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Howard Anton Calculus 10th Edition Solution Manual

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Single Variable Tenth Edition

Description

Student Solutions Manual Tamas Wiandt Rochester Institute of Technology to accompany

CALCULUS Early Transcendentals Single Variable Tenth Edition Howard Anton Drexel University

Bivens Davidson College

Stephen L

Davis Davidson College

John Wiley& Sons,

PUBLISHER ACQUISITIONS EDITOR PROJECT EDITOR ASSISTANT CONTENT EDITOR EDITORIAL ASSISTANT CONTENT MANAGER SENIOR PRODUCTION EDITOR SENIOR PHOTO EDITOR COVER DESIGNER COVER PHOTO

Laurie Rosatone David Dietz Ellen Keohane Beth Pearson Jacqueline Sinacori Karoline Luciano Kerry Weinstein Sheena Goldstein Madelyn Lesure © David Henderson/Getty Images

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All rights reserved

No part of this publication may be reproduced,

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Requests to the Publisher for permission should be addressed to the Permissions Department,

John Wiley & Sons,

Hoboken,

NJ 070305774,

ISBN 978-1-118-17381-7 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

Contents Chapter 0

Before Calculus ………

…………………………………………………………………………

Chapter 1

Limits and Continuity ………………………………………………………………………………

Chapter 2

The Derivative ……………………………………………………………………………………

Chapter 3

Topics in Differentiation ……………………………

………………………………………

Chapter 4

The Derivative in Graphing and Applications ……………………………………

Chapter 5

Integration …………………………………………………………………………………………

Chapter 6

Applications of the Definite Integral in Geometry,

Science,

Chapter 7

Principles of Integral Evaluation ………………………………………………………………

Chapter 8

Mathematical Modeling with Differential Equations …………………………………… 217

Chapter 9

Infinite Series ……………………………………………………………………………………

Chapter 10

Parametric and Polar Curves

Conic Sections ………………………………………

Appendix A

Graphing Functions Using Calculators and Computer Algebra Systems

Appendix B

Trigonometry Review ………………………………………………………………………………

Appendix C

Solving Polynomial Equations …………………………………………………………………… 297

Before Calculus Exercise Set 0

2 at x = 1

$47,700

(c) No (vertical line test fails)

(d) No (vertical line test fails)

$41,600

(c) The slope between 2000 and 2001 is steeper than the slope between 2001 and 2002,

so the median income was declining more rapidly during the first year of the 2-year period

(a) √ f2(0) = 3(0) − 2 = −2

f (3t) = 3(3t) − 2 = 27t − 2

f (3t) = 1/(3t) for t > 1 and f (3t) = 6t for t ≤ 1

Range: y = 0

Range: {1,

√ √ (c) Natural domain: x ≤ − 3 or x ≥ 3

Range: y ≥ 0

(d) x2 − 2x + 5 = (x − 1)2 + 4 ≥ 4

So G(x) is defined for all x,

Natural domain: all x

(e) Natural domain: sin x = 1,

For such x,

Range: y ≥ 2

−4 (f ) Division by 0 occurs for x = 2

For all other x,

which is nonnegative for x ≥ −2

Natural √ √ domain: [−2,

2) ∪ (2,

The range of x + 2 is [0,

But we must exclude x = 2,

Range: [0,

2) ∪ (2,

(a) The curve is broken whenever someone is born or someone dies

(b) C decreases for eight hours,

increases rapidly (but continuously),

Chapter 0

25 − x2

Exercise Set 0

The graph of an exponential function passes through (0,

but the graph of y = x3 does not

6 = (log 0

the curve passes through the origin

log(1/2) < 0 so 3 log(1/2) < 2 log(1/2)

t = −125 ln(1/5) = 125 ln 5 ≈ 201 days

Chapter 0

Let IA and IB be the intensities of the automobile and blender,

Then log10 IA /I0 = 7 and log10 IB /I0 = 9

IA = 107 I0 and IB = 109

3 ≈ 200

2) = 16

E = 1016

M2 − M1 = 1/1

Chapter 0 Review Exercises y 5

(a) If the side has length x and height h,

Then the cost C = 5x2 + 2(4)(xh) = 5x2 + 64/x

+∞) because x can be very large (just take h very small)

(a) The base has sides (10 − 2x)/2 and 6 − 2x,

so V = (6 − 2x)(5 − x)x ft3

(b) From the picture we see that x < 5 and 2x < 6,

57 ft ×3

79 ft ×1

Chapter 0 Review Exercises 11

For g(h(x)) to be defined,

we also require √ we require h(x) = 0,

For f (g(h(x))) to be defined,

So the domain of f ◦ g ◦ h consists of all x except ±1 and ± 2

For all x in the domain,

(f ◦ g ◦ h)(x) = 1/(2 − x2 )

