Final Exam Study GuideSolutions

Note: These problems are just some of the types of problems that might appear on the exam.However, to fully prepare for the exam, in addition to making sure you can do all of these prob-lems, you should review your homework and class notes as well.

Section 10.1: Introduction to Limits

1. The graph of f is shown below. Use the graph of f to evaluate the indicated limits andfunction values. If a limit or function value does not exist, write “DNE.”

(a) limx→−2+

f(x) = 4

(b) limx→−2−

f(x) = 2

(c) limx→−2

f(x) DNE

(d) f(−2) = 1

(e) limx→0

f(x) = 2

(f) f(2) = 2

(g) limx→2−

f(x) = 0

(h) limx→2+

f(x) = 2

(i) limx→2

f(x) DNE

2. Suppose limx→1

f(x) = −3 and limx→1

g(x) = 2. Find

(a) limx→1

3f(x) = 3(−3) = −9

(b) limx→1

(g(x)− f(x)

)= 2− (−3) = 5

(c) limx→1

3− f(x)

4g(x)− 2

3− (−3)

4(2)− 2=

6

6= 1

3. Find each limit

(a) limx→3

(x2 − 3

)= (3)2 − 3 = 9− 3 = 6

(b) limx→1

5 = 5

(c) limx→2

x = 2

(d) limx→−1

x2 + 1

x=

(−1)2 + 1

−1= −2

4. For each limit, is the limit a 00

indeterminant form? Find the limit or explain why it doesnot exist.

(a) limx→7

(x− 7)2

x2 − 4x− 21Yes; 0

(b) limx→2

x− 2

x + 2No; 0

1

(c) limx→4

3x + 12

x2 − 4No; DNE

(d) limx→3

x2 − 9

x− 3Yes; 6

5. Circle true or false to the following statements.

(a) If limx→1

f(x) = 0 and limx→1

g(x) = 0, then limx→1

f(x)

g(x)does not exist.

true false

(b) If limx→1

f(x) = 2 and limx→1

g(x) = 2, then limx→2

f(x)

g(x)= 1

true false

(c) In order for limx→1

f(x) to exist and be equal to 2, it must be the case that limx→1−

f(x) = 2

and limx→1+

f(x) = 2.

true false

(d) If f is a function such that f(2) exists, then limx→2

f(x) exists.

true false

(e) If f is a polynomial, then limx→c

f(x) = f(c) for every real number c.

true false

Section 10.2: Infinite Limits and Limits at Infinity

1. Suppose f(x) =1

x− 5. Find each limit. Write ∞, −∞, or DNE where appropriate.

(a) limx→5−

f(x)

Note the limit will be ±∞ since if we try plugging in x = 5 we get nonzerozero

. To de-termine which one, we plug in a number very close to 5 but to the left of 5, and we seethat we get a negative number. Thus, lim

x→5−f(x) = −∞.

(b) limx→5+

f(x) For similar reasons as in (a), limx→5+

f(x) =∞.

(c) limx→5

f(x) DNE since the left and right-sided limits are different.

2. Suppose f(x) =3− x

x + 2. Find each limit. Write ∞, −∞, or DNE where appropriate.

(a) limx→−2−

f(x)

Note the limit will be ±∞ since if we try plugging in x = −2 we get nonzerozero

. Todetermine which one, we plug in a number close to −2 but to the left of −2. For

example, we can plug in −2.1 and we get that f(−2.1) =3− (−2.1)

−2.1 + 2< 0. Thus,

limx→−2−

f(x) = −∞.

(b) limx→−2+

f(x) For similar reasons as in (a), limx→−2+

f(x) =∞.

(c) limx→−2

f(x) DNE since the left and right-sided limits are not the same

2

3. Identify all vertical and horizontal asymptotes of the following functions.

(a) f(x) =1

x + 3

Note limx→−3+

f(x) =∞. This is enough to show there is a vertical asymptote at x = −3.

There are no other zeros of the denominator other than x = −3, so this is the only

vertical asymptote. Also, there is a horizontal asymptote at x = 0 since limx→∞

1

x + 3= 0.

(b) f(x) =x2 − 2x− 3

x2 − 9

Note f(x) =(x− 3)(x + 1)

(x + 3)(x− 3). Note there is a hole in the graph at x = 3 since x = 3 is

a zero of the denominator and yet

limx→3

f(x) = limx→3

x + 1

x + 3=

4

6

So there is no vertical asymptote at x = 3. The only other zero of the denominator isx = −3. Note

limx→−3+

f(x) = limx→−3+

x + 1

x + 3=∞

and so there is a vertical asymptote at x = −3. Also, note that

limx→∞

f(x) = limx→∞

x2

x2= 1

and so there is a horizontal asymptote at y = 1.

