CALCULUS 1: HOMEWORK SOLUTIONS, FALL 2012

STEVEN HEILMAN

Contents

1. Chapter 1 21.1. Section 1.1 21.2. Section 1.2 21.3. Section 1.3 31.4. Section 1.4 51.5. Section 1.5 81.6. Section 1.6 92. Chapter 2 112.1. Section 2.1 112.2. Section 2.2 122.3. Section 2.3 122.4. Section 2.4 132.5. Section 2.5 142.6. Section 2.6 152.7. Section 2.7 162.8. Section 2.8 173. Chapter 3 173.1. Section 3.1 173.2. Section 3.2 183.3. Section 3.3 183.4. Section 3.4 193.5. Section 3.5 193.6. Section 3.6 203.7. Section 3.7 204. Chapter 4 224.1. Section 4.1 224.2. Section 4.2 224.3. Section 4.3 224.4. Section 4.4 234.5. Section 4.5 244.6. Section 4.6 254.7. Section 4.7 255. Appendix: Notation 26

Date: December 7, 2012.

1

1. Chapter 1

1.1. Section 1.1.

Exercise 1.1.1. Sketch the region in the plane that solves the inequality

|x|+ |y| ≤ 1.

We will show that this region is a diamond with corners (1, 0), (−1, 0), (0, 1), (0,−1).

Proof. We break into four cases that together encompass all (x, y) ∈ R× R.Case 1. x ≥ 0 and y ≥ 0. In this case, |x| = x and |y| = y, so we need to find (x, y) ∈ R×R

such that x + y ≤ 1. In the first quadrant, this is the region under (and including) the liney = −x+ 1.

Case 2. x ≤ 0 and y ≥ 0. In this case, |x| = −x and |y| = y, so we need to find(x, y) ∈ R× R such that −x+ y ≤ 1. In the second quadrant, this is the region under (andincluding) the line y = 1 + x.

Case 3. x ≤ 0 and y ≤ 0. In this case, |x| = −x and |y| = −y, so we need to find(x, y) ∈ R × R such that −x − y ≤ 1. In the third quadrant, this is the region above (andincluding) the line y = −x− 1.

Case 4. x ≥ 0 and y ≤ 0. In this case, |x| = x and |y| = −y, so we need to find(x, y) ∈ R × R such that x − y ≤ 1. In the fourth quadrant, this is the region above (andincluding) the line y = x− 1. �

Exercise 1.1.2. Consider the curve that satisfies the following equation for x, y ∈ Rx4 − 4x2 − x2y2 + 4x2 = 0.

Is the curve the graph of a function?

We show that this curve is not the graph of a function

Proof. If this curve were the graph of a function y(x), we would have x4−x2(y(x))2 = 0, i.e.for x 6= 0 we would have (y(x))2 = x2. But then y(x) could have the value x or −x. Sincethere are two possible values for y(x), we conclude that y(x) cannot be a single function, i.e.the given curve is not the graph of a function. �

1.2. Section 1.2.

Exercise 1.2.1. Let f(x) = m1x + b1 and g(x) = m2x + b2, m1,m2, b1, b2 ∈ R. Show thatf ◦ g is also a linear function. What is the slope of f ◦ g?

We show that f ◦ g is a linear function with slope m1m2.

Proof.(f ◦ g)(x) = f(g(x)) = m1(m2x+ b2) + b1 = [m1m2]x+ [m1b2 + b1].

Note that M = m1m2 ∈ R and B = m1b2 + b1 ∈ R, so (f ◦ g)(x) = Mx + B is a linearfunction with slope M = m1m2. �

Exercise 1.2.2. For x ∈ R, let

f0(x) =1

2− x.

For n ∈ Z, n ≥ 1, define fn+1(x) = (f0 ◦ fn)(x).

(a) Find formulas for f1, f2, f3, f4

2

(b) Find an explicit formula for fn(x).(c) Graph f0, f1, f2, f3 on the same set of axes and describe the qualitative behavior of

the functions.

Proof of (a).

f1(x) = f0(f0(x)) =1

2−(

12−x

) =2− x

2(2− x)− 1=

2− x3− 2x

f2(x) = f0(f1(x)) =1

2−(

2−x3−2x

) =3− 2x

2(3− 2x)− (2− x)=

3− 2x

4− 3x

f3(x) = f0(f2(x)) =1

2−(3−2x4−3x

) =4− 3x

2(4− 3x)− (3− 2x)=

4− 3x

5− 4x

f4(x) = f0(f3(x)) =1

2−(4−3x5−4x

) =5− 4x

2(5− 4x)− (4− 3x)=

5− 4x

6− 5x

�

Proof of (b). For n a positive integer, we have

fn(x) =(n+ 1)− nx

(n+ 2)− (n+ 1)x.

We can justify this formula by induction. We first check the base case. From part (a), weknow that

f1(x) =2− x3− 2x

.

The base case of induction is complete. We now complete the inductive step. Assume that

fn(x) = (n+1)−nx(n+2)−(n+1)x

. We will show that fn+1(x) = (n+2)−(n+1)x(n+3)−(n+2)x

. Observe

fn+1(x) = f0(fn(x)) =1

2−(

(n+1)−nx(n+2)−(n+1)x

)=

(n+ 2)− (n+ 1)x

2((n+ 2)− (n+ 1)x)− ((n+ 1)− nx)=

(n+ 2)− (n+ 1)x

(n+ 3)− (n+ 2)x

Since the inductive step has been completed, the assertion is proven. �

Description of (c). Each function has a vertical asymptote that appears to approachx = 1 as n→∞. Also, the functions seem to be approaching the value of 1 away from thesingularities.

1.3. Section 1.3.

Exercise 1.3.1. (1.3.22)

(a) Evaluate h(x) = (tan(x)− x)/x3 for x = 1, .5, .1, .05, .01, and .005.

(b) Guess the value of limx→0tan(x)−x

x3.

(c) Evaluate h(x) for successively smaller values of x until you finally reach 0 values forh(x). Are you still confident that your guess in part (b) is correct?

3

(d) Graph the function h in the viewing rectangle [−1, 1] by [0, 1]. Then zoom in to-ward the point where the graph crosses the y-axis to estimate the limit of h(x) as xapproaches 0. Continue to zoom in until you observe distortions in the graph of h.Compare with the results of part (c).

(a) h(1) = 0.55740772465490, h(.5) = 0.37041991875032, h(.1) = 0.33467208545054,h(.05) = 0.33366700431031, h(.01) = 0.33334666720702, h(.005) = 0.33333666669966.

(b) We guess that limx→0tan(x)−x

x3= 1

3.

(c) and (d) For x < 10−10, the electronic device computes an equality tan(x) = x. Specifi-cally, the electronic device is evaluating a polynomial in x of the form x+bx3+cx5+· · ·+kx19.For small x, the x3, x5 and higher order terms are so small that they contribute nothing

4

when added to x, since the electronic device only has about 13 digits of precision (in base10). So, for such x values, the electronic device will say that h(x) = 0.

Exercise 1.3.2. In your own words, explain why limx→0H(x) does not exist, according tothe intuitive definition of the limit.

When x is very close to 0 with x negative, we have H(x) = 0. However, when x is veryclose to 0 with x positive, we have H(x) = 1. So, no matter how close to zero we look, we canalways find nearby x, y with H(x) = 0 and H(y) = 1. So, it is impossible for limx→0H(x)to exist, since H(x) is not getting close to any particular value.

Exercise 1.3.3. Find two functions f and g so that neither function has a limit as x→ a,but limx→a(f + g) exists.