(a) The circle of radius 1 centered at (a,

the family of all circles of radius 1 with centers on the parabola y = x2

(b) All translates of the parabola y = x2 with vertex on the line y = x/2

20 t 100

(a) –20 3π 2π (t − 101) = ,

which is the same date as t = 9

so during the night of January (b) When 365 2 10th-11th

When x = 0 the value of the green curve is higher than that of the blue curve,

therefore the blue curve is given by y = 1 + 2 sin x

The points A,

D are the points of intersection of the two curves,

where 1+2 sin x = 2 sin(x/2)+2 cos(x/2)

Let sin(x/2) = p,

Then 2 sin x = 4 sin(x/2) cos(x/2) (basic trigonometric identity),

so the equation which yields the points of intersection becomes 1 + 4pq = 2p + 2q,

Thus A has coordinates √ √ either sin(x/2) = 1/2 or cos(x/2) √ 3),

B has coordinates (π/3,

1 + 3),

C has coordinates (2π/3,

1 + 3),

and D'has coordinates (−2π/3,

1 − 3)

f (g(x)) = x for all x in the domain of g,

and g(f (x)) = x for all x in the domain of f

They are reflections of each other through the line y = x

(c) The domain of one is the range of the other and vice versa

(d) The equation y = f (x) can always be solved for x as a function of y

Functions with no inverses include y = x2 ,

Chapter 0 (b) f (x) = (x − 1)2

f does not have an inverse because f is not one-to-one,

√ (c) x = f (y) = (ey )2 + 1

f −1 (x) = y = ln x − 1 = (d) x = f (y) =

1 − 2y y

The range of f consists of all x < 3x  3π − 2 3π +2 1−x 2 −2 the domain of f −1

Hence f −1 (x) = tan or x >

Draw right triangles of sides 5,

Then sin[cos−1 (4/5)] = 3/5,

and cos[sin−1 (5/13)] = 12/13

5 13 65 31

(b) The curve y = e−x/2 sin 2x has x−intercepts at x = −π/2,

It intersects the curve y = e−x/2 at x = π/4,

5π/4 and it intersects the curve y = −e−x/2 at x = −π/4,

Chapter 0 Review Exercises

so it is reasonable to expect it to be zero somewhere in between

(This will be established later in this book

654 and 3

(a) The functions x2 and tan x are positive and increasing on the indicated interval,

so their product x2 tan x is also increasing there

So is ln x

hence the sum f (x) = x2 tan x + ln x is increasing,

(b) The asymptotes for f (x) are x = 0,

The asymptotes for f −1 (x) are y = 0,

Chapter 0

Limits and Continuity Exercise Set 1

01 −0

001 −0

9950166 0

9995002 0

9999500 1

0000500 1

0005002 1

0050167

The limit appears to be 1

001 0 0

The limit is 1/3

Chapter 1 50

The limit is +∞

7143 −7

0111 −67

001 −667

0 −6667

The limit is −∞

25 −0

001 −0

The limit is 3

99 −0

999 −1

01 −1

2161 54

1415 −4

536 −53

19 −539

The limit does not exist

define f (x) = x for x = a and f (a) = a + 1

Then limx→a f (x) = a = f (a) = a + 1

define f (x) = 0 for x < 0 and f (x) = x + 1 for x ≥ 0

Then the left and right limits exist but are unequal

thus y − 1 = −2(x + 1) or y = −2x − 1

thus y − 1 = 4(x − 1) or y = 4x − 3

(a) The length of the rod while at rest

Exercise Set 1

The length of the rod approaches zero as its speed approaches c

The limit appears to be 3

The limit appears to be 3

The limit does not exist

Exercise Set 1

this limit is 2 + 2 · (−4) = −6

this limit is 0 − 3 · (−4) + 1 = 13

this limit is 2 · (−4) = −8

this limit is (f ) By Theorem 1

6 + 2 = 2

1 2 =−

By Theorem 1

By Theorem 1

this limit is (32 − 2 · 3)/(3 + 1) = 3/4

After simplification,

and the limit is 13 + 12 + 1 + 1 = 4

After simplification,

and the limit is (−1 + 5)/(−1 − 4) = −4/5

After simplification,

and the limit is 2 · (−1) − 1 = −3

Chapter 1

After simplification,

and the limit is (22 + 5 · 2 − 2)/(22 + 2 · 2) = 3/2

The limit is +∞

The limit does not exist

The limit is −∞

The limit is +∞

The limit does not exist

The limit is +∞

The limit is +∞

After simplification,

After simplification,

moreover one cannot subtract infinities

a x+1−a 1 − 2 = and for this to have a limit it is necessary that lim (x + 1 − a) = 0,

x→1 x−1 x −1 x2 − 1 1 1 2 x+1−2 x−1 1 1 a = 2

For this value,

x→1 x + 1 x − 1 x2 − 1 x2 − 1 x −1 x+1 2

For x = 1,

The left and/or right limits could be plus or minus infinity

or equal any preassigned real number

For example,

let q(x) = x − x0 and let p(x) = a(x − x0 )n where n takes on the values 0,

Clearly,

g(x) = [f (x) + g(x)] − f (x)

By Theorem 1

lim [f (x) + g(x)] − lim f (x) = lim [f (x) + g(x) − f (x)] = x→a

Exercise Set 1

Exercise Set 1

(a) 3 + 3 · (−5) = −12 (e)