4. Find the following limits. Write ∞ or −∞ where appropriate.

(a) limx→∞

(x2 − 2x− 1

)

limx→∞

(x2 − 2x− 1

)= lim

x→∞x2 =∞

(b) limx→∞

(− x3 − x

)

limx→∞

(− x3 − x

)= lim

x→∞(−x3) = −∞

(c) limx→−∞

(x3 + 2x + x

)

limx→−∞

(x3 + 2x + x

)= lim

x→−∞x3 = −∞

3

5. Circle true or false to the following statements.

(a) A polynomial function of degree greater than or equal to 1 has neither horizontal norvertical asymptotes.

true false

(b) A rational function always has at least one vertical asymptote.

true false

(c) A rational function has at most one horizontal asymptote.

true false

6. A theorem states that for n ≥ 1 and an 6= 0

limx→−∞

(anx

n + an−1xn−1 + · · ·+ a0

)= ±∞

What conditions must n and an satisfy for the limit to be +∞?

Since limx→∞

(anx

n + an−1xn−1 + · · · + a0

)= lim

x→∞anx

n, only an and xn determine whether

the limit is ∞ or −∞. If n is odd, then limx→−∞

xn = −∞. Thus, in this case, to make sure

that limx→−∞

anxn =∞, we would need an < 0. If n is even, then lim

x→−∞xn =∞, and so in this

case we would need an > 0. Thus, to summarize, we need n odd and an < 0, or we need neven and an > 0.

Section 10.4: The Derivative

1. Write down two interpretations of f ′(a), the derivative of a function f(x) evaluated at x = a.

Two possible interpretations of f ′(a) are

� f ′(a) is the slope of the tangent line to f(x) at x = a.

� If f(x) represents the position of an object at time x, then f ′(a) represents the instan-taneous velocity of the object at time x = a.

Other interpretations include

� f ′(a) is the slope of the function f(x) at x = a

� f ′(a) is the instantaneous rate of change of f(x) at x = a.

2. Explain what f ′(2) = 3 tells us about a function f(x).

There are several possible answers. One answer might be that f ′(2) = 3 tells us thatthe slope on the graph of f(x) at x = 2 is 3.

3. Find the equation of the tangent line to f(x) = 2− x2 at x = 1.

Note f ′(x) = −2x. Now, the tangent line to f(x) = 2 − x2 at x = 1 goes through(1, f(1)) = (1, 1) and has slope f ′(1) = −2(1) = −2. Thus, using the point-slope formula,we get that the equation of the tangent line is

y − 1 = −2(x− 1) or y = −2x + 3

Section 10.5: Basic Differentiation Properties

1. If C is any constant, what isd

dx(C)?

d

dx(C) = 0 for any constant C

2. Write down the power rule.d

dxxn = nxn−1 for any real number n

4

3. If f ′(x) = 2x and g′(x) = x2, what isd

dx

(2f(x)− 1

4g(x)

)?

d

dx

(2f(x)− 1

4g(x)

)= 2f ′(x)− 1

4g′(x) = 2(2x)− 1

4(x2) = 4x− 1

4x2

4. Find the derivative of the following functions.

(a) f(x) = 7 f ′(x) = 0

(b) f(x) = 2x + 3 f ′(x) = 2 + 0 = 2

(c) f(x) = 2x2 − x + 1 f ′(x) = 4x− 1

(d) y =√x + x3/4 dy

dx=

1

2x−1/2 +

3

4x−1/4

(e) y =1

5x3− 2x− 3

x−2/3

Note y =1

5x−3 − 2x

x−2/3+

3

x−2/3=

1

5x−3 − 2x5/3 + 3x2/3. Hence,

dy

dx= −3

5x−4 − 10

3x2/3 + 2x−1/3

Section 10.7: Marginal Analysis in Business and Economics

1. What is the relationship between the cost function C(x), the revenue function R(x), and theprofit function P (x)? P (x) = R(x)− C(x)

2. Suppose the total cost of producing x items is

C(x) = 500 + 50x− 1

2x2

and the total profit from the sale of x items is

P (x) = 40x− 1

6x2 − 100

(a) Find the marginal cost function. C ′(x) = 50− x

(b) Find the marginal cost when 10 items are produced and interpret the result.

C ′(10) = 50 − 10 = 40. Thus, at a production level of 10 items, the cost is increasingat a rate of $40/item.

(c) Find the marginal profit function. P ′(x) = 40− 13x

(d) Find the marginal profit when 10 items are produced and interpret.

