 0 there. Moreover, f is concave down on ("#, 2), since f !!(t) < 0 there. Finally, becausef !!(t) changes sign at t = 2, f (t) has a point of inflection at t = 2.

5. y = 10x3 " x5

solution Let f (x) = 10x3 " x5. Then f !(x) = 30x2 " 5x4 and f !!(x) = 60x " 20x3 = 20x(3 " x2). Now, f isconcave up for x < "

$3 and for 0 < x <

$3 since f !!(x) > 0 there. Moreover, f is concave down for "

$3 < x < 0

and for x >$

3 since f !!(x) < 0 there. Finally, because f !!(x) changes sign at x = 0 and at x = ±$

3, f (x) has a pointof inflection at x = 0 and at x = ±

$3.

6. y = 5x2 + x4

solution Let f (x) = 5x2 + x4. Then f !(x) = 10x + 4x3 and f !!(x) = 10 + 12x2 > 10 for all x. Thus, f is concaveup for all x and has no points of inflection.

7. y = ! " 2 sin ! , [0, 2" ]solution Let f (!) = ! " 2 sin ! . Then f !(!) = 1 " 2 cos ! and f !!(!) = 2 sin ! . Now, f is concave up for 0 < ! < "

since f !!(!) > 0 there. Moreover, f is concave down for " < ! < 2" since f !!(!) < 0 there. Finally, because f !!(!)

changes sign at ! = " , f (!) has a point of inflection at ! = " .

8. y = ! + sin2 ! , [0, "]solution Let f (!) = ! + sin2 ! . Then f !(!) = 1 + 2 sin ! cos ! = 1 + sin 2! and f !!(!) = 2 cos 2! . Now, f

is concave up for 0 < ! < "/4 and for 3"/4 < ! < " since f !!(!) > 0 there. Moreover, f is concave down for"/4 < ! < 3"/4 since f !!(!) < 0 there. Finally, because f !!(!) changes sign at ! = "/4 and at ! = 3"/4, f (!) has apoint of inflection at ! = "/4 and at ! = 3"/4.

9. y = x(x " 8$

x) (x % 0)

solution Let f (x) = x(x " 8$

x) = x2 " 8x3/2. Then f !(x) = 2x " 12x1/2 and f !!(x) = 2 " 6x"1/2. Now, f

is concave down for 0 < x < 9 since f !!(x) < 0 there. Moreover, f is concave up for x > 9 since f !!(x) > 0 there.Finally, because f !!(x) changes sign at x = 9, f (x) has a point of inflection at x = 9.

Figure 6: Graphs of company profits as a function of time (i)–(vi).

Circle one.

(a) The outlook is great: The growth rate keeps increasing.

(b) Were losing money, but not as quickly as before.

(c) Were losing money, and its getting worse as time goes on.

(d) Were doing well, but our growth rate is leveling off.

(e) Business had been cooling off, but now its picking up.

(f) Business had been picking up, but now its cooling off.

Solution:

(a) (ii) An increasing growth rate implies an increasing f ′, and so a graph that isconcave up.

(b) (iv) Losing money implies a downward curve. Not as fast implies that f ′ is be-coming less negative, so that f ′′ (x) > 0.

Solution:

(c) (i) Losing money implies a downward curve. Getting worse implies that f ′ isbecoming more negative, so the curve is concave down.

(d) (iii) Were doing well implies that f is increasing, but the growth rate is levelingoff implies that f ′ is decreasing, so that the graph is concave down.

(e) (vi) Cooling off generally means increasing at a decreasing rate. The use of hadimplies that only the beginning of the graph is that way. The phrase . . . now it’spicking up implies that the end of the graph is concave up.

(f) (v) Business had been picking up implies that the graph started out concave up.Thephrase . . . but now it’s cooling off implies that the graph ends up concave down.

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2. Givenf (x) = 3x3 − 4x2 + 6x+ 6,

find:

(a) Point-of-inflection.

Solution:

f (x) = 3x3 − 4x2 + 6x+ 6

f (x) = 9x2 − 8x+ 6

f (x) = 18x− 8

The point of inflection occurs at x = 4/9.

(b) Determine the intervals on which the function is concave up.

Solution: (4/9, ∞)

(c) Determine the intervals on which the function is concave down.

Solution: (−∞, 4/9)

3. Determine the intervals on which the function is concave up or concave down.

f (θ) = 11θ + 11 sin2 θ, [0, π]

Solution:

f (θ) = 11θ + 11 sin2 θ

f ′ (θ) = 11 + 22 sin θ cos θ

= 11 + 11 sin 2θ

f ′′ (θ) = 22 cos 2θ

The points of inflection occur at θ = π/4, 3π/4; concave up on [0, π/4) or (3π/4, π];concave down on (π/4, 3π/4).

4. Find the points of inflection of the function.

f (x) = x(x− 3

√x), x > 0

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Solution:

f (x) = x(x− 3

√x), x > 0

= x2 − 3x3/2, x > 0

f ′ (x) = 2x− 9x1/2

2, x > 0

f ′′ (x) = 2− 9x−1/2

4, x > 0

=8√x− 9

4√x

, x > 0


Determine the intervals on which the function is concave up or concave down.

Solution: The function is concave up on (81/64, ∞) and concave down on (0, 81/64).


f (x) = x4 + 5x3 − 36x2 + 3

Solution:

f (x) = x4 + 5x3 − 36x2 + 3

f ′ (x) = 4x3 + 15x2 − 72x

f ′′ (x) = 12x2 + 30x− 72

= 6 (2x− 3) (x+ 4)

The points of inflection occurs at x = −4 and x = 3/2.


Solution: The function is concave up on (−∞, −4) ∪ (3/2, ∞) and concave down on(−4, 3/2).


f (x) =1

x2 + 5

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Solution:

f (x) =1

x2 + 5

f ′ (x) =−2x

(x2 + 5)2

f ′′ (x) =2 (3x2 − 5)

(x2 + 5)3

The points of inflection occurs at x = ±√

5/3.


Solution: The function is concave up on(−∞, −

√5/3)∪(√

5/3, ∞)

and concave

down on(−√

5/3,√

5/3)

.


f (x) = −3xe−7x

Solution:

f (x) = −3xe−7x

f ′ (x) = −3e−7x (1− 7x)

f ′′ (x) = 21e−7x (2− 7x)



Solution: The function is concave up on (−∞, 2/7) and concave down on (2/7, ∞).

8. For the function f (x) on the graph (Figure 9, page 22) , answer the following questions.

(a) Find the points of inflection of the function.

Solution: The point of inflection occurs at x = c.

(b) Determine the intervals on which the function is concave up or concave down.

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April 2, 2011


18. y = x ! ln x (x > 0)

solution Let f (x) = x ! ln x. Then f "(x) = 1 ! 1/x and f ""(x) = x!2 > 0 for all x > 0. Thus, f is concave upfor all x > 0 and has no points of inflection.

19. The growth of a sunflower during the first 100 days after sprouting is modeled well by the logistic curvey = h(t) shown in Figure 15. Estimate the growth rate at the point of inflection and explain its significance. Then makea rough sketch of the first and second derivatives of h(t).

20 40 60 80 100

50

100

150

200

300

250

t (days)

Height (cm)

FIGURE 15

solution The point of inflection in Figure 15 appears to occur at t = 40 days. The graph below shows the logisticcurve with an approximate tangent line drawn at t = 40. The approximate tangent line passes roughly through the points(20, 20) and (60, 240). The growth rate at the point of inflection is thus

240 ! 2060 ! 20

= 22040

= 5.5 cm/day.

Because the logistic curve changes from concave up to concave down at t = 40, the growth rate at this point is themaximum growth rate for the sunflower plant.

20 40 60 80 100

50

100

150

200

300

250

t (days)

Height (cm)

Sketches of the first and second derivative of h(t) are shown below at the left and at the right, respectively.

20 40 60 80 100

1

2

3

4

6

5

t

10080604020t

h!

0.1

"0.1

h!!

20. Assume that Figure 16 is the graph of f (x). Where do the points of inflection of f (x) occur, and on which intervalis f (x) concave down?

x

y

gecba d f

FIGURE 16

solution The function in Figure 16 changes concavity at x = c; therefore, there is a single point of inflection at x = c.The graph is concave down for x < c.