(b) 0 − 4 · (−5) + 1 = 21

(h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t

00001 0

000001 1

471128 1

560797 1

569796 1

570696 1

570786 1

The limit appears to be ≈ 1

The limit is −∞,

The limit is +∞

The limit is 3/2,

The limit is 0,

The limit is 0,

The limit is −∞,

The limit is −1/7,

The limit is 3 −5/8 = − 3 5 /2,

 √ 5 − x22 2 √ 5x − 2 = when x < 0

The limit is − 5

The limit is 1/ 6

1 x3 8 x2

The limit is

lim ( x + 3 − x) √ = lim √ = 0,

x→+∞ x2 + 3 + x x→+∞ x2 + 3 + x 33

Divide the numerator and denominator by ex :

= −2x x→+∞ 1 − e 1−0 lim

The limit is −∞

x→+∞ x x xx   2x  x 2 1 1 = lim = e2

False: lim 1+ 1+ x→+∞ x→+∞ x x

Chapter 1

True: for example f (x) = sin x/x crosses the x-axis infinitely many times at x = nπ,

It appears that lim n(t) = +∞,

When n is even,

tan x and sec x at x = nπ + π/2 and cot x and csc x at x = nπ,

(a) Every value taken by ex is also taken by et : choose t = x2

As x and t increase without bound,

Thus lim ex = lim et = +∞

f (t) → −∞) then f (t) can be made arbitrarily large (resp

small) by taking t large enough

But by considering the values g(x) where g(x) > t,

we see that f (g(x)) has the limit +∞ too (resp

If f (t) has the limit L'as t → +∞ the values f (t) can be made arbitrarily close to L'by taking t large enough

But if x is large enough then g(x) > t and hence f (g(x)) is also arbitrarily close to L

(c) For lim the same argument holds with the substitutiion ”x decreases without bound” instead of ”x increases x→−∞

For lim− substitute ”x close enough to c,

Let t = ln x

Then t also tends to +∞,

Set t = −x,

From the hint,

lim bx = lim e(ln b)x x→+∞

⎧ 0 if b < 1,

⎪ ⎪ ⎨ 1 if b = 1,

  (b) lim v = 190 1 − lim e−0

so the asymptote is v = c'= 190 ft/sec

(c) Due to air resistance (and other factors) this is the maximum speed that a sky diver can attain

00001 1

000001 1

71828 2

Exercise Set 1

(b) This is evident from the lower left term in the chart in part (a)

(c) The exponents are being multiplied by a,

so x−2 The only vertical asymptote is at x = 2

After a long division,

lim (f (x) − (x + 2)) = 0 and f (x) is asymptotic to y = x + 2

so lim (f (x)−(−x2 +1)) = 0 and f (x) is asymptotic to y = −x2 +1

x→±∞ x−3 The only vertical asymptote is at x = 3

After a long division,

lim (f (x) − sin x) = 0 so f (x) is asymptotic to y = sin x

The only vertical asymptote is at x = 1

Exercise Set 1

(a) |f (x) − f (0)| = |x + 2 − 2| = |x| < 0

(b) |f (x) − f (3)| = |(4x − 5) − 7| = 4|x − 3| < 0

1)/4 = 0

(c) |f (x) − f (4)| = |x2 − 16| < if |x − 4| < δ

We get f (x) = 16 + = 16

000124998,

000124998

999874998,

000125002

Use the smaller δ: thus |f (x) − 16| < provided |x − 4| < 0

95)2 = 3

x1 = (2

05)2 = 4

Chapter 1 (b) δ = min ( |4 − 3

|4 − 4

Now x3 −4x+5 = 1

So δ = min (1

1 − 0

9558) = 0

With the TRACE feature of a calculator we discover that (to five decimal places) (0

Set x0 = 0

Since the graph of f (x) rises from left to right,

we see that if x0 < x < x1 then 1

So we can take δ = 0

1/2 = 0

05 = δ

  2     2   x − 9 − 6x + 18   x2 − 6x + 9  x − 9 =  − 6 =  11

If x = 3,

then    x − 3  = |x − 3| < 0

05 = δ

Assume δ ≤ 1

Then −1 < x − 2 < 1 means 1 < x < 3 and then |x3 − 8| = |(x − 2)(x2 + 2x + 4)| < 19|x − 2|,

     1 1   x − 5  |x − 5|   <  ,

Assume δ ≤ 1

Then −1 < x − 5 < 1 means 4 < x < 6 and then  −  =  x 5 5x  20 δ = 0

Let > 0 be given

Then |f (x) − 3| = |3 − 3| = 0 < regardless of x,

and hence any δ > 0 will work

|3x − 15| = 3|x − 5| < if |x − 5| < /3,

δ = /3

  2   2x + x  − 1 = |2x| < if |x| < /2,

δ = /2

|f (x) − 3| = |x + 2 − 3| = |x − 1| < if 0 < |x − 1| < ,

If > 0 is given,

For the first part,

Then there exists δ > 0 such that if a < x < a + δ then |f (x) − L| <

For the left limit replace a < x < a + δ with a − δ < x < a

(a) |(3x2 + 2x − 20 − 300| = |3x2 + 2x − 320| = |(3x + 32)(x − 10)| = |3x + 32| · |x − 10|