Let f(x) = H(x) and let g(x) = −H(x). As in the previous Exercise, we know thatlimx→0 f(x) DNE and limx→0 g(x) DNE. However, f(x)+g(x) = 0, so limx→0(f(x)+g(x)) =0.

Alternately, let f(x) = 1/x and let g(x) = −1/x. Then limx→0 f(x) DNE and limx→0 g(x)DNE. However, f(x) + g(x) = 0 for x 6= 0, so limx→0(f(x) + g(x)) = 0.

1.4. Section 1.4.

Exercise 1.4.1. If limx→a(f(x)+g(x)) = 2 and limx→a(f(x)−g(x)) = 1, find limx→a[f(x)g(x)].

We show that limx→a[f(x)g(x)] = 3/4.

First Proof. Write 2f = (f + g) + (f − g). Since limx→a(f + g) exists, and limx→a(f − g)exists, we can then apply our limit law to deduce that

2 limx→a

f(x) = [limx→a

(f(x) + g(x))] + [limx→a

(f(x)− g(x))] = 2 + 1 = 3.

5

So, limx→a f(x) = 3/2. Similarly, 2g = (f + g) − (f − g). Since limx→a(f + g) exists, andlimx→a(f − g) exists, we can then apply our limit law to deduce that

2 limx→a

g(x) = [limx→a

(f(x) + g(x))]− [limx→a

(f(x)− g(x))] = 2− 1 = 1.

So, limx→a g(x) = 1/2. Finally, since limx→a f(x) exists, and limx→a g(x) exists, we can applyanother limit law to see that

limx→a

[f(x)g(x)] = (limx→a

f(x))(limx→a

g(x)) = (3/2)(1/2) = 3/4.

�

Second Proof. We use the identity

fg =1

4[(f + g)2 − (f − g)2] (∗).

Then

3

4=

1

4(22 − 12) =

1

4

([limx→a

(f(x) + g(x))]2−[limx→a

(f(x)− g(x))]2)

=1

4

(limx→a

[(f(x) + g(x))2

]− lim

x→a

[(f(x)− g(x))2

]), by the limit law for products

=1

4

(limx→a

[(f(x) + g(x))2 − (f(x)− g(x))2

]), by the limit law for sums

= limx→a

[f(x)g(x)] , by (∗)

�

Exercise 1.4.2. Find all values of a such that

limx→0

√ax+ 4− 2

x= 1.

We prove that a = 4

Proof. Let a ∈ R. For |x| < 1/ |a|, we have |ax| < 1, so the triangle inequality says

|ax+ 4| ≥ 4 − |ax| ≥ 3 > 0. In particular,√

(ax+ 4)2 = |ax+ 4| = ax + 4. For x with|x| < 1/ |a|, we write

√ax+ 4− 2

x=

(√ax+ 4− 2)

x

√ax+ 4 + 2√ax+ 4 + 2

=|ax+ 4| − 4

x(√ax+ 4 + 2)

=ax

x(√ax+ 4 + 2)

=a√

ax+ 4 + 2.

So, letting x→ 0, we get

limx→0

√ax+ 4− 2

x= lim

x→0

a√ax+ 4 + 2

=a√

4 + 2=a

4.

In conclusion, we must have a = 4. �

Exercise 1.4.3. Compute

limx→0

|2x− 1| − |2x+ 1|x

.

6

We show that limx→0|2x−1|−|2x+1|

x= −4.

Proof. Let x with |x| < 1/2. Then |2x| < 1, so 2x − 1 ≤ |2x| − 1 < 0. In particular,|2x− 1| = 1−2x. Consider again x with |x| < 1/2. Then |2x| < 1, so 2x+1 ≥ − |2x|+1 > 0.In particular, |2x+ 1| = 2x+ 1. Combining these two facts, we have, for |x| < 1/2

|2x− 1| − |2x+ 1|x

=1− 2x− (2x+ 1)

x=−4x

x= −4.

As x→ 0, we only need to consider |x| < 1/2, so we conclude that

limx→0

|2x− 1| − |2x+ 1|x

= limx→0

(−4) = −4.

�

Exercise 1.4.4. Prove that limx→0+√x [1 + (sin(2π/x))2] = 0.

Proof. Recall that 0 ≤ (sin(2π/x))2 ≤ 1. Therefore, 1 ≤ 1 + (sin(2π/x))2 ≤ 2. For x > 0,√x > 0, so we can multiply our inequality by

√x to get

√x ≤√x [1 + (sin(2π/x))2] ≤ 2

√x.

We know that limx→0+√x = 0. So, by the Squeeze Theorem, we have

0 ≤ limx→0+

√x ≤ lim

x→0+

√x [1 + (sin(2π/x))2] ≤ 2 lim

x→0+

√x = 0.

We conclude that limx→0+√x [1 + (sin(2π/x))2] = 0. �

Exercise 1.4.5. Find the following limit

limx→0

sin(3x) sin(5x)

x2.

We show that limx→0sin(3x) sin(5x)

x2= 15.

Proof. First, recall that limx→0sin(x)x

= 1. In particular, limx→0sin(3x)

3xexists, limx→0

sin(5x)5x

exists, and

limx→0

sin(3x)

3x= lim

x→0

sin(5x)

5x= lim

y→0

sin(y)

y= 1.

Therefore,

15 = 15

(limx→0

sin(3x)

3x

)(limx→0

sin(5x)

5x

)= 15

(limx→0

sin(3x)

3x

sin(5x)

5x

), by the limit law for products

= limx→0

15sin(3x) sin(5x)

15x= lim

x→0

sin(3x) sin(5x)

x2.

�

Exercise 1.4.6. Show that

limθ→0

cos(θ)− 1

θ= 0.

7

Proof. Let θ 6= 0. Then

cos(θ)− 1

θ=

cos(θ)− 1

θ· cos(θ) + 1

cos(θ) + 1=

(cos(θ))2 − 1

θ(cos(θ) + 1)=−(sin(θ))2

θ(cos(θ) + 1)=− sin(θ)

θ· sin θ

cos(θ) + 1.

Now, limθ→0(− sin(θ)/θ) = −1, limθ→0 sin(θ) = 0, and limθ→0 cos(θ) = 1. So, by our limitlaw for products,

limθ→0

cos(θ)− 1

θ=

(limθ→0

− sin(θ)

θ

)(limθ→0

sin θ

cos(θ) + 1

)= (−1)

0

2= 0.

�

1.5. Section 1.5.

Exercise 1.5.1. Find all values of a, b ∈ R such that the following function is continuous.

f(x) =

ax− b , x ≤ −1

2x2 + 3ax+ b ,−1 < x ≤ 1

4 , x > 1

.

We show that a = 3/4, b = −1/4.

Proof. By the definition of f , f is a polynomial on the intervals (−∞,−1), (−1, 1), and(1,∞). That is, f is continuous on (−∞,−1) ∪ (−1, 1) ∪ (1,∞). So, in order for f to becontinuous on the entire real line, it suffices to show that limx→−1− f(x) = limx→−1+ f(x)and limx→1− f(x) = limx→1+ f(x). We therefore have the following two equations

a(−1)− b = 2 + 3a(−1) + b, and 2 + 3a+ b = 4.

That is, −2a+2b = −2 and 3a+b = 2, i.e. a−b = 1 and 3a+b = 2. Adding these equationsgives 4a = 3, so that a = 3/4. Substituting this equality into a− b = 1 gives b = −1/4. �

Exercise 1.5.2. For what values of x is the function g(x) = (sin(3x5 + 10))1/3 continuous?

We show that g is continuous for all x ∈ R.