21. Repeat Exercise 20 but assume that Figure 16 is the graph of the derivative f "(x).

solution Points of inflection occur when f ""(x) changes sign. Consequently, points of inflection occur when f "(x)

changes from increasing to decreasing or from decreasing to increasing. In Figure 16, this occurs at x = b and at x = e;therefore, f (x) has an inflection point at x = b and another at x = e. The function f (x) will be concave down whenf ""(x) < 0 or when f "(x) is decreasing. Thus, f (x) is concave down for b < x < e.

Figure 7: Partial graph of f .

Solution: The function is concave up on (c, g) and concave down on (0, c).

9. If the figure (Figure 9, page 22) is the graph of the derivative f ′ (x), answer the followingquestions.

April 2, 2011


18. y = x ! ln x (x > 0)



20 40 60 80 100

50

100

150

200

300

250

t (days)

Height (cm)

FIGURE 15


240 ! 2060 ! 20

= 22040

= 5.5 cm/day.


20 40 60 80 100

50

100

150

200

300

250

t (days)

Height (cm)


20 40 60 80 100

1

2

3

4

6

5

t

10080604020t

h!

0.1

"0.1

h!!


x

y

gecba d f

FIGURE 16





Figure 8: Partial graph of f ′.


Solution: The points of inflection occurs at x = b and x = e.


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Solution: The function is concave up on (0, b)∪(e, g) and concave down on (b, e).

10. If the figure (Figure 9, page 22) is the graph of the second derivative f ′′ (x), answer thefollowing questions.

April 2, 2011


18. y = x ! ln x (x > 0)



20 40 60 80 100

50

100

150

200

300

250

t (days)

Height (cm)

FIGURE 15


240 ! 2060 ! 20

= 22040

= 5.5 cm/day.


20 40 60 80 100

50

100

150

200

300

250

t (days)

Height (cm)


20 40 60 80 100

1

2

3

4

6

5

t

10080604020t

h!

0.1

"0.1

h!!


x

y

gecba d f

FIGURE 16





Figure 9: Partial graph of f ′′.


Solution: The points of inflection occurs at x = a, x = d and x = f .


Solution: The function is concave up on (a, d) ∪ (f, g) and concave down on(0, a) ∪ (d, f).

11. Given

y =3

sinx+ 4, [0, 2π] .

(a) Find the critical points of the function and determine whether the critical point isa local minimum or maximum (or neither).

Solution:

y =3

sinx+ 4, [0, 2π]

y′ =−3 cosx

(sinx+ 4)2, (0, 2π)

The critical points occur at x = π/2 and x = 3π/2; the local maximum occurs atx = 3π/2, and the local minimum occurs at x = π/2.

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(b) Find the points of inflection of the function.

Solution:

y =3

sinx+ 4, [0, 2π]

y′ =−3 cosx

(sinx+ 4)2, (0, 2π)

y′′ =3(2 + 4 sinx− sin2 x

)(sinx+ 4)3

, (0, 2π)

The analysis of this particular problem may be beyond the ability of some students.Yes, it does require that you really know pre-calculus well enough to deal with non-trivial solutions. The point of inflection occurs at x = π − arcsin

(2−√

6).

(c) Determine the intervals on which the function is concave up or concave down.

Solution: The analysis of this particular problem may be beyond the ability ofsome students. Yes, it does require that you really know pre-calculus well enoughto deal with non-trivial solutions. Concave up on

(0, π − arcsin

(2−√

6))

, and

concave down on(π − arcsin

(2−√

6), 2π

).

12. Givenf (x) = e−3x cos 3x,

(−π

6,π

2

),

(a) Find the critical points of the function and determine whether the critical point isa local minimum or maximum (or neither).

Solution:

f (x) = e−3x cos 3x,(−π

6,π

2

)f ′ (x) = −3e−3x (cos 3x+ sin 3x) ,

(−π

6,π

2

)The critical points occur at x = −π/12 and x = π/4; the local maximum occursat x = −π/12, and the local minimum occurs at x = π/4.

(b) Find the points of inflection of the function.

Solution:

f (x) = e−3x cos 3x,(−π

6,π

2

)f ′ (x) = −3e−3x (cos 3x+ sin 3x) ,

(−π

6,π

2

)f ′′ (x) = 18e−3x sin 3x,

(−π

6,π

2

)

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The points of inflection occur at x = 0 and x = π/3.

(c) Determine the intervals on which the function is concave up or concave down.

Solution: Concave down on (−π/6, 0) ∪ (π/3, π/2); concave up on (0, π/3)

13. Sketch the graph of an increasing function such that f ′′ (x) changes from positive tonegative at x = 2 and from negative to positive at x = 4. Do the same for a decreasingfunction.

Solution: The graph shown (Figure 10, page 24) at the left is an increasing functionwhich changes from concave up to concave down at x = 2 and from concave down toconcave up at x = 4. The graph shown (Figure 10, page 24) at the right is a decreasingfunction which changes from concave up to concave down at x = 2 and from concavedown to concave up at x = 4.

April 2, 2011

S E C T I O N 4.4 The Shape of a Graph 429

x!!!

2 , !!4"

!!4

#!!

4 , 3!4

$3!4

#3!4 , 3!

2

$

f " + 0 ! 0 +f # M $ m #

x!!!

2 , 0"

0 (0, !) !#!, 3!

2

$

f "" ! 0 + 0 !f ! I " I !

51. y = (x2 ! 2)e!x (x > 0)

solution Let f (x) = (x2 ! 2)e!x .

• Then f "(x) = !(x2 ! 2x ! 2)e!x = 0 gives x = 1 +%

3 as a candidate for an extrema.• Moreover, f ""(x) = (x2 ! 4x)e!x = 0 gives x = 4 as a candidate for a point of inflection.

x#

0, 1 +%

3$

1 +%

3#

1 +%

3, &$

f " + 0 !f # M $

x (0, 4) 4 (4, &)

f "" ! 0 +f ! I "

52. y = ln(x2 + 2x + 5)

solution Let f (x) = ln(x2 + 2x + 5). Then

f "(x) = 2x + 2

x2 + 2x + 5= 0

when x = !1. This is the only critical point. Moreover,

f ""(x) = !2(x ! 1)(x + 3)

(x2 + 2x + 5)2 ,

so x = 1 and x = !3 are candidates for inflection points.

x (!&, !1) !1 (!1, &)

f " ! 0 +f $ m #

x (!&, !3) !3 (!3, 1) 1 (1, &)

f "" ! 0 + 0 !f ! I " I !

53. Sketch the graph of an increasing function such that f ""(x) changes from + to ! at x = 2 and from ! to + at x = 4.Do the same for a decreasing function.

solution The graph shown below at the left is an increasing function which changes from concave up to concavedown at x = 2 and from concave down to concave up at x = 4. The graph shown below at the right is a decreasingfunction which changes from concave up to concave down at x = 2 and from concave down to concave up at x = 4.

x2 4

2

1

y

x2 4

6

2

4

y

Figure 10: Requested graphs.

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0.4 mth.121.04.05

1. Evaluate the limit.

limx→8

√x+ 1− 3

x3 − 63x− 8

Solution:

limx→8

√x+ 1− 3

x3 − 63x− 8H= lim

x→8

12

(x+ 1)−1/2

3x2 − 63

=1

774


limx→0

sin 6x

x2 + 5x

Solution:

limx→0

sin 6x

x2 + 5xH= lim

x→0

6 cos 6x

2x+ 5

=6

5


limx→0

3x3

sinx− x

Solution:

limx→0

3x3

sinx− xH= lim

x→0

9x2

cosx− 1H= lim

x→0

18x

− sinxH= lim

x→0

18

− cosx= −18

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limx→0

cos 4x− 1

sin 9x

Solution:

limx→0

cos 4x− 1

sin 9xH= lim

x→0

−4 sin 4x

9 cos 9x= 0


limx→0

cos 18x− sin2 2x

sin 2x

Solution: The top is going towards 1 and the bottom is going towards zero.

limx→0

cos 18x− sin2 2x

sin 2x= DNE


limx→−∞

−2x sin1

x

Solution:

limx→−∞

−2x sin1

x= −2 lim

x→−∞

sin 1x

1x

H= −2 lim

x→−∞

− 1x2

cos 1x

− 1x2

= −2 limx→−∞

cos1

x= −2


limx→16

(1√x− 4

− 8

x− 16

)

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Solution:

limx→16

(1√x− 4

− 8

x− 16

)= lim

x→16

√x− 4

x− 16

H= lim

x→16

12√x

1

=1

8

8. Evaluate the limit.limx→0

(2x)sin 9x

Solution:

y = (2x)sin 9x

ln y = sin 9x · ln 2x

ln y =ln 2x

1/ sin 9x

y = eln 2x

1/ sin 9x

limx→0

(2x)sin 9x = limx→0

eln 2x

1/ sin 9x

H= lim

x→0e

1/x−9 cot 9x csc 9x

= limx→0

esin 9x tan 9x

−9x

= limx→0

e− sin 9x

9x·tan 9x

= e0 = 1

9. Evaluate the limit.limx→π

2

(secx− tanx)

Solution:

limx→π

2

(secx− tanx) = limx→π

2

1− sinx

cosx

H= lim

x→π2

− cosx

− sinx

= 0

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limx→2

ex3 − e8

x− 2

Solution:

limx→2

ex3 − e8

x− 2H= lim

x→2

3x2ex3

1

= 12e8

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0.5 mth.121.04.06

1. You are given the following graph (Figure 11, page 29) of the function f (x):

April 2, 2011


2. If the sign combination of f ! and f !! changes from ++ to +" at x = c, then (choose the correct answer):(a) f (c) is a local min (b) f (c) is a local max(c) c is a point of inflection

solution Because the sign of the second derivative changes at x = c, the correct response is (c): c is a point ofinflection.