P ′(10) = 40 − 103

= 1103≈ 36.67. Thus, at a production level of 10 items, the profit is

increasing at a rate of about $36.67 per item.

(e) Find the revenue function.

Since P (x) = R(x)− C(x), we get that

R(x) = P (x) + C(x) = 40x− 1

6x2 − 100 + 500 + 50x− 1

2x2

= 90x− 2

3x2 + 400

(f) Find the revenue gained from selling 10 items.

R(10) = 90(10)− 2

3(102) + 400

= 1300− 200

3

=3900− 200

3

=3700

3≈ 1233.33

Thus, the revenue gained from selling 10 items is about $1233.33.

5

3. A company is producing couches and their total cost function is C(x) = 100, 000 + 50x.Their price-demand equation is x = 8000− 40p.

(a) Express the price p as a function of the demand x, then find the domain of this function.

Solving the price-demand equation for p gives p = − 1

40x + 200. Setting p ≥ 0 we

get that x ≤ 8000. Also x ≥ 0. Thus, the domain is 0 ≤ x ≤ 8000.

(b) Find the marginal cost. C ′(x) = 50

(c) Find the revenue function as a function of x and state its domain.

The revenue function is R(x) = xp(x) = x(− 1

40x + 200

)= − 1

40x2 + 200x. The

domain is the same as the cost function so 0 ≤ x ≤ 8000.

(d) Find the marginal revenue. R′(x) = − 1

20x + 200

(e) Find R′(2000) and R′(5000) and interpret.

R′(2000) = − 1

20(2000) + 200 = −100 + 200 = 100. This means that at a produc-

tion level of 2000 couches, the revenue is increasing at a rate of $100 per couch. Also,R′(5000) = − 1

20(5000) + 200 = −250 + 200 = −50. Thus, at a production level of 5000

couches, the revenue is decreasing at a rate of $50 per couch.

(f) Graph the cost function and the revenue function. Indicate the break-even points onthe graph.

The black line is the cost function and the red line is the revenue function. The break-even points are the (x, y) coordinates which are plotted on the graph. (Note: On atest a rough sketch would be appropriate and you would not need to find the actualbreak-even points, just indicate where they are on the graph.)

(g) The break-even points are about (764, 138197) and (5236, 361803). Use these to writedown the intervals of loss and the interval of profit.

The interval of loss would be (0, 764) and (5236, 8000) (we can see on the graph thatfor any x in these intervals, the cost function is above the revenue function and so weare experiencing loss). The interval of profit would be (764, 5236) (we can see for any xin this interval, the revenue function is above the cost function so we are experiencingprofit).

(h) Find the profit function P (x).

P (x) = R(x)− C(x) = − 1

40x2 + 200x−

(100, 000 + 50x

)= − 1

40x2 + 150x− 100, 000

6

Section 11.2: Derivatives of Exponential and Logarithmic Functions

1. Findd

dx

(5ex − ln(x) + 2

).

d

dx

(5ex − ln(x) + 2

)= 5ex − 1

x

2. Use log properties to find derivatives of the following functions. (Note: If you use the logproperties you won’t need to use the chain rule.)

(a) f(x) = x4 − ln(x5)

First, note f(x) = x4 − 5 ln(x). Thus, f ′(x) = 4x3 − 5x.

(b) y = ln(xex)

First, note y = ln(x) + ln(ex) = ln(x) + x. Thus,dy

dx=

1

x+ 1.

(c) y = ln( x

5√x

)First, note

y = ln(x)− ln(5√x)

= ln(x)− (ln(5) + ln(√x)

= ln(x)− ln(5)− 1

2ln(x)

=1

2ln(x)− ln(5)

Alternatively, note y = ln(1

5x1/2) = ln(

1

5) +

1

2ln(x). Thus,

dy

dx=

1

2x.

Section 11.3: Derivatives of Products and Quotients

1. Fill in the blanks:

(a) The product rule says that given two differentiable functions F (x) and S(x), the deriva-tive of their product is

d

dx

(F (x)S(x)

)= F ′(x)S(x) + F (x)S ′(x)

(b) The quotient rule says that given two differentiable functions T (x) and B(x), the deriva-tive of their quotient is

d

dx

(T (x)

B(x)

)=

T ′(x)B(x)− T (x)B′(x)

(B(x))2

2. Findd

dx

(x2ex

).

d

dx

(x2ex

)= 2xex + x2ex.