Proof. Let f(x) = x1/3, let h(x) = sin(x), and let k(x) = 3x5 + 10. Then g(x) = f(h(k(x))).Also, k is a polynomial, so it is continuous with domain all real numbers. And h is continuouswith domain all real numbers. And f is continuous with domain all real numbers. Therefore,g has domain all real numbers, and it is continuous since it is the composition of 3 continuousfunctions. �

Exercise 1.5.3. Use the Intermediate Value Theorem to show that the following equationhas a root in the given interval.

2 sin(x) = 3− 2x, (0, π/2).

Proof. Let h(x) = 2 sin(x) − 3 + 2x. Since h is the sum of continuous functions, h is itselfcontinuous. Also, h(0) = −3 and h(π/2) = 2 − 3 + π > 2. So, by the IntermediateValue Theorem, there exists y ∈ (0, π/2) such that 2 sin(y) − 3 + 2y = h(y) = 0, i.e.2 sin(y) = 3− 2y. �

8

1.6. Section 1.6.

Exercise 1.6.1. Find

limu→∞

4u4 + 5

(u2 − 2)(2u2 − 1).

We show that limu→∞4u4+5

(u2−2)(2u2−1) = 2.

Proof. Let u 6= 0. Then

4u4 + 5

(u2 − 2)(2u2 − 1)=

4u4 + 5

2u4 − 5u2 + 2=

4 + 5u4

2− 5u2

+ 2u4

.

So, letting u→∞ and using our limit law for quotients,

limu→∞

4u4 + 5

(u2 − 2)(2u2 − 1)= lim

u→∞

4 + 5u4

2− 5u2

+ 2u4

=limu→∞

(4 + 5

u4

)limu→∞

(2− 5

u2+ 2

u4

) =4

2= 2.

�

Exercise 1.6.2. Find all horizontal and vertical asymptotes of the following two functions.

(a) y(x) = (cos(x))2

x2.

(b) y(x) =√x2 + x+ 1−

√x2 − x.

Proof of (a). We show that y(x) has a vertical asymptote at x = 0 and a horizontal asymp-tote of 0. Recall that cos(0) = 1. Since cos(x) is continuous, there exists δ > 0 such that:if |x| < δ, then |cos(x)− 1| < 1/2. Since cos(x) ≤ 1 for all x ∈ R, we conclude: if |x| < δ,then 1/2 ≤ cos(x) ≤ 1. Squaring these inequalities: if |x| < δ, then 1/4 ≤ (cos(x))2 ≤ 1.Then, dividing by x2:

if 0 < |x| < δ, then1

4x2≤ (cos(x))2

x2≤ 1

x2.

As x→ 0, we need only consider 0 < |x| < δ. So, we conclude by the Squeeze Theorem that

∞ = limx→0

1

4x2≤ lim

x→0

(cos(x))2

x2≤ lim

x→0

1

x2=∞.

That is, limx→0(cos(x))2

x2= ∞. On the set (−∞, 0) ∪ (0,∞), y(x) is continuous, since it is

the quotient of two continuous functions with a nonzero denominator. Therefore, the onlyvertical asymptote occurs at x = 0.

We now treat the horizontal asymptote. Recall that −1 ≤ cos(x) ≤ 1 for all x ∈ R.Squaring this inequality gives 0 ≤ (cos(x))2 ≤ 1 for all x ∈ R. Then, dividing by x2 forx 6= 0, we have

0 ≤ (cos(x))2

x2≤ 1

x2.

As x→∞, we need only consider x 6= 0. So, we conclude by the Squeeze Theorem that

0 = limx→∞

0 ≤ limx→∞

(cos(x))2

x2≤ lim

x→∞

1

x2= 0.

So, limx→∞(cos(x))2

x2= 0, i.e. 0 is a horizontal asymptote of y(x). Similarly,

0 = limx→−∞

0 ≤ limx→−∞

(cos(x))2

x2≤ lim

x→−∞

1

x2= 0.

9

So, 0 is the only horizontal asymptote of y(x). �

Proof of (b). We show that y(x) has no vertical asymptotes, though y(x) has horizontalasymptotes of 1 and −1. We first discuss the horizontal asymptotes. First, note that thedomain of y(x) is (−∞, 0] ∪ [1,∞). Specifically, by the quadratic formula, x2 + x + 1 hasno real roots, and it has value 1 at x = 0. Therefore, x2 + x + 1 > 0 for all x ∈ R. Also,x2 − x = x(x− 1) which is nonnegative if and only if: x ≤ 0 or x ≥ 1.

Now, for x ∈ (−∞, 0] ∪ [1,∞), write

√x2 + x+ 1−

√x2 − x = (

√x2 + x+ 1−

√x2 − x)

√x2 + x+ 1 +

√x2 − x√

x2 + x+ 1 +√x2 − x

=(x2 + x+ 1)− (x2 − x)√x2 + x+ 1 +

√x2 − x

=2x+ 1√

x2 + x+ 1 +√x2 − x

=x(2 + 1

x)√

x2(1 + 1x

+ 1x2

) +√x2(1− 1

x)

=x(2 + 1

x)

|x|(√

1 + 1x

+ 1x2

+√

1− 1x

)=

(x

|x|

) 2 + 1x√

1 + 1x

+ 1x2

+√

1− 1x

.

We now apply the limit law for products to get

1 = 1 · 1 =

(limx→∞

x

|x|

) limx→∞

2 + 1x√

1 + 1x

+ 1x2

+√

1− 1x

= lim

x→∞

( x

|x|

) 2 + 1x√

1 + 1x

+ 1x2

+√

1− 1x

= lim

x→∞

(√x2 + x+ 1−

√x2 − x

).

So, 1 is a horizontal asymptote of y(x). Again applying the limit law for products,

−1 = (−1) · 1 =

(lim

x→−∞

x

|x|

) limx→−∞

2 + 1x√

1 + 1x

+ 1x2

+√

1− 1x

= lim

x→−∞

( x

|x|

) 2 + 1x√

1 + 1x

+ 1x2

+√

1− 1x

= lim

x→−∞

(√x2 + x+ 1−

√x2 − x

).

So, −1 is also a horizontal asymptote of y(x).We now show that there are no vertical asymptotes of y(x). Since limx→∞ y(x) = 1 and

limx→−∞ y(x) = −1, the definition of the limit says: there exists N > 1 such that: if

10

x ∈ (−∞,−N) ∪ (N,∞), then |y(x)| ≤ 2. For x ∈ [−N, 0] ∪ [1, N ], we use the triangle

inequality and also the inequality√|y| ≤ 1 + y, y ∈ R to get

|y(x)| ≤√x2 + x+ 1 +

√x2 − x ≤ 2 +

∣∣x2 + x+ 1∣∣+∣∣x2 − x∣∣

≤ 2 +N2 +N + 1 +N2 +N = 2N2 + 2N + 3.

So, for all x ∈ (−∞, 0] ∪ [1,∞), we have |y(x)| ≤ 2N2 + 2N + 3, so it is impossible for avertical asymptote to exist. �

2. Chapter 2

2.1. Section 2.1.

Exercise 2.1.1. (2.1.30) Find f ′(a) for f(x) = 4√1−x .

We show that f ′(a) = 2(1−a)3/2 for a < 1.

Proof. Let a < 1. Then 1 − a > 0, and for h with 0 < |h| < 1 − a we have 1 − a − h >1− a− |h| > 0. So,

f(a+ h)− f(a)

h=

4√1−a−h −

4√1−a

h=

4√1−a√

1−a−h√1−a −

4√1−a−h√

1−a√1−a−h

h=

4(√

1− a−√

1− a− h)

h√

1− a− h√

1− a

=4(√

1− a−√

1− a− h)

h√

1− a− h√

1− a·√

1− a+√

1− a− h√1− a+

√1− a− h

=4((1− a)− (1− a− h))

h√

1− a− h√

1− a (√

1− a+√

1− a− h )

=4h

h√

1− a− h√

1− a (√

1− a+√

1− a− h )

=4√

1− a− h√

1− a (√

1− a+√

1− a− h ).