3. The second derivative of the function f (x) = (x " 4)"1 is f !!(x) = 2(x " 4)"3. Although f !!(x) changes sign atx = 4, f (x) does not have a point of inflection at x = 4. Why not?

solution The function f does not have a point of inflection at x = 4 because x = 4 is not in the domain of f.

Exercises1. Determine the sign combinations of f ! and f !! for each interval A–G in Figure 16.

CB D E F GAx

y

y = f (x)

FIGURE 16solution

• In A, f is decreasing and concave up, so f ! < 0 and f !! > 0.• In B, f is increasing and concave up, so f ! > 0 and f !! > 0.• In C, f is increasing and concave down, so f ! > 0 and f !! < 0.• In D, f is decreasing and concave down, so f ! < 0 and f !! < 0.• In E, f is decreasing and concave up, so f ! < 0 and f !! > 0.• In F, f is increasing and concave up, so f ! > 0 and f !! > 0.• In G, f is increasing and concave down, so f ! > 0 and f !! < 0.

2. State the sign change at each transition point A–G in Figure 17. Example: f !(x) goes from + to " at A.

Ax

y

y = f (x)

CB D E F G

FIGURE 17

solution

• At B, the graph changes from concave down to concave up, so f !! goes from " to +.• At C, the graph changes from decreasing to increasing, so f ! goes from " to +.• At D, the graph changes from concave up to concave down, so f !! goes from + to ".• At E, the graph changes from increasing to decreasing, so f ! goes from + to ".• At F, the graph changes from concave down to concave up, so f !! goes from " to +.• At G, the graph changes from decreasing to increasing, so f ! goes from " to +.

In Exercises 3–6, draw the graph of a function for which f ! and f !! take on the given sign combinations.

3. ++, +", ""solution This function changes from concave up to concave down at x = "1 and from increasing to decreasing atx = 0.

x

y

!1

0 1!1

Figure 11: Graph is f .

Find the point(s) where the second derivative changes sign from negative to positive.

Solution: At D.

2. Given the graph (Figure 12, page 30) of

y = x3 − 6x2 + 17x,

indicate the transition points.2

Solution:

y = x3 − 6x2 + 17x

y = 3x2 − 12x+ 17

y = 6x− 12

No local extrema; inflection point at x = 2.


y = cosx+x

2, [0, π]


2Local minima, local maxima, and inflection point.3Local minima, local maxima, and inflection point.

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-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-500

-400

-300

-200

-100

100

200

300

400

500

Figure 12: Partial graph of y = x3 − 6x2 + 17x.

0 1 2 3

1

Figure 13: Graph of y = cosx+x

2, on [0, π].

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Solution:

y = cosx+x

2, [0, π]

y′ = − sinx+1

2, (0, π)

y′′ = − cosx, (0, π)

Local maximum at x = π/6; local minimum at x = 5π/6; inflection point at x = π/2.


y = xe−7x2

,


-1 0 1

-0.2

-0.1

0.1

0.2

Figure 14: Partial graph of y = xe−7x2.

Solution:

y = xe−7x2

y′ = e−7x2 (

1− 14x2)

y′′ = −14xe−7x2 (

3− 14x2)

Local maximum at x = 1/√

14; local minimum at x = −1/√

14; inflection points atx = 0 and x = ±

√3/14.

4Local minima, local maxima, and inflection point.

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y = (3x2 − 1)e−x2

,


-3 -2 -1 0 1 2 3

-1

1

Figure 15: Partial graph of y = (3x2 − 1) e−x2.

Solution:

y =(3x2 − 1

)e−x

2

y′ = 2x(4− 3x2

)e−x

2

y′′ = 2(6x4 − 17x2 + 4

)e−x

2

Local maximum at x = ±2/√

3; local minimum at x = 0; inflection points at:

x = ±

√17±

√193

12.


y = x− 2 ln(x2 + 1

),


5Local minima, local maxima, and inflection point.6Local minima, local maxima, and inflection point.

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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

-5

-4

-3

-2

-1

1

2

3

Figure 16: Partial graph of y = x− 2 ln (x2 + 1).

Solution:

y = x− 2 ln(x2 + 1

)y′ =

x2 − 4x+ 1

x2 + 1

y′′ =4 (x+ 1) (x− 1)

(x2 + 1)2

Local maximum at x = 2 −√

3; local minimum at x = 2 +√

3; inflection points atx = ±1.


y = 3x+1

x,


Solution:

y = 3x+1

x

y =3x2 − 1

x2

y =2

x3

Local maximum at x = −1/√

3; local minimum at x = 1/√

3; no inflection points.


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-3 -2 -1 0 1 2 3

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

Figure 17: Partial graph of y = 3x+1

x.


y =3

x+

3

x− 1,


-3 -2 -1 0 1 2 3

-20

-15

-10

-5

5

10

15

20

Figure 18: Partial graph of y =3

x+

3

x− 1.


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Solution:

y =3

x+

3

x− 1

y′ = −3 (2x2 − 2x+ 1)

x2 (x− 1)2

y′′ =6 (2x3 − 3x2 + 3x− 1)

x3 (x− 1)3

No extrema; inflection point at x = 1/2. You’ll need to use the graph and the rationalroot theorem (MTH 119) to determine this.


y = 1− 9

x+

3

x3,


-10 -5 0 5 10

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

Figure 19: Partial graph of y = 1− 9

x+

3

x3.

Solution:

y = 1− 9

x+

3

x3

y′ = 9x−4(x2 − 1

)y′′ = −18x−5

(x2 − 2

)Local maximum at x = −1; local minimum at x = 1; inflection points at x = ±

√2.


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y =1√

x2 + 2,


-10 -5 0 5 10

1

Figure 20: Partial graph of y =1√

x2 + 2.

Solution:

y =1√

x2 + 2

y′ = −x(x2 + 2

)−3/2y′′ =

(x2 − 1

) (x2 + 2

)−5/2Local maximum at x = 0; inflection points at x = ±1.


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0.6 mth.121.04.07

1. Wire of length 40 meters is divided into two pieces and the pieces are bent into a squareand a circle. How many meters of wire should be allotted to the circle so that the sumof the areas is minimized?

Solution: Let x be the circle’s circumference, and 40− x the square’s perimeter. Theradius of the circle is x/ (2π), and the side of the square is (40− x) /4. The functionwe want to minimized is area, which is a function of x, where x ∈ (0, 40).

A (x) =

(40− x

4

)2

+ π( x

2π

)2A′ (x) =

x− 40

8+

x

2π

The critical value x = 40π/ (π + 4π). Simple sign-analysis indicates that this is aminimum. So we need to allot 40π/ (π + 4π) meters to the circle.

2. Find a positive number x such that the sum of 25x and1

xis as small as possible.

Solution: Here x > 0 and we’re trying to minimize f .

f (x) = 25x+1

x

f ′ (x) =25x2 − 1

x2

The critical value x = 1/5. Simple sign-analysis indicates that this is a minimum.

3. Find the rectangle of maximum area that can be inscribed in a right triangle with legsof length a = 5 and b = 8 if the sides of the rectangle x, y are parallel to the legs of thetriangle, as in the figure (Figure 21, page 38).

Solution: Let x ∈ (0, 5) be the base of the rectangle and y ∈ (0, 8) its height. Settingup similar triangles gives the following relationship.