3. If f(x) =x2

4x− 3, find f ′(x). f ′(x) =

2x(4x− 3)− x2(4)

(4x− 3)2=

4x2 − 6x

(4x− 3)2

4. Find f ′(z) if f(z) =z ln(z)

z3. Do not simplify.

f ′(z) =

ddz

(z ln z

)z3 − z ln z d

dz

(z3)

(z3)2=

(ln z + z 1

z

)z3 − z ln z(3z2)

(z3)2

7

5. Findd

dt

(et√t− 2

3 ln(t) + t3/2

). Do not simplify.

d

dt

(et√t− 2

3 ln(t) + t3/2

)=

ddt

(et√t− 2

) (3 ln t + t3/2

)−(et√t− 2

)ddt

(3 ln t + t3/2

)(3 ln t + t3/2)

2

=

(et√t + 1

2ett−1/2

) (3 ln t + t3/2

)−(et√t− 2

) (3t

+ 32t1/2)

(3 ln t + t3/2)2

6. Find f ′(x) if f(x) =ex − ln(x)

1− ex. Do not simplify.

f ′(x) =

ddx

(ex − ln(x)

)(1− ex)−

(ex − ln(x)

)ddx

(1− ex

)(1− ex)2

=

(ex − 1

x

)(1− ex)−

(ex − ln(x)

)(−ex)

(1− ex)2

Section 11.4: The Chain Rule

1. Find the indicated derivatives. You don’t have to simplify.

(a) f ′(x) if f(x) = (3x2 − 2x)10 f ′(x) = 10(3x2 − 2x)9(6x− 2)

(b)dy

dxif y = ln(x2 + 1)

dy

dx=

1

x2 + 1(2x)

(c) y′ if y = ex4

y′ = ex4

(4x3)

(d) f ′(x) if f(x) = (x2 lnx)5

f ′(x) = 5(x2 lnx

)4 d

dx

(x2 lnx

)= 5

(x2 lnx

)4(2x lnx + x2

(1

x

))

(e) f ′(w) if f(w) =1

3√

(2w − ew)4

First, note that f(w) = (2w − ew)−4/3. Thus,

f ′(w) = −4

3(2w − ew)−7/3(2− ew)

(f)d

dt

((1− t)2et

3−t)

d

dt

((1− t)2et

3−t)

= 2(1− t)(−1)et3−t + (1− t)2et

3−t(3t2 − 1)

Section 12.1: First Derivative and Graphs

1. For each function, find the intervals on which f(x) is increasing and the intervals on whichf(x) is decreasing. Find any local maxes or mins.

(a) f(x) = 2x2 − 8x + 9

Note f ′(x) = 4x − 8 = 4(x − 2). Thus, x = 2 is the only partition number forf ′(x) (it is also a critical point for f(x)). We make a sign chart for f ′(x).

f ′(x) −−−−−−−−−−−0+++++++++++

2

Thus, f is increasing on (2,∞) and decreasing on (−∞, 2). There is a local min at(2, f(2)) = (2, 1). There is no local max.

8

(b) f(x) = −x4 + 50x2

Note f ′(x) = −4x3 + 100x = −4x(x2 − 25) = −4x(x + 5)(x − 5). Thus, the parti-tion numbers of f ′(x) (and the critical points of f(x)) are x = 0, x = −5, and x = 5.The sign chart for f ′(x) is given below.

f ′(x) ++++++++0−−−−−−−−0++++++++0−−−−−−−−

−5 0 5

Thus, we see that f is increasing on (−∞,−5) and on (0, 5) and decreasing on (−5, 0)and (5,∞). There is a local max at (−5, f(−5)) = (−5, 625) and (5, f(5)) = (5, 625).There is a local min at (0, 0).

(c) f(x) = x3 + 3x2 + 3x

Note f ′(x) = 3x2 + 6x + 3 = 3(x2 + 2x + 1) = 3(x + 1)2. Hence the only parti-tion number for f ′(x) is x = −1. Here is a sign chart for f ′(x):

f ′(x) ++++++++0++++++++

−1

Thus, f is increasing on (−∞,∞). There are no local maxes or mins.

2. Use the given info to sketch the graph of f . Assume its domain is (−∞,∞).

� f(−2) = 4, f(0) = 0, f(2) = −4

� f ′(−2) = 0, f ′(0) = 0, f ′(2) = 0

� f ′(x) > 0 on (−∞,−2) and (2,∞)

� f ′(x) < 0 on (−2, 0) and (0, 2)

Section 12:2: Second Derivative and Graphs

1. Find the inflection points of f(x) = ln(x2 − 4x + 5).

This is a problem in your textbook. See the full solution on p. 653. The inflection pointsare (1, ln 2) and (3, ln 2).

2. Suppose f(x) = x1/3. Find the inflection point(s) of f(x). Also, find the intervals on whichf(x) is concave up and the intervals on which f(x) is concave down.