(b) If |x − 10| < 1 then |3x + 32| < 65,

If δ < 1 then |2x2 − 2| = 2|x − 1||x + 1| < 6|x − 1| < if |x − 1| < /6,

If δ < 1/2 and |x − (−2)| < δ then −5/2 < x < −3/2,

then    |x + 2|  1    x + 1 − (−1) = |x + 1| < 2|x + 2| < if |x + 2| < /2,

Exercise Set 1

    √ √ √ x + 2   x − 4  1 √ 35

| x − 2| = ( x − 2) √ = < |x − 4| < if |x − 4| < 2 ,

Let > 0 be given and take δ =

If |x| < δ,

then |f (x) − 0| = 0 < if x is rational,

and |f (x) − 0| = |x| < δ =

(a) We have to solve the equation 1/N 2 = 0

1 here,

(b) This will happen when N/(N + 1) = 0

(c) Because the function 1/x3 approaches 0 from below when x → −∞,

we have to solve the equation 1/N 3 = −0

(d) The function x/(x + 1) approaches 1 from above when x → −∞,

so we have to solve the equation N/(N + 1) = 1

We obtain N = −101

1 + x22

x2      x   1     < 0

N = 999

N = −202

x+2      4x − 1   11   < 0

N = −57

 − 2 =  2x + 5 2x + 5    1 1 1 51

      11   4x − 1    < if |2x+5| > 11 ,

 √      2 2  2 x   2  √ √ 2 2 2 2     55

−1 1 < −1000 if |x − 3| < √ 2 (x − 3) 10 10

1 1 1 ,

|x| < < −10000 if x4 < 4 x 10000 10

If M > 0 then

1 1 1 1 ,

> M when 0 < (x − 3)2 < (x − 3)2 M M M

If M > 0 then

Chapter 1

If M < 0 then −

If x > 2 then |x + 1 − 3| = |x − 2| = x − 2 < if 2 < x < 2 + ,

If x > 4 then

√ x − 4 < if x − 4 < 2 ,

If x > 2 then |f (x) − 2| = |x − 2| = x − 2 < if 2 < x < 2 + ,

(a) Definition: For every M < 0 there corresponds a δ > 0 such that if 1 < x < 1 + δ then f (x) < M

In our case 1 1 1 1 < M ,

we want 1−x M M M (b) Definition: For every M > 0 there corresponds a δ > 0 such that if 1 − δ < x < 1 then f (x) > M

In our case 1 1 1 1 we want > M ,

there corresponds an N > 0 such that if x > N then f (x) > M ,

there corresponds an N < 0 such that if x < N then f (x) < M ,

5 + δ 7

5 − δ

Exercise Set 1

(a) No: lim f (x) does not exist

No: f (3) is not defined

(a) No: f (1) and f (3) are not defined

No: lim f (x) does not exist

Exercise Set 1

(b) One second could cost you one dollar

this is a continuous function on the real numbers

this is a continuous function on the real numbers

The function is not continuous at x = −1/2 and x = 0

The function is not continuous at x = 0,

this is a continuous function on the real numbers

this is a continuous function on the real numbers

f (x) = 2x + 3 is continuous on x < 4 and f (x) = 7 + is continuous on 4 < x

lim f (x) = lim f (x) = f (4) = 11 so f is continuous at x = 4

(a) f is continuous for x < 1,

so if k = 5 then f is continuous for x→1

(b) f is continuous for x < 2,

m = 5/3 and f is continuous everywhere if k = 4

lim− f (x) = −1 = +1 = lim+ f (x) so the discontinuity is not removable

define f (−3) = −3 = lim f (x),

then the discontinuity is removable

(c) f is undefined at x = ±2

so define f (2) = 1 and f becomes continuous there

so the discontinuity is not removable

Chapter 1 y

Discontinuity at x = 1/2,

(b) 2x2 + 5x − 3 = (2x − 1)(x + 3) 39

Write f (x) = x3/5 = (x3 )1/5 as the composition (Theorem 1

Since f and g are continuous at x = c'we know that lim f (x) = f (c) and lim g(x) = g(c)

In the following we use x→c x→c Theorem 1

(a) f (c) + g(c) = lim f (x) + lim g(x) = lim (f (x) + g(x)) so f + g is continuous at x = c

(b) Same as (a) except the + sign becomes a − sign

(c) f (c)g(c) = lim f (x) lim g(x) = lim f (x)g(x) so f g is continuous at x = c

Then by Theorem 1

lim f (h + c) = f ( lim (h + c)) = f (c)

lim g(h) = lim f (c + h) = f (c) = g(0),

so g(h) is continuous at h = 0

That is,

Of course such a function must be discontinuous

Let f (x) = 1 on 0 ≤ x < 1,

and f (x) = −1 on 1 ≤ x ≤ 2

If f (x) = x3 + x2 − 2x − 1,

The Intermediate Value Theorem gives us the result

For the negative root,

use intervals on the x-axis as follows: [−2,

2) > 0,

For the positive root use the interval [0,

8) > 0,

75 of [0

For the positive root,

use intervals on the x-axis as follows: [2,

3) > 0,

Since f (2

235 of [2

Consider the function f (θ) = T (θ + π) − T (θ)