8

5=

y

5− x

y = 8− 8

5x

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April 2, 2011


solution Let x be the width of the corral and therefore the diameter of the semicircle, and let y be the height of therectangular section. Then the perimeter of the corral can be expressed by the equation 2y + x + !

2 x = 2y + (1 + !2 )x =

600 m or equivalently, y = 12

!600 ! (1 + !

2 )x". Since x and y must both be nonnegative, it follows that x must

be restricted to the interval [0, 6001+!/2 ]. The area of the corral is the sum of the area of the rectangle and semicircle,

A = xy + !8 x2. Making the substitution for y from the constraint equation,

A(x) = 12x

#600 ! (1 + !

2)x

$+ !

8x2 = 300x ! 1

2

#1 + !

2

$x2 + !

8x2.

Now, A"(x) = 300 !!1 + !

2"x + !

4 x = 0 implies x = 300!1+ !

4" # 168.029746 m. With A(0) = 0 m2,

A

%300

1 + !/4

&# 25204.5 m2 and A

%600

1 + !/2

&# 21390.8 m2,

it follows that the corral of maximum area has dimensions

x = 3001 + !/4

m and y = 1501 + !/4

m.

8. What is the maximum area of a rectangle inscribed in a right triangle with 5 and 8 as in Figure 10. The sides of therectangle are parallel to the legs of the triangle.

5

8

FIGURE 10

solution Position the triangle with its right angle at the origin, with its side of length 8 along the positive y-axis, andside of length 5 along the positive x-axis. Let x, y > 0 be the lengths of sides of the inscribed rectangle along the axes.By similar triangles, we have 8

5 = y5!x or y = 8 ! 8

5x. The area of the rectangle is thus A(x) = xy = 8x ! 85x2. To

guarantee that both x and y remain nonnegative, we must restrict x to the interval [0, 5]. Solve A"(x) = 8 ! 165 x = 0 to

obtain x = 52 . Since A(0) = A(5) = 0 and A( 5

2 ) = 10, the maximum area is A#

52

$= 10 when x = 5

2 and y = 4.

9. Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r = 4 (Figure 11).

r

FIGURE 11

solution Place the center of the circle at the origin with the sides of the rectangle (of lengths 2x > 0 and 2y > 0)

parallel to the coordinate axes. By the Pythagorean Theorem, x2 + y2 = r2 = 16, so that y ='

16 ! x2. Thus the areaof the rectangle is A(x) = 2x · 2y = 4x

'16 ! x2. To guarantee both x and y are real and nonnegative, we must restrict

x to the interval [0, 4]. Solve

A"(x) = 4'

16 ! x2 ! 4x2'

16 ! x2= 0

for x > 0 to obtain x = 4$2

= 2$

2. Since A(0) = A(4) = 0 and A(2$

2) = 32, the rectangle of maximum area has

dimensions 2x = 2y = 4$

2.

10. Find the dimensions x and y of the rectangle inscribed in a circle of radius r that maximizes the quantity xy2.

solution Place the center of the circle of radius r at the origin with the sides of the rectangle (of lengths x > 0 andy > 0) parallel to the coordinate axes. By the Pythagorean Theorem, we have ( x

2 )2 + (y2 )2 = r2, whence y2 = 4r2 ! x2.

Let f (x) = xy2 = 4xr2 ! x3. Allowing for degenerate rectangles, we have 0 % x % 2r . Solve f "(x) = 4r2 ! 3x2 forx & 0 to obtain x = 2r$

3. Since f (0) = f (2r) = 0, the maximal value of f is f ( 2r$

3) = 16

9$

3r3 when x = 2r$3

and

y = 2(

23 r .

Figure 21: Rectangle inscribed in a right triangle.

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We’re trying to maximize the area.

A = xy

A (x) = x

(8− 8

5x

)A′ (x) = 8− 16

5x

The critical value x = 5/2. Simple sign-analysis indicates that this is a maximum. Therectangle is 2.5 by 4.

4. Find the dimensions of the rectangle of maximum area that can be inscribed in a circleof radius r = 3.

April 2, 2011


solution Let x be the width of the corral and therefore the diameter of the semicircle, and let y be the height of therectangular section. Then the perimeter of the corral can be expressed by the equation 2y + x + !

2 x = 2y + (1 + !2 )x =

600 m or equivalently, y = 12

!600 ! (1 + !

2 )x". Since x and y must both be nonnegative, it follows that x must

be restricted to the interval [0, 6001+!/2 ]. The area of the corral is the sum of the area of the rectangle and semicircle,

A = xy + !8 x2. Making the substitution for y from the constraint equation,

A(x) = 12x

#600 ! (1 + !

2)x

$+ !

8x2 = 300x ! 1

2

#1 + !

2

$x2 + !

8x2.

Now, A"(x) = 300 !!1 + !

2"x + !

4 x = 0 implies x = 300!1+ !

4" # 168.029746 m. With A(0) = 0 m2,

A

%300

1 + !/4

&# 25204.5 m2 and A

%600

1 + !/2

&# 21390.8 m2,

it follows that the corral of maximum area has dimensions

x = 3001 + !/4

m and y = 1501 + !/4

m.

8. What is the maximum area of a rectangle inscribed in a right triangle with 5 and 8 as in Figure 10. The sides of therectangle are parallel to the legs of the triangle.

5

8

FIGURE 10

solution Position the triangle with its right angle at the origin, with its side of length 8 along the positive y-axis, andside of length 5 along the positive x-axis. Let x, y > 0 be the lengths of sides of the inscribed rectangle along the axes.By similar triangles, we have 8

5 = y5!x or y = 8 ! 8

5x. The area of the rectangle is thus A(x) = xy = 8x ! 85x2. To

guarantee that both x and y remain nonnegative, we must restrict x to the interval [0, 5]. Solve A"(x) = 8 ! 165 x = 0 to

obtain x = 52 . Since A(0) = A(5) = 0 and A( 5

2 ) = 10, the maximum area is A#

52

$= 10 when x = 5

2 and y = 4.

9. Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r = 4 (Figure 11).

r

FIGURE 11

solution Place the center of the circle at the origin with the sides of the rectangle (of lengths 2x > 0 and 2y > 0)

parallel to the coordinate axes. By the Pythagorean Theorem, x2 + y2 = r2 = 16, so that y ='

16 ! x2. Thus the areaof the rectangle is A(x) = 2x · 2y = 4x

'16 ! x2. To guarantee both x and y are real and nonnegative, we must restrict

x to the interval [0, 4]. Solve

A"(x) = 4'

16 ! x2 ! 4x2'

16 ! x2= 0

for x > 0 to obtain x = 4$2

= 2$

2. Since A(0) = A(4) = 0 and A(2$

2) = 32, the rectangle of maximum area has

dimensions 2x = 2y = 4$

2.

10. Find the dimensions x and y of the rectangle inscribed in a circle of radius r that maximizes the quantity xy2.

solution Place the center of the circle of radius r at the origin with the sides of the rectangle (of lengths x > 0 andy > 0) parallel to the coordinate axes. By the Pythagorean Theorem, we have ( x

2 )2 + (y2 )2 = r2, whence y2 = 4r2 ! x2.

Let f (x) = xy2 = 4xr2 ! x3. Allowing for degenerate rectangles, we have 0 % x % 2r . Solve f "(x) = 4r2 ! 3x2 forx & 0 to obtain x = 2r$

3. Since f (0) = f (2r) = 0, the maximal value of f is f ( 2r$

3) = 16

9$

3r3 when x = 2r$3

and

y = 2(

23 r .

Figure 22: Rectangle inscribed in a circle.

Solution: I suggest you center the circle at the origin and you’ll see that x2 + y2 = 9.Let x, y ∈ (0, 3) represent a point on the circle in the first quadrant.

x2 + y2 = 9

y2 = 9− x2

y =√

9− x2

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We’re trying to maximize the area.

A = 2x · 2yA (x) = 4x

√9− x2

A′ (x) =4 (9− 2x2)√

9− x2

The critical value x = 3/√

2. Simple sign-analysis indicates that this is a maximum.The rectangle is 3

√2 by 3

√2.