Note f ′(x) =1

3x−2/3. So f ′′(x) = −2

9x−5/3 = − 2

93√x5

. Thus, the only partition num-

ber of f ′′(x) is at x = 0. Here’s a sign chart for f ′′(x):

f ′′(x) ++++++++0−−−−−−−−

0

Since f ′′(x) changes sign at x = 0 and since x = 0 is in the domain of f(x), (0, 0) isan inflection point of f(x). Also, notice f is concave up on (−∞, 0) and concave down on(0,∞).

9

3. Suppose f(x) = x4 − 2x3. Find

(a) The domain of f(x).

(b) The intercepts of f(x).

(c) Make a sign chart for f ′(x). Find its critcal numbers. Find any local extrema. List theintervals on which f is increasing and the intervals on which f is decreasing.

(d) Make a sign chart for f ′′(x). Find any inflection point(s) and list the intervals on whichf(x) is concave up or concave down.

(e) Sketch the graph of f .

The solution for this problem is given in your book on pages 655-656.

Section 12.3: L’Hopital’s Rule

1. For each limit, determine if the limit is a0

0indeterminate form, a

∞∞

indeterminate form,

or neither. Can L’Hopital’s Rule be applied to find the limit? If not, explain why not. If itcan be applied, use L’Hopital’s Rule to find the limit.

(a) limx→0

ln(1 + x2)

x4Solution in book on p. 668

(b) limx→1

lnx

xSolution in book on p. 669

(c) limx→−∞

ln(1 + 2ex)

ex

This is Matched Problem 8 in your book on p. 671. The solution is given on p.674

(d) limx→∞

e−x

lnx

Note that limx→∞

lnx = ∞; however, limx→∞

e−x = 0. Thus, this is not a∞∞

indetermi-

nate form. Thus, L’Hopital’s Rule can not be used.

(e) limx→0

3x + 1− e3x

x2

Note that limx→0

(3x + 1 − e3x

)= 0 + 1 − e0 = 1 − 1 = 0 and lim

x→0x2 = 0. Thus,

the limit is a0

0indeterminate form and hence L’Hopital’s Rule can be used. We get

that

limx→0

3x + 1− e3x

x2= lim

x→0

3− 3e3x

2x

This limit is still a0

0indeterminate form and we use L’Hopital’s Rule again to get that

limx→0

3x + 1− e3x

x2= lim

x→0

−9e3x

2= −9

2

2. For each limit listed, first (a) find the limit using previous methods, and then (b) useL’Hopital’s Rule to find the limit.

(a) limx→∞

2x2 − 1

x2 + 1

Using Theorem 4 on p. 509, we get that

limx→∞

2x2 − 1

x2 + 1= lim

x→∞

2x2

x2= 2

Using L’Hopital’s Rule (note the limit is a∞∞

indeterminate form), we get the sameanswer:

limx→∞

2x2 − 1

x2 + 1= lim

x→∞

4x

2x= 2

10

(b) limx→∞

3x− 1

x2 − 4

Again we can use Theorem 4 on p. 509 to get

limx→∞

3x− 1

x2 − 4= lim

x→∞

3x

x2= lim

x→∞

3

x= 0

Using L’Hopital’s Rule (note the limit is a∞∞

indeterminate form):

limx→∞

3x− 1

x2 − 4= lim

x→∞

3

2x= 0

(c) limx→−3

x2 − 9

x + 3

We can evaluate this limit using previous methods by factoring:

limx→−3

x2 − 9

x + 3= lim

x→−3

(x + 3)(x− 3)

x + 3= lim

x→−3(x− 3) = −3− 3 = −6

Using L’Hopital’s Rule (note the limit is a0

0indeterminate form):

limx→−3

x2 − 9

x + 3= lim

x→−3

2x

1= 2(−3) = −6

Section 12.4: Curve Sketching Techniques

1. Use the graphing strategy taught in this section to analyze and graph the following functions.

(a) g(x) =4x + 3

x2

This is Matched Problem 2 on p. 677 of your book. The solution is given on p.686.

(b) f(x) = xe−0.5x

This is Matched Problem 3 in your book on p. 679. The solution is given on pages686-687.

(c) f(x) = x lnx

This is Matched Problem 4 in your book. The solution is given on p. 687.