So, applying our limit laws for products, sums, and quotients,

2

(1− a)3/2=

4

(1− a)2√

1− a=

4√1− a

√1− a (

√1− a+

√1− a )

=4

(limh→0

√1− a− h )

√1− a (

√1− a+ (limh→0

√1− a− h ))

=4

limh→0

[√1− a− h

√1− a (

√1− a+

√1− a− h )

]= lim

h→0

[4√

1− a− h√

1− a (√

1− a+√

1− a− h )

]= lim

h→0

f(a+ h)− f(a)

h= f ′(a).

�

11

2.2. Section 2.2.

Exercise 2.2.1. (2.2.46)

(a) If g(x) = x2/3, show that g′(0) does not exist.(b) If a 6= 0, find g′(a).(c) Show that y = x2/3 has a vertical tangent line at (0, 0).(d) Illustrate part (c) by graphing y = x2/3.

(a) Let x = 0 and let h 6= 0. Then

g(x+ h)− g(x)

h=h2/3

h= h−1/3.

So, g′(0) = limh→0g(x+h)−g(x)

h= limh→0 h

−1/3, and the latter limit does not exist.

(b) For a 6= 0, we show that g′(a) = (2/3)a−1/3.

Proof. Let a 6= 0. We use the formula x− y = (x1/3 − y1/3)(x2/3 + x1/3y1/3 + y2/3). Observe

(a+ h)2/3 − a2/3

h=

(a+ h)2/3 − a2/3

h· (a+ h)4/3 + (a+ h)2/3a2/3 + a4/3

(a+ h)4/3 + (a+ h)2/3a2/3 + a4/3

=(a+ h)2 − a2

h[(a+ h)4/3 + (a+ h)2/3a2/3 + a4/3]

=a2 + 2ah+ h2 − a2

h[(a+ h)4/3 + (a+ h)2/3a2/3 + a4/3]

=2a+ h

(a+ h)4/3 + (a+ h)2/3a2/3 + a4/3.

So, letting h→ 0 and using our limit laws, we get

2

3

1

a1/3=

2a

a4/3 + a4/3 + a4/3=

limh→0 (2a+ h)

limh→0 ((a+ h)4/3 + (a+ h)2/3a2/3 + a4/3)

= limh→0

2a+ h

(a+ h)4/3 + (a+ h)2/3a2/3 + a4/3= lim

h→0

(a+ h)2/3 − a2/3

h= g′(a).

�

(c) From part (b), g′(a) = (2/3)a−1/3. So, lima→0 |g′(a)| = (2/3) lima→0 |a|−1/3 =∞.(d)

2.3. Section 2.3.

Exercise 2.3.1. (2.3.50) If a ball is thrown vertically upward with a velocity of 80 ft/s, thenits height after t seconds is s(t) = 80t− 16t2.

(a) What is the maximum height reached by the ball?(b) What is the velocity of the ball when it is 96 ft above the ground on its way up? On

its way down?

(a) We show that the maximum height of the ball is 100 feet.

12

Proof. Note that s′(t) = 80− 32t. Let t such that s′(t) = 0. Then 80− 32t = 0, i.e. t = 5/2.For t ∈ (0, 5/2), g′(t) > 0, i.e. the ball is rising in the air. For t > 5/2, g′(t) < 0, i.e. the ballis falling in the air. So, the maximum height is reached when t = 5/2, i.e. when the heightof the ball is s(5/2) = 80(5/2)− 16(5/2)2 = 200− 100 = 100. �

(b) We show that the velocity at 96 ft on the way up is 16 ft/s, and the velocity at 96 fton the way down is −16 ft/s.

Proof. Let t with s(t) = 96. Then 80t − 16t2 = 96, so 5t − t2 = 6, so t2 − 5t + 6 = 0, i.e.(t− 2)(t− 3) = 0. For t = 2, s(t) = 96 and s′(t) = 80− 64 = 16 > 0, so the ball is rising att = 2 with velocity 16 ft/s. For t = 3, s(t) = 96 and s′(t) = 80 − 96 = −16, so the ball isfalling at t = 2 with velocity −16 ft/s. �

Exercise 2.3.2. (2.3.66) Suppose the curve y(x) = x4 + ax3 + bx2 + cx + d has a tangentline when x = 0 with equation y = 2x + 1 and a tangent line when x = 1 with equationy = 2− 3x. Find the values of a, b, c, and d.

We show that (a, b, c, d) = (1,−6, 2, 1).

Proof. Note that y′(x) = 4x3 + 3ax2 + 2bx + c. At x = 0, y′(x) = c, and we have a tangentline with slope 2, so y′(0) = c = 2. Also, since the tangent line is y = 2x + 1 at x = 0, wemust have y(0) = 1, so that y(0) = d = 1. So, we have determined the values of c and d. Atx = 1, y′(x) = 4+3a+2b+c = 4+3a+2b+2 = 6+3a+2b. Also, at x = 1 we have a tangentline with slope −3, so y′(1) = 6 + 3a + 2b = −3, i.e. 3a + 2b = −9. Also, since the tangentline is y = 2− 3x at x = 1, we must have y(1) = −1, so that y(1) = 1 + a+ b+ c+ d = −1,i.e. 1 + a + b + 2 + 1 = −1, i.e. a + b = −5, i.e. −2a − 2b = 10. Adding the equations3a+ 2b = −9 and −2a− 2b = 10, we get a = 1. Finally, since a+ b = −5, we conclude thatb = −6. �

2.4. Section 2.4.

13

Exercise 2.4.1. (2.4.48) Find equations of the tangent lines to the curve y = x−1x+1

that areparallel to the line x− 2y = 2.

We show the tangent lines are given by y = (1/2)x− 1/2 and y = (1/2)x+ 7/2.

Proof. Let x 6= −1. Then y′(x) = (x+1)−(x−1)(x+1)2

= 2/(x + 1)2. The line x − 2y = 2 has slope

1/2, so we solve the equation y′(x) = 1/2. We get (x+ 1)2 = 4, so |x+ 1| = 2, so x = 1,−3.Note that y(1) = 0 and y(−3) = 2. So, the two tangent lines are y = (1/2)(x − 1) andy = (1/2)(x+ 3) + 2. �

2.5. Section 2.5.

Exercise 2.5.1. (2.5.74) Suppose y = f(x) is a curve that always lies above the x-axis andnever has a horizontal tangent, where f is differentiable everywhere. For what value of y isthe rate of change of y5 with respect to x eighty times the rate of change of y with respectto x?

We show that y = 2.

Proof. We are asked to find y such that (d/dx)(y5) − 80(d/dx)(y) = 0. Applying the chainrule, we need y such that 0 = 5y4y′ − 80y′ = y′(5y4 − 80). Since y never has a horizontaltangent, y′(x) 6= 0 for all x, so we must have 5y4 = 80, i.e. y4 = 16. Since y is always abovethe x-axis, y(x) > 0 for all x, so we must have y = 2. �

Exercise 2.5.2. (2.5.76) If F (x) = f(xf(xf(x))), where f(1) = 2, f(2) = 3, f ′(1) =4, f ′(2) = 5, and f ′(3) = 6, find F ′(1).

We show that F ′(1) = 198.

Proof.