5. Find the point on the line y = 7x closest to the point (1, 0).

Solution: Here I am minimize the square of the distance, and I need to minimize thisfunction.

d2 = (x− 1)2 + (y − 0)2

d2 = (x− 1)2 + (7x− 0)2

f (x) = (x− 1)2 + (7x− 0)2

f ′ (x) = 2 (x− 1) + 98x

= 100x− 2

The critical value x = 1/50. Simple sign-analysis indicates that this is a minimum.The point is (1/50, 7/50). x = 1/50, y = 7/50

6. Find the equation of the line through P = (5, 4) such that the triangle bounded by thisline and the axes in the first quadrant has minimal area.11

Solution: You may need to draw a picture to see this. Let x > 5 and y > 4 be the

11Use the standard coordinate variables x and y.

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values of the x and y-intercepts for the line. The slope of this line is −y/x.

−yx

=4− 0

5− x

y =4x

x− 5

A =1

2· x · y

A (x) =1

2· x · 4x

x− 5

=2x2

x− 5

A′ (x) =2x (x− 10)

(x− 5)2

The critical value x = 10. Simple sign-analysis indicates that this is a minimum. Sox = 10, y = 8 and the area is 40. The equation of the line is:

y − 4 = − 8

10(x− 5)

y = −4

5x+ 8

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0.7 mth.121.04.08

1. Apply Newton’s Method to f (x) and initial guess x0 to calculate x1, x2, and x3. (Roundyour answers to seven decimal places.)

f (x) = x2 − 18, x0 = 4

Solution:

x0 = 4 ≈ 4.0000000

x1 =17

4≈ 4.2500000

x2 =577

136≈ 4.2426471

x3 =665857

156944≈ 4.2426407

2. The figure (Figure 23, page 42) is the graph of f (x) = x3+2x+4. Use Newton’s methodto approximate the unique real root of f (x) = 0.12

-1.25 -1 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1

0.8

1.6

2.4

3.2

4

Figure 23: Partial graph of f (x) = x3 + 2x+ 4.

Solution: Initial guess from graph.

x0 = −1.17000

x1 ≈ −1.17956

x2 ≈ −1.17951

3. Use Newton’s method to find the solution13 of 5 sinx = 3 cos 2x in [0, π/2].14

12Round your answer to five decimal places.13Graph is provided (Figure 24, page 43) that may help in determining an initial guess.14Round your answer to three decimal places.

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0 1

-4

-3

-2

-1

1

2

3

4

5

6

Figure 24: Partial graph of y = 5 sin x and y = 3 cos 2x.

Solution: Initial guess from graph.

x0 = 0.42

x1 ≈ 0.416

4. Find the x-coordinate and y-coordinate of the point P in the figure (Figure 25, page 43)where the tangent line to the graph (Figure 25, page 43) of y = 6 cosx passes throughthe origin.15

0 1 2 3

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

P

Figure 25: Partial graph of y = 6 cos x and a tangent line at P .

15Round your answers to three decimal places.

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Solution:

x ≈ 2.798

y ≈ −5.650

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0.8 mth.121.04.09

1. Find the general antiderivative of f (x) and check your answer by differentiating.16

f (x) = 15x2

Solution:

f (x) = 15x2

F (x) = 5x3 + C

f (x) = F ′ (x)


f (x) = 11x−3/5

Solution:

f (x) = 11x−3/5

F (x) =55x2/5

2+ C

f (x) = F ′ (x)


f (x) = 13 cosx− 8 sinx

Solution:

f (x) = 13 cos x− 8 sinx

F (x) = 13 sinx+ 8 cosx+ C

f (x) = F ′ (x)

16Use C for the constant of integration.17Use C for the constant of integration.18Use C for the constant of integration.

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f (x) = 18ex − 10x−2

Solution:

f (x) = 18ex − 10x−2

F (x) = 18ex + 10x−1 + C

f (x) = F ′ (x)

5. Solve the initial value problem.

dy

dx= 6x3, y (0) = 9

Solution:

dy

dx= 6x3

y (x) =3

2x4 + C

y (0) =3

2(0)4 + C = 9

y =3

2x4 + 9


dy

dt= 3√t, y (1) = 7

Solution:

dy

dt= 3

√t

y (t) = 2t3/2 + C

y (1) = 2 (1)3/2 + C = 7

y = 2t3/2 + 5

19Use C for the constant of integration.

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dz

dt= t−3/2, z (25) = −15

Solution:

dz

dt= t−3/2

z (t) = −2t−1/2 + C

z (25) = −2 (25)−1/2 + C = −15

z = −2t−1/2 − 73

5


dy

dx= 3 sin x, y

(π2

)= 9

Solution:

dy

dx= 3 sinx

y (x) = −3 cosx+ C

y(π

2

)= −3 cos

(π2

)+ C = 9

y = −3 cosx+ 9


dy

dx= ex, y (4) = 0

Solution:

dy

dx= ex

y (x) = ex + C

y (4) = e4 + C = 0

y = ex − e4

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10. Find f ′ (x) and f (x).

f ′′ (x) = 54x, f ′ (0) = 9, f (0) = 5

Solution:

f ′′ (x) = 54x

f ′ (x) = 27x2 + C1

f ′ (x) = 27 (0)2 + C1 = 9

f (x) = 9x3 + 9x+ C2

f (0) = 9 (0)3 + 9 (0) + C2 = 5

y (x) = 9x3 + 9x+ 5

11. Find f ′ and then find f .

f ′′ (x) = 140x3 − 84x2 + 18x− 12, f ′ (1) = 8, f (1) = 3

Solution:

f ′′ (x) = 140x3 − 84x2 + 18x− 12,

f ′ (x) = 35x4 − 28x3 + 9x2 − 12x+ C1

f ′ (x) = 35 (1)4 − 28 (1)3 + 9 (1)2 − 12 (1) + C1 = 8

f (x) = 7x5 − 7x4 + 3x3 − 6x2 + 4x+ C2

f (x) = 7 (1)5 − 7 (1)4 + 3 (1)3 − 6 (1)2 + 4 (1) + C2 = 3

f ′ (x) = 27 (0)2 + C1 = 9

f (x) = 7x5 − 7x4 + 3x3 − 6x2 + 4x+ 2

12. Given thatf ′′ (x) = cos x, f ′

(π2

)= 4, f

(π2

)= 13.

(a) f ′ (x) =

Solution:

f ′′ (x) = cosx

f ′ (x) = sinx+ C1

f ′(π

2

)= sin

(π2

)+ C1 = 4

f ′ (x) = 3 + sinx

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(b) f (x) =

Solution:

f ′ (x) = 3 + sinx

f (x) = 3x− cosx+ C2

f(π

2

)= 3

(π2

)− cos

(π2

)+ C2 = 13

f (x) = 3x− cosx+ 13− 3π

2

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0.9 mth.121.05.01

1. The figure (Figure 26, page 50) shows the velocity of an object over a 3-minute interval.

S E C T I O N 5.1 Approximating and Computing Area 531

Exercises1. An athlete runs with velocity 4 mph for half an hour, 6 mph for the next hour, and 5 mph for another half-hour.

Compute the total distance traveled and indicate on a graph how this quantity can be interpreted as an area.

SOLUTION The figure below displays the velocity of the runner as a function of time. The area of the shaded regionequals the total distance traveled. Thus, the total distance traveled is (4) (0.5) + (6) (1) + (5) (0.5) = 10.5 miles.

0.5 1

Vel

ocity

(mph

)

1.5Time (hours)

2

123456

2. Figure 14 shows the velocity of an object over a 3-min interval. Determine the distance traveled over the intervals[0, 3] and [1, 2.5] (remember to convert from miles per hour to miles per minute).

3 min

mph

21

20

30

10

FIGURE 14

SOLUTION The distance traveled by the object can be determined by calculating the area underneath the velocity graphover the specified interval. During the interval [0, 3], the object travels

!1060

"!12

"+

!2560

"(1) +

!1560

"!12

"+

!2060

"(1) = 23

24! 0.96 mile.

During the interval [1, 2.5], it travels!

2560

"!12

"+

!1560

"!12

"+

!2060

" !12

"= 1

2= 0.5 mile.

3. A rainstorm hit Portland, Maine, in October 1996, resulting in record rainfall. The rainfall rate R(t) on October 21is recorded, in inches per hour, in the following table, where t is the number of hours since midnight. Compute the totalrainfall during this 24-hour period and indicate on a graph how this quantity can be interpreted as an area.

t 0–2 2–4 4–9 9–12 12–20 20–24

R(t) 0.2 0.1 0.4 1.0 0.6 0.25

SOLUTION Over each interval, the total rainfall is the time interval in hours times the rainfall in inches per hour. Thus

R = 2(.2) + 2(.1) + 5(.4) + 3(1.0) + 8(.6) + 4(.25) = 11.4 inches.