2. Sketch a function which satisfies the following properties.

� f(−3) = −1, f(0) = 0, f(3) = 1

� f ′(x) < 0 on (−∞,−2) and (2,∞)

� f ′(x) > 0 on (−2, 2)

� f ′′(x) < 0 on (∞,−2) and (−2, 0)

� f ′′(x) > 0 on (0, 2) and (2,∞)

� vertical asymptotes: x = −2, x = 2

� horizontal asymptotes: y = 0

11

Section 12.5: Absolute Maxima and Minima

1. Find the absolute maximum and absolute minimum of f(x) = x3 − 12x on each of thefollowing intervals:

(a) [−3, 3]

Note f is continuous on [−3, 3]. Thus, we just need to evaluate f at any criticalpoints and at the endpoints. The absolute max will be the maximum of these values,and the absolute min will be the minimum of these values. First, we find the criticalpoints. Note f ′(x) = 3x2−12 = 3(x2−4) = 3(x+2)(x−2). Thus, f ′(x) = 0 at x = −2and x = 2. The function values at these critical points are

f(−2) = (−2)3 − 12(−2) = −8 + 24 = 16

f(2) = (2)3 − 12(2) = 8− 24 = −16

Now we evaluate f at the endpoints. Note

f(−3) = (−3)3 − 12(−3) = −27 + 36 = 9

f(3) = (3)3 − 12(3) = 27− 36 = −9

Thus, the absolute max is 16 and the absolute min is −16.

(b) [−3, 1]

Note

f(−3) = (−3)3 − 12(−3) = −27 + 36 = 9

f(1) = (1)3 − 12(1) = −11

We also know f evaluated at x = −2 from part (a):

f(−2) = (−2)3 − 12(−2) = −8 + 24 = 16

Note the critical number x = 2 is outside of the interval [−4, 1] so we don’t care aboutf evaluated at this critical number. Thus, the absolute max is 16 and the absolute minis −11.

2. Find the absolute extrema of each function on (0,∞).

(a) f(x) = 12− x− 5

x(b) f(x) = f ln(x)− x

This is Matched Problem 3 from your book. The answers are on p. 695.

12

Section 12.6: Optimization

1. Find the greatest possible product of two numbers given that the sum of the two numbersequals 28.

Let x and y be the two numbers. We must have that x+y = 28. So y = 28−x. The functionwe want to maximize is f(x) = xy = x(28 − x) = 28x − x2. Note f ′(x) = 28 − 2x. Thus,f ′(x) = 0 when x = 14. A sign chart for f ′(x) would show that there is an absolute maximumat x = 14. Thus, one of the numbers is 14. The other number is y = 28− x = 28− 14 = 14.Thus, the two numbers are 14 and 14.

2. Find two numbers whose difference is 80 and whose product is a minimum.

Let x and y be the two numbers. We must have x − y = 80. So y = x − 80. We want tomaximize f(x) = xy = x(x − 80) = x2 − 80x. Note f ′(x) = 2x − 80. So f ′(x) = 0 whenx = 40. A sign chart for f ′(x) would show that there is an absolute minimum at x = 40.Thus, one of the numbers is 40. The other number is y = x − 80 = 40 − 80 = −40. Thus,the two numbers are 40 and −40.

3. Find the area of the largest rectangle that can be made with a perimeter of 56 ft.

Let x and y be the lengths of the two sides of the rectangle. The perimeter is 56 ft, sowe have the equation 2x + 2y = 56. This simplifies to x + y = 28. Thus, y = 28 − x. Wewant to maximize A(x) = xy = x(28−x) = 28x−x2. Note A′(x) = 28−2x. Thus, A′(x) = 0when x = 14. A sign chart would show that there is an absolute max at x = 14. Thus, oneof the sides has length 14 ft. Note when x = 14, y = 28 − x = 28 − 14 = 14, so the otherside also has length 14 ft. This means the area would be (14)2 = 196 ft2.

4. Find the smallest possible perimeter of a rectangle whose area is 400 ft2.

Let x and y be the lengths of the two sides of the rectangle. The area of the rectan-

gle is 400 ft2, so we have the equation xy = 400. So y =400

x. We want to maximize

P (x) = 2x + 2y = 2x + 2

(400

x

)= 2x +

800

x. Note P ′(x) = 2 − 800

x2=

2x2 − 800

x2. Thus,

P ′(x) = 0 when 2x2 − 800 = 0. Note

2x2 − 800 = 0 ⇒ 2(x2 − 400) = 0

⇒ 2(x− 20)(x + 20) = 0

⇒ x = 20 or x = −20

Since a length can’t be negative, x = 20. A sign chart for P ′(x) would show that a minimumoccurs at x = 20. Thus, the smallest possible perimeter occurs when x = 20. Note P (20) =2(20) + 800

20= 40 + 40 = 80. So the largest possible perimeter of a rectangle with area 400

ft2 is 80 ft.