F ′(x) = f ′(xf(xf(x))) · ddx

[xf(xf(x))], chain rule

= f ′(xf(xf(x))) ·(xd

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

), product rule

= f ′(xf(xf(x))) ·(x(f ′(xf(x)) · d

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

), chain rule

= f ′(xf(xf(x))) · [x(f ′(xf(x)) · [xf ′(x) + f(x)]) + f(xf(x))] , product rule

So, plugging in the given values of f and its derivatives,

F ′(1) = f ′(f(f(1))) · [f ′(f(1)) · [f ′(1) + f(1)] + f(f(1))]

= f ′(f(2)) · [f ′(2) · [4 + 2] + f(2)] = f ′(3) · [5 · 6 + 3] = 6 · [33] = 198

�

14

2.6. Section 2.6.

Exercise 2.6.1. (2.6.12) Suppose y(x) satisfies√x+ y = 1 + x2y2. We find dy/dx by

implicit differentiation.

We show that dy/dx = (−1 + 4xy2√x+ y)/(1− 4yx2

√x+ y).

Proof. Differentiating√x+ y = 1 + x2y2 yields

1 + y′(x)

2√x+ y

= x22yy′(x) + 2xy2.

So, for x+y > 0, we have−1+(x22yy′(x)+2xy2)2√x+ y = y′(x), so y′(x)[1−4yx2

√x+ y ] =

−1 + 4xy2√x+ y, so for 1− 4yx2

√x+ y 6= 0,

y′(x) =−1 + 4xy2

√x+ y

1− 4yx2√x+ y

.

�

Exercise 2.6.2. (2.6.30)

(a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find anequation of the tangent line to this curve at the point (1,−2).

(b) At what points does this curve have a horizontal tangent?(c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common

screen.

(a) We show that the tangent line is given by y = (−9/4)x+ 1/4.

Proof of (a). We first differentiate implicitly in x. The result is 2yy′ = 3x2 + 6x. So, fory 6= 0, we have

y′ =3x2 + 6x

2y.

So, at the point (1,−2), we have y′(1) = −9/4. The tangent line is therefore given by(y + 2) = (−9/4)(x− 1). �

(b) We show that the curve has horizontal tangents only at the points (−2, 2) and (−2,−2).

Proof of (b). From part (a), we solve the equation 0 = y′(x) = (3x2 + 6x)/2y for y 6= 0. So,3x(x+ 2) = 0, so x = 0,−2. For x = −2, the equation for the curve says y2 = x2(x+ 3) = 4,so |y(x)| = 2, and we have two points with horizontal tangents: (−2, 2) and (−2,−2).

For x = 0, the equation for the curve says that y2 = x3 + 3x2, so y(0) = 0. But thederivative formula we used assumed that y 6= 0, so we will need to calculate the derivativein another way, if y = 0. Since y2 = x3 + 3x2, the Tschirnhausen curve can be expressed asthe graph of two functions: y1(x) :=

√x3 + 3x2 and y2(x) := −

√x3 + 3x2.

Since we have checked for horizontal tangents where y 6= 0, it remains to check thederivatives of the points where y = 0. If y = 0, we have x2(x + 3) = 0. So, we have eitherx = 0 or x = −3. We first check the case x = 0. In this case,

y1(h)− y1(0)

h=

√h3 + 3h2

h=|h|h

(h+ 3).

15

So, limh→0+y1(h)−y1(0)

h= 3 and limh→0−

y1(h)−y1(0)h

= −3. Similarly,

y2(h)− y2(0)

h=−√h3 + 3h2

h= −|h|

h(h+ 3).

So, limh→0+y2(h)−y2(0)

h= −3 and limh→0−

y2(h)−y2(0)h

= 3. In order for the curve to have ahorizontal tangent at x = 0, at least one of these limits would need to zero. Since this is notthe case, y does not have a horizontal tangent line at x = 0.

In the case x = −3, first note that the domains of y1 and y2 require that x ≥ −3, so wecan only compute the difference quotients as limits from the right. So, for h > 0

y1(−3 + h)− y1(−3)

h=

√(−3 + h)3 + 3(−3 + h)2

h

=

√−27 + 27h− 9h2 + h3 + 3(9− 6h+ h2)

h

=

√−27 + 27h− 9h2 + h3 + 27− 18h+ 3h2

h

=

√9h− 6h2 + h3

h=

√h(9− 6h+ h2)

h=

√h(h− 3)2

h= |h− 3|h−1/2.

So, limh→0+y1(−3+h)−y1(−3)

h= ∞. Similarly, limh→0+

y2(−3+h)−y2(−3)h

= −∞. Since both ofthese limits are nonzero, we conclude that y does not have a horizontal tangent at x = −3.

Since we have considered all points in the domain of the curve, we conclude that the onlypoints with a horizontal tangent are (−2, 2) and (−2,−2). �

(c)

2.7. Section 2.7.

16

2.8. Section 2.8.

Exercise 2.8.1. (a) Show that the local linear approximation of the function f(x) =(1 + x)15 is 1 + 15x near x = 0, meaning (1 + x)15 ≈ 1 + 15x when x is near zero.

(b) Find an approximate value of cos(31o) using the linear approximation method. (Re-call that 1o = π/180 radians.)

(a) Since f ′(x) = 15(1 + x)14, we have f ′(0) = 15. So, near x = 0,

f(x) ≈ f(0) + f ′(0)x = 1 + 15x.

(b) We show that cos(31o) ≈ (√

3/2)− π/360.

Proof. Recall that cos(30o) = cos(π/6) =√

3/2. Let g(x) = cos(x). Then g′(x) = − sin(x),so g′(π/6) = −1/2. Using the linear approximation for x near π/6, we have

g(x) ≈ g(π/6) + (x− π/6)g′(π/6) =√

3/2− (1/2)(x− π/6)

So, g((π/6) + π/180) ≈√

3/2− (1/2)(π/180) �

3. Chapter 3

3.1. Section 3.1.

Exercise 3.1.1. (3.1.16) Find the domain of the following functions

(a) f(t) = sin(e−t).(b) g(t) =

√1− 2t.

We show that the domain of f is (−∞,∞), and the domain of g is (−∞, 0].

Proof. Let k(t) = e−t and let h(t) = sin(t). Then f = h ◦ k. Also, the domain of k is(−∞,∞), and the range of k is (0,∞). And the domain of h is (−∞,∞). Since the domainof h contains the range of k, we conclude that the domain of f is equal to the domain of k.Since the domain of k is (−∞,∞), we conclude that the domain of f is (−∞,∞).

We now examine g. Let r(t) = 2t, and let t ≤ 0. Since r is increasing, r(t) ≤ r(0) = 20 = 1.So, for t ≤ 0, −2t ≥ −20 = −1, and then 1− 2t ≥ 1− 1 = 0. So, for t ≤ 0, 1− 2t is in thedomain of the square root function. In conclusion, the domain of g contains (−∞, 0]. Wenow show that (0,∞) is not in the domain of g. Let t > 0. Since r is strictly increasing,r(t) > r(0) = 20 = 1. So, for t > 0, −2t < −20 = −1, and then 1− 2t < 1 − 1 = 0. So, fort > 0, 1 − 2t is not in the domain of the square root function. We conclude that (0,∞) isnot in the domain of g. In summary, the domain of g is exactly (−∞, 0]. �

Exercise 3.1.2. (3.1.31) Prove that the following function is odd: f(x) = 1−e1/x1+e1/x

.

Proof. Let x 6= 0. Then

f(−x) =1− e1/(−x)

1 + e1/(−x)=

1− e−1/x

1 + e−1/x=

1− e−1/x

1 + e−1/x· e

1/x

e1/x=e1/x − 1

e1/x + 1= −f(x).

�

17

3.2. Section 3.2.

Exercise 3.2.1. (3.2.26) Find a formula for the inverse of the function y(x) = ex

1+2ex.

We show that y−1(x) = log(x/(1− 2x)).