Note that T has period 2π,

T (θ + 2π) = T (θ),

so that f (θ + π) = T (θ + 2π) − T (θ + π) = −(T (θ + π) − T (θ)) = −f (θ)

Now if f (θ) ≡ 0,

Otherwise,

there exists θ such that f (θ) = 0 and then f (θ + π) has an opposite sign,

and thus there is a t0 between θ and θ + π such that f (t0 ) = 0 and the statement follows

Since R and L'are arbitrary,

we can introduce coordinates so that L'is the x-axis

Let f (z) be as in Exercise 54

Then for large z,

By the Intermediate Value Theorem there is a z1 such that f (z1 ) = half of the area of the ellipse

Exercise Set 1

This is a composition of continuous functions,

so it is continuous everywhere

Discontinuities at x = nπ,

Exercise Set 1

Discontinuities at x = nπ,

Discontinuities at x =

π 5π + 2nπ,

sin−1 u is continuous for −1 ≤ u ≤ 1,

3) ∪ (3,

lim cos x→+∞ x→+∞ x x 19

θ→0 θ→0 3θ θ

θ→0+

tan 7x tan 7x 7 sin 7x 3x 7 7 = · · ,

x→0 sin 3x sin 3x 3 cos 7x 7x sin 3x 3·1 3

√ sin x sin x 1 √ = lim x lim = 0

  sin x2 sin x2 = lim x lim = 0

t2 = 33

θ sin θ

θ2 = (1)2 · 2 = 2

θ→0 1 − cos θ

Note that = · = = x x x x x 1 + cos 3x x(1 + cos 3x) sin 3x sin 3x ·

Thus x 1 + cos 3x lim

x→0 x→0 x x 1 + cos 3x x 1 + cos 4x

Chapter 1 4 4

093497 0

100932 0

100842 0

098845 0

091319 0

The limit appears to be 0

Then t → 0 as x → 5 and lim

sin(x − 5) 1 sin t 1 1 = lim lim = ·1=

x→5 x + 5 t→0 t x2 − 25 10 10

True: let > 0 and δ =

Then if |x − (−1)| = |x + 1| < δ then |f (x) + 5| <

(a) The student calculated x in degrees rather than radians

Thus lim = (b) sin x◦ = sin t where x◦ is measured in degrees,

t is measured in radians and t = ◦ x →0 x◦ 180 sin t π lim =

t→0 (180t/π) 180 sin kx 1 = k,

and the nonzero solution is k =

π−x t = lim = 1

sin(πx) = sin(πt + π) = − sin πt

sin(πx) sin πt = − lim = −π

cos(t + π/4) = ( 2/2)(cos t − sin t),

sin(t + π/4) = ( 2/2)(sin t + cos t),

so x − π/4 √ √ √ 2 sin t cos x − sin x sin t

− t→0 t t x − π/4 x→π/4 57

which gives the desired result

Since lim sin(1/x) does not exist,

lim f (x) = 0 by the Squeezing Theorem

Chapter 1 Review Exercises

By the IVT there must be a solution of f (x) = 0

5 y=x 1 0

(a) Gravity is strongest at the poles and weakest at the equator

80 f 30

(b) Let g(φ) be the given function

Then g(38) < 9

so by the Intermediate Value Theorem there is a value c'between 38 and 39 for which g(c) = 9

8 exactly

Chapter 1 Review Exercises 1

The limit is

Chapter 1 3 3 3x + 9 = with limit −

If x = −3 then

By the highest degree terms,

32 25 =

If x = 0,

If x = 0,

As t → π/2+ ,

048/8) = 0

  2   4x − 9 − 6 < 0

(b)  2x − 3 (c) |x2 − 16| < 0

if δ < 1 then |x + 4| < 9 if |x − 4| < 1

then |x2 − 16| = |x − 4||x + 4| ≤ 9|x − 4| < 0

then |x2 − 16| < 9|x − 4| < 9(1/9000) = 1/1000 = 0

Let = f (x0 )/2 > 0

then there corresponds a δ > 0 such that if |x − x0 | < δ then |f (x) − f (x0 )| < ,

f (x) > f (x0 ) − = f (x0 )/2 > 0,

(a) f is not defined at x = ±1,

(c) f is not defined at x = 0 and x = −3,

For x < 2 f is a polynomial and is continuous

for x > 2 f is a polynomial and is continuous

At x = 2,

f (2) = −13 = 13 = lim+ f (x),

Chapter 1 Making Connections 35

(d) (3 − 18)/(4 − 2) = −7

It is a straight line with slope equal to the velocity

Chapter 2 f (x1 ) − f (x0 ) 2x21 − 2x20 = lim = lim (2x1 + 2x0 ) = 4x0 x1 →x0 x1 − x0 x1 →x0 x1 − x0