5. Suppose you want to fence a rectangular area with one side against a barn (so no fencingneeds to be used for that side). If the amount of fencing to be used is 40 ft, find the dimen-sions of the rectangle that has the maximum area.

Drawing a picture here would be a good idea. Let x represent the length of each sideof the rectangle neither of which is opposite the barn. Let y represent the length of theside of the rectangle which is opposite the barn. The amount of fencing to be used is40 ft, so we must have that 2x + y = 40. So y = 40 − 2x. We want to maximizeA(x) = xy = x(40 − 2x) = 40x − 2x2. Note A′(x) = 40 − 4x. So A′(x) = 0 whenx = 10. A sign chart would show that an absolute max occurs at x = 10. When x = 10,y = 40− 2x = 40− 2(10) = 20. Thus, the dimensions are 10 ft × 20 ft (with the side of 20ft of fencing being opposite the barn).

Section 13.1: Antiderivatives and Indefinite Integrals

1. What does it mean for F (x) to be an antiderivative of f(x)?

It means F ′(x) = f(x).

13

2. If n is any real number, what do the family of antiderivatives of xn look like?

If n 6= −1, then the family of antiderivatives of xn look like 1n+1

xn+1 + C. If n = −1,then the family of antiderivatives of xn = x−1 look like ln |x|+ C.

3. Find an antiderivative of f(x) = 3x2.

An antiderivative of f(x) = 3x2 is F (x) = x3 + 2. (One could also have chosen any functionof the form F (x) = x3 + C.)

4. Find an antiderivative of f(x) = 2x−1.

An antiderivative of f(x) = 2x−1 is F (x) = 2 ln |x|. (One could also have chosen anyfunction of the form F (x) = 2 ln |x|+ C.)

5. Find each indefinite integral.

(a)

∫ (x2 − 5

x+ 4ex

)dx

∫ (x2 − 5

x+ 4ex

)dx =

1

3x3 − 5 ln |x|+ 4ex + C

(b)

∫ (√x− 2

x2

)dx

∫ (√x− 2

x2

)dx =

∫ (x1/2 − 2x−2

)dx =

2

3x3/2 + 2x−1 + C

(c)

∫ ( 1√x− x5

)dx

∫ ( 1√x− x5

)dx =

∫ (x−1/2 − x5

)dx = 2x1/2 − 1

6x6 + C

(d)

∫x(x2 + 1) dx

∫x(x2 + 1) dx =

∫ (x3 + x

)dx =

1

4x4 +

1

2x2 + C

(One could also do a u-substitution with u = x2 + 1.)

6. Circle true or false to the following statements.

(a) Every function has an infinite number of antiderivatives.

true false

(b) An antiderivative of f(x) = 1x

is F (x) = ln |x|+ 10.

true false

(c) An antiderivative of f(x) = 2x is x2 + x.

true false

(d) The constant function f(x) = 0 is an antiderivative of itself.

true false

(e) The function f(x) = 5ex is an antiderivative of itself.

14

true false

(f) If n is an integer, then1

n + 1xn+1 is an antiderivative of xn.

true false The integer could be −1

Section 13.2: Integration by Substitution

1. Use integration by substitution to find each integral.

(a)

∫ √2− x dx

Let u = 2− x. Then du = −dx and∫ √2− x dx =

∫u1/2 (−du) = −2

3u3/2 + C = −2

3(2− x)3/2 + C

(b)

∫x√x2 + 1 dx

Let u = x2 + 1, du = 2x dx. Then∫x√x2 + 1 dx =

1

2

∫u1/2 du =

1

2

(2

3u3/2

)+ C =

1

3

(x2 + 1

)3/2+ C

(c)

∫x2

√x3 + 5

dx

Let u = x3 + 5, du = 3x2 dx. Then∫x2

√x3 + 5

dx =

∫ 13du√u

=1

3

∫u−1/2 du =

1

3

(2u1/2

)+ C =

2

3

(x3 + 5

)1/2+ C

2. Integrate.

(a)

∫e−3x dx

(b)

∫x

x2 − 9dx

(c)

∫5t2(t3 + 4)−2 dt

This is Matched Problem 5 on p. 731. The answers are on p. 736.

Section 13.4: The Definite Integral

1. Find the area under the graph of f(x) =√x but above the x-axis between x = 1 and x = 9.

Note√x lies completely above the x-axis between x = 1 and x = 9. Thus, the area is

given by∫ 9

1

√x dx =

∫ 9

1

x1/2 dx =2

3x3/2

∣∣∣91

=2

3

((9)3/2 − (1)3/2

)=

2

3

(27− 1

)=

52

3

2. Consider the graph of f(x) shown below, where A,B, and C represent the shaded regionsshown.

Suppose

area of A = 1.5 area of B = 2.5 area of C = 0.5

Find the following definite integrals.