Proof. First, note that the domain of y is (−∞,∞). Note that (1+2ex)y = ex, so ex(1−2y) =y, so ex = y/(1− 2y), so x = log(y/(1− 2y)). To verify this formula, observe that

y(log(x/(1− 2x))) =elog(x/(1−2x)

1 + 2elog(x/(1−2x)=

x/(1− 2x)

1 + 2x/(1− 2x)=

x

(1− 2x) + 2x= x.

Also

log(y(x)/(1− 2y(x))) = log(y(x))− log(1− 2y(x)) = log

(ex

1 + 2ex

)− log

(1− 2

ex

1 + 2ex

)= log

(ex

1 + 2ex

)− log

((1 + 2ex)− 2ex

1 + 2ex

)= x− log(1 + 2ex)− log(1) + log(1 + 2ex) = x.

�

Exercise 3.2.2. (3.2.40) Suppose f−1 is the inverse of a differentiable function f and letG(x) = 1/f−1(x). If f(3) = 2 and f ′(3) = 1/9, find G′(2).

We show that G′(2) = −1.

Proof. Observe,

d

dx[1/f−1(x)] =

− ddx

(f−1(x))

(f−1(x))2= − 1

f ′(f−1(x))

1

(f−1(x))2.

Also, since f(3) = 2, we have f−1(2) = 3, so G′(2) = [−1/(f ′(3))][1/(32)] = (−9)(1/9) =−1. �

Exercise 3.2.3. (3.2.66) Solve the following equations for x:

(a) log(log(x)) = 1.(b) eax = Cebx, for a 6= b.

For part (a), we show that x = ee. For part (b), we show that x = log(C)/(a− b).

Proof. Suppose log(log(x)) = 1. By exponentiating both sides of this equation, we getlog(x) = e. And then by exponentiating both sides of the equation log(x) = e, we getx = ee.

Suppose eax = Cebx for a 6= b. Taking log of both sides gives ax = log(C) + bx, sox = − log(C)/(b− a). �

3.3. Section 3.3.

Exercise 3.3.1. (3.3.28) Differentiate the following function: y = eu−e−u

eu+e−u .

We show that y′(u) = 4e2u+e−2u+2

.

18

Proof.

y′(u) =(eu + e−u)(eu + e−u)− (eu − e−u)(eu − e−u)

(eu + e−u)2

=e2u + 1 + 1 + e−2u − [e2u − 1− 1 + e−2u]

e2u + 1 + 1 + e−2u

=4

e2u + e−2u + 2.

In the language of hyperbolic trigonometric functions, ddu

tanh(u) = (cosh(u))−2. �

Exercise 3.3.2. (3.3.54) Find the derivative of the function y =√x ex

2−x(x+ 1)2/3.

We show that y′(x) =√x ex

2−x(x+ 1)2/3[

12x

+ 2x− 1 + 23(x+1)

].

Proof. Let u(x) = log(y(x)). Then u′(x) = y′(x)/y(x), so y′(x) = y(x)u′(x). We thereforecompute u′(x).

u′(x) =d

dx[log(√x ex

2−x(x+ 1)2/3)]

=d

dx[(1/2) log(x) + x2 − x+ (2/3) log(x+ 1)]

= (1/2)(1/x) + 2x− 1 + (2/3)(1/(x+ 1)).

�

Exercise 3.3.3. (3.3.67) Show that the function y = Ae−x +Bxe−x satisfies the differentialequation y′′ + 2y′ + y = 0.

Proof.

y′′ + 2y′ + y

= [Ae−x +Bxe−x −Be−x −Be−x] + 2[−Ae−x −Bxe−x +Be−x] + [Ae−x +Bxe−x]

= [A+Bx−B −B − 2A− 2Bx+ 2B + A+Bx]e−x = 0.

�

3.4. Section 3.4.

3.5. Section 3.5.

Exercise 3.5.1. Let g(x) = x cos−1(x/4)−√

16− x2. Find g′(2).

We show that g′(2) = π/3.

Proof. Let x in the domain of g. Recall that (d/dx) cos−1(x) = −1/√

1− x2. Then

g′(x) =−x

4√

1− (x/4)2+ cos−1(x/4)− (−2x)

2√

16− x2.

So, g′(2) = −(1/2)(2/√

3) + cos−1(1/2) + 2/√

12 = −(1/√

3) + (π/3) + (1/√

3) = π/3. �

Exercise 3.5.2. (3.5.34) Find an equation of the tangent line to the curve y = 3 cos−1(x/2)at the point (1, π).

19

We show that the tangent line is given by y = (−√

3)x+ π +√

3.

Proof. Note that y′(x) = −(3/2)/√

1− (x/2)2, so y′(1) = −(3/2)(2/√

3) = −3/√

3 = −√

3.

So, the tangent line is given by (y − π) = −√

3(x− 1). �

Exercise 3.5.3. (3.5.36) Find limx→∞ cos−1(

1+x2

1+2x2

).

We show that limx→∞ cos−1(

1+x2

1+2x2

)= π/3.

Proof. First, note that

limx→∞

1 + x2

1 + 2x2= lim

x→∞

(1/x2) + 1

(1/x2) + 2=

1

2.

Since the domain of the inverse cosine function is [−1, 1], note that 1/2 is in the interior of thedomain of the inverse cosine function. Also, since the inverse cosine function is continuouson its domain, we can move the limit inside the inverse cosine function. That is,

limx→∞

cos−1(

1 + x2

1 + 2x2

)= cos−1

(limx→∞

1 + x2

1 + 2x2

)= cos−1(1/2) = π/3

�

3.6. Section 3.6.

3.7. Section 3.7.

Exercise 3.7.1. (3.7.26) Find limx→0+ sin(x) log(x).

We show that limx→0+ sin(x) log(x) = 0.

Proof. Let x > 0. Then write sin(x) log(x) = log(x)1/ sin(x)

. We will now check the assumptions of

L’Hopital’s rule. For x with 0 < x < 1, the function 1/ sin(x) has derivative− cos(x)/ sin2(x).So, for 0 < x < 1, the function 1/ sin(x) has nonzero derivative. Both log(x) and 1/ sin(x)are differentiable for 0 < x < 1. Also, limx→0+ log(x) = −∞, and limx→0+(1/ sin(x)) = ∞.In conclusion, L’Hopital’s rule applies. We conclude that

limx→0+

sin(x) log(x) = limx→0+

log(x)

1/ sin(x)= lim

x→0+

1/x

− cos(x)/ sin2(x)= lim

x→0+

− sin2(x)

x cos(x)(∗)

We now need to apply L’Hopital’s rule again. We will once again check the assump-tions of L’Hopital’s rule. For x with 0 < x < 1/10, the function x cos(x) has derivativex(− sin(x)) + cos(x). So, for 0 < x < 1/10, the function x cos(x) has nonzero derivative.Both sin2(x) and x cos(x) are differentiable for 0 < x < 1/10. Also, limx→0+ sin2(x) = 0, andlimx→0+ x cos(x) = 0. In conclusion, L’Hopital’s rule applies. We conclude that

limx→0+

− sin2(x)

x cos(x)= lim

x→0+

−2 sin(x) cos(x)

−x sin(x) + cos(x)= lim

x→0+

−2 sin(x)

−x sin(x)cos(x)

+ 1=

0

1= 0. (∗∗)

Combining (∗) and (∗∗) completes the exercise. �

Exercise 3.7.2. (3.7.30) Find limx→0(1

sin(x)− cos(x)

sin(x)).

We show that limx→0(1

sin(x)− cos(x)

sin(x)) = 0.