x1 →x0

Tangent 1

(d) The tangent line is the x-axis

x1 →2

f (x1 ) − f (2) 1/x1 − 1/2 2 − x1 −1 1 = lim = lim = lim =− x →2 x →2 x →2 x1 − 2 x1 − 2 2x1 (x1 − 2) 2x1 4 1 1 1

x1 →x0

f (x1 ) − f (x0 ) 1/x1 − 1/x0 x0 − x1 −1 1 = lim = lim = lim =− 2 x1 →x0 x1 →x0 x0 x1 (x1 − x0 ) x1 →x0 x0 x1 x1 − x0 x1 − x0 x0

Tangent

x1 →x0

f (x1 ) − f (x0 ) (x21 − 1) − (x20 − 1) (x21 − x20 ) = lim = lim = lim (x1 + x0 ) = 2x0 x1 →x0 x1 →x0 x1 − x0 x1 →x0 x1 − x0 x1 − x0

(x1 + f (x1 ) − f (x0 ) = lim = lim x1 →x0 x →x x1 − x0 1 0

x1 →x0

Let x = 1 + h

Velocity represents the rate at which position changes

(b) About (67 − 43)/6 = 4◦ F/h

(c) Decreasing most rapidly at about 9 P

rate of change of temperature is about −7◦ F/h (slope of estimated tangent line to curve at 9 P

(a) During the first year after birth

(b) About 6 cm/year (slope of estimated tangent line at age 5)

Exercise Set 2

(c) The growth rate is greatest at about age 14

Growth rate (cm/year)

(b) vave = 19,200/40 = 480 ft/s

3t3 = 1000

938 ≈ 66

943 ft/s

3 · 403 0

The instantaneous velocity at t = 1 equals the limit as h → 0 of the average velocity during the interval between t = 1 and t = 1 + h

Exercise Set 2

(a) f  (a) is the slope of the tangent line

y = 5x − 16 f (x + h) − f (x) 2(x + h)2 − 2x2 4xh + 2h2 = lim = lim = 4x

f  (1) = 4 so the tangent line is given h→0 h→0 h→0 h h h by y − 2 = 4(x − 1),

f (x + h) − f (x) (x + h)3 − x3 = lim = lim (3x2 + 3xh + h2 ) = 3x2

f  (0) = 0 so the tangent line is h→0 h→0 h→0 h h given by y − 0 = 0(x − 0),

√ √ √ √ √ √ f (x + h) − f (x) x+1+h− x+1 x+1+h− x+1 x+1+h+ x+1 √ √ = = lim = lim h→0 h→0 h→0 h h h x+1+h+ x+1 √ h 1 1 √

f (8) = 8 + 1 = 3 and f  (8) = lim √ so the tangent line is given by = √ h→0 h( x + 1 + h + 6 2 x+1 x + 1)

Chapter 2 y−3=

x − (x + Δx) 1 1 − −Δx 1 1 x(x + Δx) = lim = lim − = − 2

f  (x) = lim x + Δx x = lim Δx→0 Δx→0 Δx→0 xΔx(x + Δx) Δx→0 Δx Δx x(x + Δx) x (x + Δx)2 − (x + Δx) − (x2 − x) 2xΔx + (Δx)2 − Δx = lim = lim (2x − 1 + Δx) = 2x − 1

Δx→0 Δx→0 Δx→0 Δx Δx

Δx→0

= lim √ √ √ Δx→0 2x x x + Δx( x + x + Δx) f (t + h) − f (t) [4(t + h)2 + (t + h)] − [4t2 + t] 4t2 + 8th + 4h2 + t + h − 4t2 − t = lim = lim = h→0 h→0 h→0 h h h 2 8th + 4h + h lim = lim (8t + 4h + 1) = 8t + 1

If the tangent line is horizontal then f  (a) = 0

|x| is continuous but not differentiable at x = 0

 (1 − (x + h)2 ) − (1 − x2 ) −2xh − h2 dy  dy = lim = lim = lim (−2x − h) = −2x,

h→0 h→0 dx h→0 h h dx x=1 5

Exercise Set 2

2 − 2

34 − 2

34 − 0

(b) The tangent line at x = 1 appears to have slope about 0

so f (3) − f (1) gives the worst

f (2) − f (0) gives the best approximation and 2−0

(a) dollars/ft (b) f  (x) is roughly the price per additional foot

(c) If each additional foot costs extra money (this is to be expected) then f  (x) remains positive