The figure below is a graph of the rainfall as a function of time. The area of the shaded region represents the total rainfall.

5 10

Rai

nfal

l(i

nche

s pe

r hou

r)

15Time (hours)

20 25

0.2

0.4

0.6

0.8

1

4. The velocity of an object is v(t) = 32t ft/s. Use Eq. (2) and geometry to find the distance traveled over the timeintervals [0, 2] and [2, 5].

Figure 26: Velocity as a function of time.

(a) Determine the distance traveled over the interval [0, 3]. (Remember to convertfrom mi/hr to mi/min. Round your answer to two decimal places.)

Solution: Here d = r · t.

1

2· 10

60+ 1 · 25

60+

1

2· 15

60+ 1 · 20

60=

11

6≈ 1.83

1.83 miles.

(b) Determine the distance traveled over the interval [1, 2.5]. (Remember to convertfrom mi/hr to mi/min. Round your answer to two decimal places.)

Solution: Here d = r · t.

1

2· 25

60+

1

2· 15

60+

1

2· 20

60=

1

2

0.50 miles.

2. A rainstorm hit a city resulting in record rainfall. The rainfall rate R (t) is recorded, incentimeters per hour, in the following table (Figure 27, page 51), where t is the numberof hours since midnight. Compute the total rainfall during this 24-hour period.

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S E C T I O N 5.1 Approximating and Computing Area 531

Exercises1. An athlete runs with velocity 4 mph for half an hour, 6 mph for the next hour, and 5 mph for another half-hour.

Compute the total distance traveled and indicate on a graph how this quantity can be interpreted as an area.

SOLUTION The figure below displays the velocity of the runner as a function of time. The area of the shaded regionequals the total distance traveled. Thus, the total distance traveled is (4) (0.5) + (6) (1) + (5) (0.5) = 10.5 miles.

0.5 1

Vel

ocity

(mph

)

1.5Time (hours)

2

123456

2. Figure 14 shows the velocity of an object over a 3-min interval. Determine the distance traveled over the intervals[0, 3] and [1, 2.5] (remember to convert from miles per hour to miles per minute).

3 min

mph

21

20

30

10

FIGURE 14

SOLUTION The distance traveled by the object can be determined by calculating the area underneath the velocity graphover the specified interval. During the interval [0, 3], the object travels

!1060

"!12

"+

!2560

"(1) +

!1560

"!12

"+

!2060

"(1) = 23

24! 0.96 mile.

During the interval [1, 2.5], it travels!

2560

"!12

"+

!1560

"!12

"+

!2060

" !12

"= 1

2= 0.5 mile.

3. A rainstorm hit Portland, Maine, in October 1996, resulting in record rainfall. The rainfall rate R(t) on October 21is recorded, in inches per hour, in the following table, where t is the number of hours since midnight. Compute the totalrainfall during this 24-hour period and indicate on a graph how this quantity can be interpreted as an area.

t 0–2 2–4 4–9 9–12 12–20 20–24

R(t) 0.2 0.1 0.4 1.0 0.6 0.25

SOLUTION Over each interval, the total rainfall is the time interval in hours times the rainfall in inches per hour. Thus

R = 2(.2) + 2(.1) + 5(.4) + 3(1.0) + 8(.6) + 4(.25) = 11.4 inches.

The figure below is a graph of the rainfall as a function of time. The area of the shaded region represents the total rainfall.

5 10

Rai

nfal

l(i

nche

s pe

r hou

r)

15Time (hours)

20 25

0.2

0.4

0.6

0.8

1

4. The velocity of an object is v(t) = 32t ft/s. Use Eq. (2) and geometry to find the distance traveled over the timeintervals [0, 2] and [2, 5].

Figure 27: Table for rainfall.

Solution:

2 (0.2) + 2 (0.1) + 5 (0.4) + 3 (1.0) + 8 (0.6) + 4 (0.25) = 11.4

11.4 centimeters.

3. Compute R6, L6, and M3 to estimate the distance traveled over [0, 3] if the velocity athalf-second intervals is as follows (Figure 28, page 51):

532 C H A P T E R 5 THE INTEGRAL

SOLUTION The total distance traveled is given by the area under the graph of v = 32t . From the figure below, we seethat the region under the velocity graph over the interval [0, 2] is a right triangle with base 2 and height 64. The areaunder the graph is 1

2 (2)(64) = 64, so the object travels 64 feet from t = 0 to t = 2.The region under the velocity graph over the interval [2, 5] is a trapezoid with height 3 and bases 64 and 160. The

area under the graph is 12 (3)(64 + 160) = 336, so the object travels 336 feet from t = 2 to t = 5.

t

v(t)

160

20406080

100120140

010 2 3 4 5

5. Compute R6, L6, and M3 to estimate the distance traveled over [0, 3] if the velocity at half-second intervals is asfollows:

t (s) 0 0.5 1 1.5 2 2.5 3

v (ft/s) 0 12 18 25 20 14 20

SOLUTION For R6 and L6, !t = 3!06 = .5. For M3, !t = 3!0

3 = 1. Then

R6 = 0.5 sec (12 + 18 + 25 + 20 + 14 + 20) ft/sec = .5(109) ft = 54.5 ft,

L6 = 0.5 sec (0 + 12 + 18 + 25 + 20 + 14) ft/sec = .5(89) ft = 44.5 ft,

and

M3 = 1 sec (12 + 25 + 14) ft/sec = 51 ft.

6. Use the following table of values to estimate the area under the graph of f (x) over [0, 1] by computing the averageof R5 and L5.

x 0 0.2 0.4 0.6 0.8 1

f (x) 50 48 46 44 42 40

SOLUTION !x = 1!05 = .2. Thus,

L5 = .2 (50 + 48 + 46 + 44 + 42) = .2(230) = 46,

and

R5 = .2 (48 + 46 + 44 + 42 + 40) = .2(220) = 44.

The average is

46 + 442

= 45.

This estimate is frequently referred to as the Trapezoidal Approximation.

7. Consider f (x) = 2x + 3 on [0, 3].(a) Compute R6 and L6 over [0, 3].(b) Find the error in these approximations by computing the area exactly using geometry.

SOLUTION Let f (x) = 2x + 3 on [0, 3].

(a) We partition [0, 3] into 6 equally-spaced subintervals. The left endpoints of the subintervals are!

0, 12 , 1, 3

2 , 2, 52

"

whereas the right endpoints are!

12 , 1, 3

2 , 2, 52 , 3

".

• Let a = 0, b = 3, n = 6, !x = (b ! a) /n = 12 , and xk = a + k!x , k = 0, 1, . . . , 5 (left endpoints). Then

L6 =5#

k=0f (xk)!x = !x

5#

k=0f (xk ) = 1

2(3 + 4 + 5 + 6 + 7 + 8) = 16.5.

Figure 28: Table of velocities.

Solution:

R6 =1

2· [12 + 18 + 25 + 20 + 14 + 20] =

109

2= 54.5 feet

L6 =1

2· [0 + 12 + 18 + 25 + 20 + 14] =

89

2= 44.5 feet

M3 = 1 · [12 + 25 + 14] = 51 feet

4. Consider f (x) = 3x+ 4 on [0, 3]. Compute L6 and R6 over [0, 3]. Use geometry to findthe exact area A

(a) L6 =

Solution:

1

2[f (0) + f (0.5) + f (1) + f (1.5) + f (2) + f (2.5)] =

93

4

(b) R6 =

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Solution:

1

2[f (0.5) + f (1) + f (1.5) + f (2) + f (2.5) + f (3)] =

111

4

(c) A =

Solution:

3 · 4 +1

2· 3 · (13− 4) =

51

2

5. Consider the following figure (Figure 29, page 52).

534 C H A P T E R 5 THE INTEGRAL

M6 = 14

6!

k=1f"

xk ! 12!x

#= 1

4(2.4 + 2.3 + 2.2 + 1.85 + 1.45 + 0.8) = 2.75.

10. Estimate R2, M3, and L6 for the graph in Figure 16.

21

1

1.5

0.5

13

23

43

53

x

y

FIGURE 16

SOLUTION Let f (x) on [0, 2] be given by Figure 16. To calculate R2, we take !x = 2!02 = 1 and evaluate the

function at the right endpoints. This gives

R2 = !x( f (1) + f (2)) = 1(1.125 + 1) = 2.125.