20

Proof. We will apply L’Hopital’s rule to the function (1− cos(x))/ sin(x). We first check theassumptions of L’Hopital’s rule. For x with −1 < x < 1, the function sin(x) has derivativecos(x). So, for −1 < x < 1, the function sin(x) has nonzero derivative. Both (1−cos(x)) andsin(x) are differentiable for −1 < x < 1. Also, limx→0(1−cos(x)) = 0, and limx→0 sin(x) = 0.In conclusion, L’Hopital’s rule applies to the function (1− cos(x))/ sin(x). Now, observe

limx→0

(1

sin(x)− cos(x)

sin(x)

)= lim

x→0

1− cos(x)

sin(x)= lim

x→0

sin(x)

cos(x)= lim

x→0tan(x) = 0.

�

Exercise 3.7.3. (3.7.54) Find a, b such that limx→0

(sin(2x)x3

+ a+ bx2

)= 0.

We show that (a, b) = (4/3,−2).

Proof. Suppose a, b ∈ R satisfy limx→0

(sin(2x)x3

+ a+ bx2

)= 0. Let x 6= 0. We first write

sin(2x)

x3+ a+

b

x2=

sin(2x) + ax3 + bx

x3.

Since limx→0 x3 = 0, we must have limx→0(sin(2x)+ax3+bx) = 0 as well. (If limx→0(sin(2x)+

ax3 + bx) 6= 0, then our assumption is violated.) Now, are in a position to apply L’Hopital’srule to the function (sin(2x) + ax3 + bx)/x3, which gives

limx→0

sin(2x) + ax3 + bx

x3= lim

x→0

2 cos(2x) + 3ax2 + b

3x2. (∗)

Now, whenever we apply L’Hopital’s rule as an equality of the form limx→0 f(x)/g(x) =limx→0 f

′(x)/g′(x), the limit on the right must exist. (This is an assumption that we need inorder to apply L’Hopital’s rule.) Since limx→0 3x2 = 0, in order for the limit on the right of(∗) to exist, we must have limx→0(2 cos(2x) + 3ax2 + b) = 0. That is, we must have b = −2.In this case, we can then apply L’Hopital’s rule twice to the right side of (∗) to get

limx→0

2 cos(2x) + 3ax2 + b

3x2= lim

x→0

−4 sin(2x) + 6ax

6x

= limx→0

−8 cos(2x) + 6a

6= lim

x→0

−4 + 3a

3. (∗∗)

Combining (∗), (∗∗) and our assumption, we have −4 + 3a = 0, so that a = 4/3. �

Exercise 3.7.4. Find limx→∞ e−x sin(x).

We show that limx→∞ e−x sin(x) = 0.

Proof. Let x ∈ R. Recall that −1 ≤ sin(x) ≤ 1. Multiplying these inequalities by e−x, wehave −e−x ≤ e−x sin(x) ≤ e−x. So, by the Squeeze Theorem, we have

0 = limx→∞−e−x ≤ lim

x→∞e−x sin(x) ≤ lim

x→∞e−x = 0.

That is, limx→∞ e−x sin(x) = 0. �

21

4. Chapter 4

4.1. Section 4.1.

Exercise 4.1.1. (4.1.51) Let a, b > 0. Find the maximum value of f(x) = xa(1−x)b on theinterval [0, 1].

We show that the maximum value is

f(a/(a+ b)) = (a/(a+ b))a(1− a/(a+ b))b = aabb/(a+ b)a+b.

Proof. Observe

f ′(x) = −xab(1− x)b−1 + axa−1(1− x)b

= xa−1(1− x)b−1(−bx+ a(1− x)) = xa−1(1− x)b−1(a− x(a+ b)).

So, the critical numbers of f are x = 0, 1, a/(a + b). Note that f is continuous on theinterval [0, 1], since it is a product of continuous functions. So, by the Closed IntervalMethod, the maximum value of f must occur at its critical numbers. Since f(0) = f(1) = 0,and since f(x) > 0 for 0 < x < 1, we conclude that the maximum value of f occurs at thepoint x = a/(a+ b). �

4.2. Section 4.2.

4.3. Section 4.3.

Exercise 4.3.1. (4.3.42) Let f(x) = x − x2/6 − (2/3) log(x). For this function, do thefollowing.

(a) Find the vertical and horizontal asymptotes.(b) Find the intervals of increase or decrease.(c) Find the local maxima and minima.(d) Find the intervals of concavity and the inflection points.(e) Sketch the graph of f using (a)-(d).

(a) The only asymptote of f is a vertical asymptote at x = 0.(b) f increases on (1, 2) and decreases on (0, 1) ∪ (2,∞).(c) f has a local maximum at x = 2 and a local minimum at x = 1.(d) f is concave up on (

√2,∞), it is concave down on (0,

√2), and it has an inflection

point at x =√

2.

Proof of (a). First of all, the domain of f is (0,∞). To see this, note that f is the sum ofa polynomial (which has domain (−∞,∞)) and the natural logarithm (which has domain(0,∞)). Now, limx→∞ f(x) = −∞, limx→0 f(x) =∞, and f has no other asymptotes, sincef is continuous on (0,∞). Recall that f is the sum of functions that are continuous on(0,∞), so f itself is continuous on (0,∞). In summary, f has no horizontal asymptotes, andit has a single vertical asymptote at x = 0.

To see that limx→∞ f(x) = −∞, observe

limx→∞

f(x)

log(x)= lim

x→∞

x− x2/6log(x)

− 2/3 = limx→∞

1− x/3x−1

− 2/3 = limx→∞

x− x2/3− 2/3 = −∞.

Since limx→∞ log(x) =∞, we therefore conclude that limx→∞ f(x) = −∞. �

22

Proof of (b). Let x > 0. Then f ′(x) = 1 − x/3 − 2/(3x). Suppose x satisfies f ′(x) = 0.

Then −x2/3 + x− 2/3 = 0, so x = (−1±√

1− 8/9)/(−2/3) = (−1± (1/3))/(−2/3). Thatis, x = 1, 2. Since f ′ is a continuous function on (0,∞), we conclude that there are onlythree distinct regions of increase and decrease. Specifically, for 0 < x < 1, f ′(x) < 0, for1 < x < 2, f ′(x) > 0, and for 2 < x <∞, f ′(x) < 0. �

Proof of (c). From part (b), there are only two critical numbers for f in the interval (0,∞).At x = 1, f ′(x) = 0. Also, f is decreasing for 0 < x < 1 and f is increasing for 1 < x < 2,so x = 1 is a local minimum. At x = 2, f ′(x) = 0. Also, f is increasing for 1 < x < 2, andf is decreasing for 2 < x <∞. Therefore, x = 2 is a local maximum. �

Proof of (d). Let x > 0. Then f ′′(x) = −1/3 + (2/3)x−2. Suppose x satisfies f ′′(x) = 0.Then −x2/3 + 2/3 = 0, so x = ±

√2. Since x > 0, we conclude that x =

√2. Note that

f ′′ is continuous on the interval (0,∞). For 0 < x <√

2, f ′′(x) > 0, and for√

2 < x < ∞,f ′′(x) < 0. So, x =

√2 is an inflection point. �

4.4. Section 4.4.

Exercise 4.4.1. (4.4.36) We describe how to sketch the curve y = e2x − ex.

We show that y(x) = 0 only at x = 0, y′(x) = 0 only at x = log(1/2), y is decreasing on(−∞, log(1/2)), y is increasing on (log(1/2),∞), y is concave down on (−∞, log(1/4)), y isconcave up on (log(1/4),∞), and y has an inflection point only at x = log(1/4). Moreover,limx→∞ y(x) =∞, and limx→−∞ y(x) = 0.

Proof. We first find the points where y(x) = 0. Suppose y(x) = 0. Then e2x = ex, so ex = 1,so x = 0. That is, y(x) = 0 only for x = 0.