15

(a)

∫ 0

−4f(x) dx = −B = −2.5

(b)

∫ 1

−4f(x) dx = −B − C = −2.5− 0.5 = −3

(c)

∫ −4−5

f(x) dx = A = 1.5

(d)

∫ 1

−5f(x) dx = A−B − C = 1.5− 2.5− 0.5 = −1.5

(e)

∫ −50

f(x) dx −∫ 0

−5f(x) dx = −A + B = −1.5 + 2.5 = 1

3. Find the following definite integrals.

(a)

∫ 2

1

(x3 − x

)dx

∫ 2

1

(x3 − x

)dx =

(1

4x4 − 1

2x2)∣∣∣2

1

=1

4(2)4 − 1

2(2)2 −

(1

4(1)4 − 1

2(1)2

)= 4− 2−

(−1

4

)= 2 +

1

4

=9

4

(b)

∫ 3

1

(6x2 − 1

x

)dx

∫ 3

1

(6x2 − 1

x

)dx =

(2x3 − ln |x|

)∣∣∣31

= 2(3)3 − ln |3| −(

2(1)3 − ln |1|)

= 54− ln(3)− 2 + 0

= 52− ln(3)

(c)

∫ 1

0

x

x2 − 2dx

16

Let u = x2 − 2, du = 2x dx. Note when x = 0, u = 02 − 2 = −2 and when x = 1,u = (1)2 − 2 = −1. Thus,∫ 1

0

x

x2 − 2dx =

1

2

∫ −1−2

1

udu

=1

2ln |u|

∣∣∣−1−2

=1

2

(ln | − 1| − ln | − 2|

)= −1

2ln(2)

(d)

∫ 2

−1

√4− x dx

Let u = 4 − x, du = −dx. When x = −1, u = 4 − (−1) = 5. When x = 2,u = 4− 2 = 2. Thus, ∫ 2

−1

√4− x dx =

∫ 2

5

√u (−du)

= −2

3u3/2

∣∣∣25

= −2

3

((2)3/2 − (5)3/2

)

(e)

∫ 3

1

2x√x2 − 1

dx

Let u = x2 − 1, du = 2x dx. Note when x = 1, u = (1)2 − 1 = 0, and whenx = 3, u = (3)2 − 1 = 8. Thus,∫ 3

1

2x√x2 − 1

dx =

∫ 8

0

du√u

=

∫ 8

0

u−1/2 du

= 2u1/2∣∣∣80

= 2(√

8−√

0)

= 2√

8 = 2(2√

2) = 4√

2

17

Section 14.1: Area Between Curves

1. Find the area bounded by the graphs of f(x) = x and g(x) = −x2 + 4x as shown below.

The area is given by∫ 3

0

(− x2 + 4x− x

)dx =

∫ 3

0

(− x2 + 3x

)dx

=(− 1

3x3 +

3

2x2)∣∣∣3

0

= −1

3(3)3 +

3

2(3)2 − (0 + 0)

= −9 +27

2= −18

2+

27

2=

9

2

18

2. Find the area between the graph of f(x) = x2−x−2 and the x-axis over [−2, 2]. The graphof f(x) has been graphed below. The area will be the sum of the areas of the two shadedregions A and B as shown.

We need to evaluate two separate integrals here, one to find the area of region A and one tofind the area of region B. We can then add these together to get the total area.

First, we find the area of region A. This is given by∫ −1−2

(x2 − x− 2− 0

)dx =

(1

3x3 − 1

2x2 − 2x

)∣∣∣−1−2

=1

3(−1)3 − 1

2(−1)2 − 2(−1)−

(1

3(−2)3 − 1

2(−2)2 − 2(−2)

)= −1

3− 1

2+ 2−

(− 8

3− 2 + 4

)= −1

3− 1

2+ 2 +

8

3− 2

=7

3− 1

2

=14

6− 3

6

=11

6

Now we find the area of region B. This is given by∫ 2

−1

(0−

(x2 − x− 2

))dx =

∫ 2

−1

(−x2 + x + 2

)dx

=

(−1

3x3 +

1

2x2 + 2x

) ∣∣∣2−1

= −1

3(2)3 +

1

2(2)2 + 2(2)−

(−1

3(−1)3 +

1

2(−1)2 + 2(−1)

)= −8

3+ 2 + 4−

(1

3+

1

2− 2

)= −8

3+ 6− 1

3− 1

2+ 2

= −3 + 6− 1

2+ 2

= 5− 1

2

=9

2

Thus, the total area is11

6+

9

2=

19

3.

19