To calculate M3, we take !x = 2!03 = 2

3 and evaluate the function at the subinterval midpoints. This gives

M3 = !x"

f"

13

#+ f (1) + f

"53

##= 2

3(0.625 + 1.125 + 0.75) = 1.667.

Finally, for L6, we take !x = 2!06 = 1

3 and evaluate the function at the left endpoints. This gives

L6 = 13(0.5 + 0.625 + 0.9 + 1.125 + 1 + 0.75) = 1.633.

11. Let f (x) =$

x2 + 1 and !x = 13 . Sketch the graph of f (x) and draw the rectangles whose area is represented by

the sum%6

i=1 f (1 + i!x)!x .

SOLUTION Because the summation index runs from i = 1 through i = 6, we will treat this as a right-endpoint

approximation to the area under the graph of y =$

x2 + 1. With !x = 13 , it follows that the right endpoints of the

subintervals are x1 = 43 , x2 = 5

3 , x3 = 2, x4 = 73 , x5 = 8

3 and x6 = 3. The sketch of the graph with the rectangles

represented by the sum%6

i=1 f (1 + i!x)!x is given below.

0.5

0.51

1.52

2.53

1 1.5 2 2.5 3x

y

12. Calculate the area of the shaded rectangles in Figure 17. Which approximation do these rectangles represent?

1 32!1!3 !2x

y

y =1 + x24 ! x

FIGURE 17

SOLUTION Each rectangle in Figure 17 has a width of 1 and the height is taken as the value of the function at themidpoint of the interval. Thus, the area of the shaded rectangles is

1"

2629

+ 2213

+ 185

+ 145

+ 1013

+ 629

#= 18784

1885" 9.965.

Because there are six rectangles and the height of each rectangle is taken as the value of the function at the midpoint ofthe interval, the shaded rectangles represent the approximation M6 to the area under the curve.

Figure 29: y with shaded rectangles.

(a) Calculate the area of the shaded rectangles. (Round your answer to three decimalplaces.)

Solution: Here f (x) = y.

1 · [f (−2.5) + f (−1.5) + f (−0.5) + f (0.5) + f (1.5) + f (2.5)] ≈ 9.965

(b) Which approximation do these rectangles represent? Circle one.

L5 M6 M5 R6 L6

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Solution: M6 should be circled.

6. Calculate the R5 approximation for f (x) = 3x2 + 5x on [−1, 1].

Solution:

R5 =2

5

[f

(−3

5

)+ f

(−1

5

)+ f

(1

5

)+ f

(3

5

)+ f

(5

5

)]=

2

5

[−48

25− 22

25+

28

25+

102

25+

200

25

]=

104

25

7. Write the sum2

6 · 7+

3

7 · 8+ · · ·+ n

(n+ 4) · (n+ 5)

in summation notation in terms of k.

Solution:

2

6 · 7+

3

7 · 8+ · · ·+ n

(n+ 4) · (n+ 5)=

n∑k=2

k

(k + 4) · (k + 5)

8. Calculate300∑j=101

j

by writing it as a difference of two sums and using the power sums formulas.

Solution:

300∑j=101

j =300∑j=1

j −100∑j=1

j

=300 · 301

2− 100 · 101

2= 40100

9. Calculate30∑j=1

(8j + 1)2

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by expanding and using the power sums formulas.

Solution:

30∑j=1

(8j + 1)2 = 6430∑j=1

j2 + 1630∑j=1

j +30∑j=1

1

= 64 · 30 · 31 · 61

6+ 16 · 30 · 31

2+ 1 · 30

= 612590


limN→∞

N∑s=1

3s2 − 5s− 5

N3

Solution:

limN→∞

N∑s=1

3s2 − 5s− 5

N3= lim

N→∞

1

N3

[3N (N + 1) (2N + 1)

6− 5N (N + 1)

2− 5N

]= lim

N→∞

[1

2

(1 +

1

N

)(2 +

1

N

)− 5 (N + 1)

2N2− 5

N2

]= 1

11. Evaluate the limit if f (x) = 71x+ 142 on [1, 4].

limN→∞

RN =

Solution:

limN→∞

3

N

N∑j=1

f

(1 +

3j

N

)= lim

N→∞

3

N

N∑j=1

[71

(1 +

3j

N

)+ 142

]

= limN→∞

3

N

N∑j=1

[213 +

213j

N

]= lim

N→∞

3

N

[213N +

213

N· N (N + 1)

2

]= lim

N→∞

[639 +

639

2·(

1 +1

N

)]= 639 +

639

2=

1917

2

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12. Evaluate the limit if f (x) = 48x2 + 64x on [0, 2].

limN→∞

RN =

Solution:

limN→∞

2

N

N∑j=1

f

(2j

N

)= lim

N→∞

2

N

N∑j=1

[48

(2j

N

)2

+ 64

(2j

N

)]

= limN→∞

[384

N3

N∑j=1

j2 +256

N2

N∑j=1

j

]

= limN→∞

[384

N3· N (N + 1) (2N + 1)

6+

256

N2· N (N + 1)

2

]= lim

N→∞

[64 (N + 1) (2N + 1)

N2+

128 (N + 1)

N

]= lim

N→∞

[64

(1 +

1

N

)(2 +

1

N

)+ 128

(1 +

1

N

)]= 128 + 128 = 256

13. Describe20 the area represented by the limit.

limN→∞

1

N

N∑j=1

(j

N

)9

The limit represents the area between the graph of f (x) = and the x-axis over

the interval[

,].

Solution:

f (x) = x9, [0, 1]

20Fill in the boxes.

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0.10 mth.121.05.02

1. Draw a graph of the signed area represented by the integral and compute it using geom-etry. ∫ 4

−42x dx

Solution: Your graph (Figure 30, page 56) should look similar to mine.∫ 4

−42x dx = 0

-20 -15 -10 -5 0 5 10 15 20

-10

-5

5

10

Figure 30: Partial graph of integrated area.


−25 dx


−25 dx = 8 · 5 = 40

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-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5


3. Draw a graph of the signed area represented by the integral and compute it using geom-etry. ∫ 3π/2

π/2

2 sinx dx

Solution: Your graph (Figure 32, page 58) should look similar to mine.∫ 3π/2

π/2

2 sinx dx = 0


0

√81− x2 dx


0

√81− x2 dx =

1

4· π · 92 =

81π

4

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-1 0 1 2 3 4 5 6 7 8

-3

-2

-1

1

2

3


-10 -7.5 -5 -2.5 0 2.5 5 7.5

2.5

5

7.5

10


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−83 |x| dx


−83 |x| dx =

1

2· 8 · 24 +

1

2· 7 · 21

=339

2

-15 -10 -5 0 5 10 15 20

5

10

15

20


6. Consider the figure (Figure 35, page 60). Evaluate.

(a)

∫ 3

0

g (x) dx

Solution: ∫ 3

0

g (x) dx = −1

2+

4

2=

3

2

(b)

∫ 5

3

g (x) dx

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S E C T I O N 5.2 The Definite Integral 559

(b) The definite integral! 6

0 f (x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis and asemicircle of radius 2 which lies above the x-axis. Therefore,

" 6

0f (x) dx = 1

2! (2)2 ! 1

2! (1)2 = 3!

2.

(c) The definite integral! 4

1 f (x) dx is the signed area of one-quarter of a circle of radius 1 which lies below the x-axisand one-quarter of a circle of radius 2 which lies above the x-axis. Therefore,

" 4

1f (x) dx = 1

4! (2)2 ! 1

4! (1)2 = 3

4!.

(d) The definite integral! 6

1 | f (x)| dx is the signed area of one-quarter of a circle of radius 1 and a semicircle of radius2, both of which lie above the x-axis. Therefore,

" 6

1| f (x)| dx = 1

2! (2)2 + 1

4! (1)2 = 9!

4.

In Exercises 14–15, refer to Figure 15.

1 2 3 4 5

2

1

"1

"2

y = g (t)

t

y

FIGURE 15

14. Evaluate" 3

0g(t) dt and

" 5

3g(t) dt .

SOLUTION

• The region bounded by the curve y = g(x) and the x-axis over the interval [0, 3] is comprised of two righttriangles, one with area 1

2 below the axis, and one with area 2 above the axis. The definite integral is therefore equal

to 2 ! 12 = 3

2 .• The region bounded by the curve y = g(x) and the x-axis over the interval [3, 5] is comprised of another two right

triangles, one with area 1 above the axis and one with area 1 below the axis. The definite integral is therefore equalto 0.