Now, note that y′(x) = 2e2x − ex. We will find the zeros of y′. Suppose y′(x) = 0. Then2e2x = ex, so 2ex = 1, so x = log(1/2). So, y′(x) = 0 only for x = log(1/2) < 0. Since y′(x)

23

is a continuous function, it therefore only intersects the x-axis once. Now, y′(0) = 2− 1 = 1,and y′(−2 log(2)) = 2e−4 log(2) − e−2 log(2) = 21−4 − 2−2 = (1/8) − (1/4) < 0. Therefore,for x < log(1/2), y′(x) < 0, and for x > log(1/2), y′(x) > 0. That is, y is decreasing for∞ < x < log(1/2) and y is increasing for log(1/2) < x <∞.

We now find the inflection points of y. Note that y′′(x) = 4e2x − ex. Suppose y′′(x) = 0.Then 4e2x = ex, so 4ex = 1, so x = log(1/4) < log(1/2) < 0. As before, y′′(x) is acontinuous function, so it only intersects the x-axis once. Now, y′′(0) = 4 − 1 = 3, andy′′(−2 log(4)) = 4e−4 log(4) − e−2 log(4) = 41−4 − 4−2 = (1/64) − (1/16) < 0. Therefore, forx < log(1/4), y′′(x) < 0, and for x > log(1/4), y′′(x) > 0. So, y has an inflection point atx = log(1/4).

Finally, we describe the asymptotes of y. Since y is continuous on (−∞,∞), it has novertical asymptotes. We now describe the horizontal asymptotes. Note that y(x) = ex(ex−1).Since limx→∞ e

x =∞, and limx→∞(ex − 1) =∞, we conclude that limx→∞ y(x) =∞. Also,since limx→−∞ e

2x = 0, and limx→−∞ ex = 0, we conclude that limx→−∞ y(x) = 0. �

4.5. Section 4.5.

Exercise 4.5.1. (4.5.24) Let a, b > 0. Find the area of the largest rectangle that can beinscribed in the ellipse x2/a2 + y2/b2 = 1. You may assume that the rectangle is alignedwith the axes, and that its vertices touch the ellipse.

We show that the largest area rectangle has area 2ba.

Proof. Suppose the upper right point of the rectangle has coordinates (x, y) with x, y > 0,

−a ≤ x ≤ a, −b ≤ y ≤ b. Then y2 = b2(1− x2/a2). Since y > 0, we have y = b√

1− x2/a2.Then the rectangle has width 2x and height 2b

√1− x2/a2, so its volume V (x) is given by

V (x) = 4bx√

1− x2/a2.

24

We therefore try to maximize V (x) on the interval [0, a]. Note that V (x) is a continuousfunction of x on the interval [0, a], since it is the composition of continuous functions. So,by the Closed Interval Method, it suffices to check the value of V on the endpoints of theinterval [0, a] along with the critical numbers of V .

For x ∈ (0, a), note that

V ′(x) = −4ba−2x2(1− x2/a2)−1/2 + 4b(1− x2/a2)1/2

= (1− x2/a2)−1/2(−4bx2/a2 + 4b(1− x2/a2)).

So, V ′(x) = 0 only when −x2(2/a2) + 1 = 0, i.e. when x2 = a2/2. Since x > 0, we concludethat x = a/

√2. Finally, since V (0) = V (a) = 0, and since V (x) > 0 for 0 < x < a, we

conclude that V (x) is maximized on the interval [0, a] when x = a/√

2. �

4.6. Section 4.6.

4.7. Section 4.7.

Exercise 4.7.1. (4.7.46) Two balls are thrown upward from the edge of a cliff of height 432feet. The first ball is thrown upward with a speed of 48 ft/s, and the other ball is thrownupward a second later with a speed of 24 ft/s. Do the balls ever pass each other?

Yes, the balls pass each other five seconds after the first ball is thrown.

Proof. Let t denote time in seconds, with t = 0 denoting the time that the first ball is thrown.Let a1(t), v1(t), s1(t) denote the acceleration, velocity, and position of the first ball, and leta2(t), v2(t), s2(t) denote the acceleration, velocity, and position of the second ball. All unitsuse feet and seconds. It is given that a1 = a2 = −32, v1(0) = 48, v2(1) = 24, s1(0) = 432,and s2(1) = 432. Since v′1(t) = a1(t), and since all trajectories are assumed to be continuousand differentiable, we know that v1(t) = −32t + c for some c ∈ R. Then v1(0) = 48 = c, sov1(t) = −32t + 48. Since s′1(t) = v1(t), we conclude that s1(t) = −16t2 + 48t + d for somed ∈ R. Then s1(0) = 432 = d, so

s1(t) = −16t2 + 48t+ 432.

We now again use antiderivatives for the trajectory of the second ball. Since v′2(t) = a2(t),we have v2(t) = −32t + e for some e ∈ R. Then v2(1) = 24 = −32 + e, so e = 56, sov2(t) = −32t+ 56. Since s′2(t) = v2(t), we have s2(t) = −16t2 + 56t+ g for some g ∈ R. Ands2(1) = 432 = −16 + 56 + g, so g = 392, so

s2(t) = −16t2 + 56t+ 392.

We now check whether or not the balls pass each other. Let f(t) = s1(t) − s2(t). Tocomplete the exercise it suffices to find t > 0 where f(t) changes sign. Suppose t satisfiesf(t) = 0. Then 48t+ 432− 56t− 392 = 0, so −8t = −40, so t = 5. So, f(5) = 0. Also, notethat s1(5) = s2(5) = 272 > 0, so at t = 5, both balls are still above the ground, and theyhave the same height above the ground. We now check the derivative of f at 5. Note thatf ′(t) = 48 − 56 = −8 < 0. In conclusion, for 0 < t < 5, s1(t) − s2(t) = f(t) > 0, and fort > 5, s1(t)− s2(t) = f(t) < 0. So, the balls do pass each other at the time t = 5. �

25

5. Appendix: Notation

Z = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .}, the set of integers

N = {1, 2, 3, 4, 5, . . .}, the set of natural numbers

Q = {. . . , 1/2,−2/3, 0, . . .} = {m/n : m,n ∈ Z, n 6= 0}, the set of rational numbers

R = {. . . , 1/2, π,√

3,−.24534,−1, 4, 7.2, . . .}, the set of real numbers

log denotes the natural logarithm. In the notation of the book, log = ln.

∅ denotes the empty set, the set consisting of zero elements

∈ means “is an element of.” For example, 2 ∈ Z is read as “2 is an element of Z.”

∀ means “for all”

∃ means “there exists”

Remark 5.1. In the book, there are several expressions of the form

f(x) =cos 2x− x

x2.

When the cosine is written without parentheses in the argument, it is usually understoodthat the very first thing that is written after the cosine (in this case 2x) is the argumentof the cosine. The next minus or plus sign that appears is assumed to occur outside of theparentheses (that have been omitted). That is, the expression above can be equivalentlywritten as follows

f(x) =cos(2x)− x

x2.

Remark 5.2. The following two expressions are equal:

cos2(x) = (cos(x))2.

Remark 5.3. Let f : R → R be a function with an inverse function f−1 : R → R. We usethe notation f−1 for the inverse function of f . In general, f−1 is not equal to 1/f .

Remark 5.4. Whenever a fraction has a radical of a number in the denominator, we preferto move the radical to the numerator, as in the following example:

1√5

=1√5

√5√5

=

√5

5

If a variable occurs in the radical, then we usually leave the radical in the denominator.

Courant Institute, New York University, New York NY 10012E-mail address: heilman@cims.nyu.edu

26