(d) From the approximation 1000 = f  (300) ≈ foot will cost around $1000

dF/dθ ≈ 50

f (301) − f (300) we see that f (301) ≈ f (300) + 1000,

(b) μ = (dF/dθ)/F ≈ 50/200 = 0

dT /dt ≈ −3

(b) k = (dT /dt)/(T − T0 ) ≈ (−3

lim− f (x) = lim+ f (x) = f (1),

f (1 + h) − f (1) [(1 + h)2 + 1] − 2 = lim = − h h h→0

f (1 + h) − f (1) 2(1 + h) − 2 lim (2 + h) = 2

− + + h h h→0 h→0 h→0 h→0+ y 5

Since −|x| ≤ x sin(1/x) ≤ |x| it follows by the Squeezing Theorem (Theorem 1

The x→0

f (x) − f (0) derivative cannot exist: consider = sin(1/x)

This function oscillates between −1 and +1 and does x not tend to any number as x tends to zero

   f (x) − f (x0 )    51

Let = |f (x0 )/2|

Then there exists δ > 0 such that if 0 < |x − x0 | < δ,

Since x − x0 

Chapter 2 f  (x0 ) > 0 and = f  (x0 )/2 it follows that x = x2 > x0 then f (x2 ) > f (x0 )

If x = x1 < x0 then f (x1 ) < f (x0 ) and if x − x0

Since m =  0,

Since f (0) = f  (0) = 0 we know there exists δ > 0 such that    f (0 + h) − f (0)   < whenever 0 < |h| < δ

It follows that |f (h)| < 1 |hm| for 0 < |h| < δ

Replace h with x to  2   h get the result

Moreover |mx| = |mx − f (x) + f (x)| ≤ |f (x) − mx| + |f (x)|,

which yields |f (x) − mx| ≥ |mx| − |f (x)| > 12 |mx| > |f (x)|,

(c) If any straight line y = mx + b is to approximate the curve y = f (x) for small values of x,

The inequality |f (x) − mx| > |f (x)| can also be interpreted as |f (x) − mx| > |f (x) − 0|,

the line y = 0 is a better approximation than is y = mx

See discussion around Definition 2

Exercise Set 2

2 and 2

24x7 + 2,

3 and 2

3 and 2

2t − 1,

2 and 2

dy/dx = 1 + 2x + 3x2 + 4x3 + 5x4 ,

y = (1 − x2 )(1 + x2 )(1 + x4 ) = (1 − x4 )(1 + x4 ) = 1 − x8 ,

The estimate will depend on your graphing utility and on how far you zoom in

Since f  (x) = 1 − value is f  (1) = 0

2 and 2

3πr2 ,

2 and 2

Exercise Set 2

By Theorems 2

4 and 2

d [f (x) − 8g(x)] = f  (x) − 8g  (x)

substitute x = 2 to get the result

   d'3  [4f (x) + x ] = (4f  (x) + 3x2 )x=2 = 4f  (2) + 3 · 22 = 32 35

 dV  (b) = 4π(5)2 = 100π dr r=5

(d) dy/dx = 175x4 − 48x2 − 3,

y  = −210x−8 + 60x2 (b) y = x−1 ,

y  = −6x−4 (c) y  = 3ax2 + b,

 dy d2 y d2 y  3 = 30x4 − 8x,

 d' −3  d2  −3  d3  −3  d4  −3  d4  −3  −5 −6 −7 x = −3x−4 ,

x = 12x x = −60x x = 360x x ,

= 360  dx dx2 dx3 dx4 dx4 x=1

and y  = 6 so y  + xy  − 2y  = 6 + x(6x) − 2(3x2 + 3) = 6 + 6x2 − 6x2 − 6 = 0

but = x2 − 3x + 2 = (x − 1)(x − 2) = 0 if x = 1,

dx dx The corresponding values of y are 5/6 and 2/3 so the tangent line is horizontal at (1,

The graph has a horizontal tangent at points where 1

The y-intercept is −2 so the point (0,

The x-intercept is 1 so the point (1,0) is on the graph

The slope is dy/dx = 2ax + b

at x = 0 the slope is b so b = −1,

The function is y = 3x2 − x − 2

The points (−1,

1) and (2,

The slope of the tangent line to y = x2 is y  = 2x so 2x = 1,

The point√(2,

Use the quadratic formula to get √ √ √ √ √ 4 ± 16 − 4 x0 = = 2 ± 3

The points are (2 + 3,

Chapter 2

the tangent line at x = x0 is y − y0 = (3ax20 + b)(x − x0 ) where y0 = ax30 + bx0

Solve with y = ax3 + bx to get (ax3 + bx) − (ax30 + bx0 ) = (3ax20 + b)(x − x0 ) ax3 + bx − ax30 − bx0 = 3ax20 x − 3ax30 + bx − bx0 x3 − 3x20 x + 2x30 = 0 (x − x0 )(x2 + xx0 − 2x20 ) = 0 (x − x0 )2 (x + 2x0 ) = 0,