15. Find a, b, and c such that" a

0g(t) dt and

" c

bg(t) dt are as large as possible.

SOLUTION To make the value of" a

0g(t) dt as large as possible, we want to include as much positive area as possible.

This happens when we take a = 4. Now, to make the value of" c

bg(t) dt as large as possibe, we want to make sure to

include all of the positive area and only the positive area. This happens when we take b = 1 and c = 4.

16. Describe the partition P and the set of intermediate points C for the Riemann sum shown in Figure 16. Compute thevalue of the Riemann sum.

y

x1 32.5 3.220.5 4.5 5

34.25

2015

8

FIGURE 16

SOLUTION The partition P is defined by

x0 = 0 < x1 = 1 < x2 = 2.5 < x3 = 3.2 < x4 = 5

The set of intermediate points is given by C = {c1 = 0.5, c2 = 2, c3 = 3, c4 = 4.5}. Finally, the value of the Riemannsum is

34.25(1 ! 0) + 20(2.5 ! 1) + 8(3.2 ! 2.5) + 15(5 ! 3.2) = 96.85.

Figure 35: Partial graph of g.

Solution: ∫ 5

3

g (x) dx =2

2− 2

2= 0

7. Consider the Riemann sum shown in the figure (Figure 36, page 61).

(a) x0 = < x1 = < x2 = < x3 = < x4 = 21

Solution:

P = x0 = 0 < x1 = 1 < x2 = 2.5 < x3 = 3.2 < x4 = 5

(b) Describe the set of sample points C.

Solution:

C = {0.5, 2, 3, 4.5}

(c) Compute the value of the Riemann sum.

21Fill in the boxes.

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S E C T I O N 5.2 The Definite Integral 559

(b) The definite integral! 6

0 f (x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis and asemicircle of radius 2 which lies above the x-axis. Therefore,

" 6

0f (x) dx = 1

2! (2)2 ! 1

2! (1)2 = 3!

2.

(c) The definite integral! 4

1 f (x) dx is the signed area of one-quarter of a circle of radius 1 which lies below the x-axisand one-quarter of a circle of radius 2 which lies above the x-axis. Therefore,

" 4

1f (x) dx = 1

4! (2)2 ! 1

4! (1)2 = 3

4!.

(d) The definite integral! 6

1 | f (x)| dx is the signed area of one-quarter of a circle of radius 1 and a semicircle of radius2, both of which lie above the x-axis. Therefore,

" 6

1| f (x)| dx = 1

2! (2)2 + 1

4! (1)2 = 9!

4.

In Exercises 14–15, refer to Figure 15.

1 2 3 4 5

2

1

"1

"2

y = g (t)

t

y

FIGURE 15

14. Evaluate" 3

0g(t) dt and

" 5

3g(t) dt .

SOLUTION

• The region bounded by the curve y = g(x) and the x-axis over the interval [0, 3] is comprised of two righttriangles, one with area 1

2 below the axis, and one with area 2 above the axis. The definite integral is therefore equal

to 2 ! 12 = 3

2 .• The region bounded by the curve y = g(x) and the x-axis over the interval [3, 5] is comprised of another two right

triangles, one with area 1 above the axis and one with area 1 below the axis. The definite integral is therefore equalto 0.

15. Find a, b, and c such that" a

0g(t) dt and

" c

bg(t) dt are as large as possible.

SOLUTION To make the value of" a

0g(t) dt as large as possible, we want to include as much positive area as possible.

This happens when we take a = 4. Now, to make the value of" c

bg(t) dt as large as possibe, we want to make sure to

include all of the positive area and only the positive area. This happens when we take b = 1 and c = 4.

16. Describe the partition P and the set of intermediate points C for the Riemann sum shown in Figure 16. Compute thevalue of the Riemann sum.

y

x1 32.5 3.220.5 4.5 5

34.25

2015

8

FIGURE 16

SOLUTION The partition P is defined by

x0 = 0 < x1 = 1 < x2 = 2.5 < x3 = 3.2 < x4 = 5

The set of intermediate points is given by C = {c1 = 0.5, c2 = 2, c3 = 3, c4 = 4.5}. Finally, the value of the Riemannsum is

34.25(1 ! 0) + 20(2.5 ! 1) + 8(3.2 ! 2.5) + 15(5 ! 3.2) = 96.85.

Figure 36: Riemann sum

Solution:

= (1− 0) · f (0.5) + (2.5− 1) · f (2) + (3.2− 2.5) · f (3) + (5− 3.2) · f (4.5)

= 1 · 34.25 + 1.5 · 20 + 0.7 · 8 + 1.8 · 15

= 96.85

8. Calculate the Riemann sum R (f, P, C) for the function f (x) = x2 + 5x, partitionP = {4, 6, 9, 13} and sample points C = {4.5, 8.5, 10.5}.

Solution:

R (f, P, C) = (6− 4) · f (4.5) + (9− 6) · f (8.5) + (13− 9) · f (10.5)

= 1080.75

9. Determine the sign of the integral

∫ 2π

0

4x sinx dx without calculating it. Use the graph

(Figure 37, page 62). if necessary.

Solution: Negative. Incidentally, this integral evaluates to −8π.

10. Calculate the following integral, assuming that

∫ 5

0

f (x) dx = 2 and

∫ 5

0

g (x) dx = 9.

∫ 5

0

[f (x) + g (x)] dx

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-1 0 1 2 3 4 5 6 7

-20

-16

-12

-8

-4

4

8

Figure 37: Partial graph of f (x) = 4x sinx, with shading from 0 to 2π.

Solution: ∫ 5

0

[f (x) + g (x)] dx =

∫ 5

0

f (x) dx+

∫ 5

0

g (x) dx

= 2 + 9 = 11

11. Assuming that

∫ 1

0

f (x) dx = 4,

∫ 2

0

f (x) dx = 5, and

∫ 4

1

f (x) dx = 7, calculate∫ 4

2

f (x) dx.

Solution: ∫ 4

2

f (x) dx =

∫ 4

1

f (x) dx−∫ 2

1

f (x) dx

=

∫ 4

1

f (x) dx−[∫ 2

0

f (x) dx−∫ 1

0

f (x) dx

]= 7− [5− 4] = 6

12. Calculate the integral. ∫ 4

0

|2− x| dx

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Solution: Here you are expected to visualize the area and use simple geometry toevaluate the integral. ∫ 4

0

|2− x| dx =1

2· 2 · 2 +

1

2· 2 · 2 = 4

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0.11 mth.121.05.03

1. Use FTC I to find the area of the region under the graph of the function f (x) = 8x−x2on [0, 8].

Solution: ∫ 8

0

8x− x2 dx =

(4x2 − x3

3

)∣∣∣∣80

=256

3

2. Evaluate the integral using the FTC I.

∫ 4

1

(3x5 + 2x2 − x

)dx

Solution: ∫ 4

1

(3x5 + 2x2 − x

)dx =

(x6

2+

2x3

3− x2

2

)∣∣∣∣41

= 2082

3. Evaluate the integral using the FTC I.

∫ 5

1

1

x5dx

Solution: ∫ 5

1

1

x5dx =

∫ 5

1

x−5 dx

=

(− 1

4x4

)∣∣∣∣51

=156

625

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4. Evaluate the integral using the FTC I.∫ 5

1

4x− 3√x

dx

Solution: ∫ 5

1

4x− 3√x

dx =

∫ 5

1

4x1/2 − 3x−1/2 dx

=

(8x3/2

3− 6x1/2

)∣∣∣∣51

=22√

5 + 10

3

5. Evaluate the integral using the FTC I.∫ 4

0

∣∣x2 − 4∣∣ dx

Solution: ∫ 4

0

∣∣x2 − 4∣∣ dx =

∫ 2

0

4− x2 dx+

∫ 4

2

x2 − 4 dx

=

(4x− x3

3

)∣∣∣∣20

+

(x3

3− 4x

)∣∣∣∣42

= 16

6. Evaluate the integral

∫ 3

−2f (x) dx, when f (x) =

{2− x2 x ≤ 1x2 x > 1

.

Solution: ∫ 3

−2f (x) dx =

∫ 1

−22− x2 dx+

∫ 3

1

x2 dx

=

(2x− x3

3

)∣∣∣∣1−2

+

(x3

3

)∣∣∣∣31

=35

3

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contents 0.1 mth.121.04.02 ...faculty.essex.edu/~bannon/m121.r/notes/mth.121.review.03.pdfron bannon...

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