1 1 x 2

the tangent line at x = x0 is y − y0 = − 2 (x − x0 ),

The tangent line crosses the 2 x x0 x0 x0 1 x-axis at 2x0 ,

so that the area of the triangle is (2/x0 )(2x0 ) = 2

F = GmM r−2 ,

dF 2GmM = −2GmM r−3 = − dr r3 6

f  (x) = 1 + 1/x2 > 0 for all x = 0

f is continuous at 1 because lim− f (x) = lim+ f (x) = f (1)

also lim− f  (x) = lim− (2x + 1) = 3 and lim+ f  (x) = x→1

lim 3 = 3 so f is differentiable at 1,

f (x) − f (1) equals the derivative of x2 x−1  √ f (x) − f (1) 1  x = 1,

while lim equals the derivative of x at x = 1,

namely √  = x−1 2 x x=1 x−>1+ Since these are not equal,

f is not differentiable at x = 1

f is continuous at 1 because lim− f (x) = lim+ f (x) = f (1)

(a) f (x) = 3x − 2 if x ≥ 2/3,

f (x) = −3x + 2 if x < 2/3 so f is differentiable everywhere except perhaps at 2/3

also lim f  (x) = lim (−3) = −3 and lim f  (x) = lim (3) = 3 so f is not x→2/3−

(b) f (x) = x2 − 4 if |x| ≥ 2,

f (x) = −x2 + 4 if |x| < 2 so f is differentiable everywhere except perhaps at ±2

f is continuous at −2 and 2,

also lim− f  (x) = lim− (−2x) = −4 and lim+ f  (x) = lim+ (2x) = 4 so f is not x→2

Similarly,

f is not differentiable at x = −2

      d'd d2 d'd d2 d'd [cf (x)] = [f (x)] = c'[f (x)] = c'2 [f (x)] c'[cf (x)] = 2 dx dx dx dx dx dx dx dx

Exercise Set 2

    d'd d2 d'd d2 d'd2 [f (x) + g(x)] = [f (x)] + [g(x)] = 2 [f (x)] + 2 [g(x)] [f (x) + g(x)] = 2 dx dx dx dx dx dx dx dx (b) Yes,

by repeated application of the procedure illustrated in part (a)

f  (x) = n(n − 1)(n − 2)xn−3 ,

f (n) (x) = n(n − 1)(n − 2) · · · 1 (b) From part (a),

f (k) (x) = k(k − 1)(k − 2) · · · 1 so f (k+1) (x) = 0 thus f (n) (x) = 0 if n > k

f (n) (x) = an n(n − 1)(n − 2) · · · 1

Let g(x) = xn ,

Use Exercise 52 in Section 2

If x0 = mx1 + b ,

then Exercise 52 says that f is differentiable at x1 and f  (x1 ) = mg  (x0 )

Since g  (x0 ) = nxn−1 0 77

f (x) = 27x3 − 27x2 + 9x − 1 so f  (x) = 81x2 − 54x + 9 = 3 · 3(3x − 1)2 ,

f (x) = 3(2x + 1)−2 so f  (x) = 3(−2)2(2x + 1)−3 = −12/(2x + 1)3

so f  (x) = −2(x + 1)−3 = −2/(x + 1)3

Exercise Set 2

(b) f  (x) = (x + 1) · (2) + (2x − 1) · (1) = 4x + 1

(b) f  (x) = (x2 + 1) · (2x) + (x2 − 1) · (2x) = 4x3

   1 3 1 1 d'(3x2 + 6) = (3x2 + 6)(2) + 2x − 2x − + 2x − (6x) = 18x2 − x + 12 4 4 dx 4 2

d d'(2x−3 + x−4 ) + (2x−3 + x−4 ) (x3 + 7x2 − 8) = (x3 + 7x2 − 8)(−6x−4 − 4x−5 )+ dx dx +(2x−3 + x−4 )(3x2 + 14x) = −15x−2 − 14x−3 + 48x−4 + 32x−5

f  (x) = 1 · (x2 + 2x + 4) + (x − 2) · (2x + 2) = 3x2 11

d d'(x2 + 1) dx (3x + 4) − (3x + 4) dx (x2 + 1) (x2 + 1) · 3 − (3x + 4) · 2x −3x2 − 8x + 3 = = (x2 + 1)2 (x2 + 1)2 (x2 + 1)2

d d'(3x − 4) dx (x2 ) − x2 dx (3x − 4) (3x − 4) · 2x − x2 · 3 3x2 − 8x = = (3x − 4)2 (3x − 4)2 (3x − 4)2

so x+3 d'd (x + 3) dx (2x3/2 + x − 2x1/2 − 1) − (2x3/2 + x − 2x1/2 − 1) dx (x + 3) = 2 (x + 3)

(x + 3) · (3x1/2 + 1 − x−1/2 ) − (2x3/2 + x − 2x1/2 − 1) · 1 x3/2 + 10x1/2 + 4 − 3x−1/2 = 2 (x + 3) (x + 3)2

This could be computed by two applications of the product rule,

but it’s simpler to expand f (x): f (x) = 14x + 21 + 7x−1 + 2x−2 + 3x−3 + x−4 ,

so f  (x) = 14 − 7x−2 − 4x−3 − 9x−4 − 4x−5

Chapter 2     d' d' d' d' g(x)2 = 2g(x)g  (x) and g(x)3 = g(x)2 g(x) = g(x)2 g  (x) + g(x) g(x)2 = g(x)2 g  (x) + dx dx dx dx g(x) · 2g(x)g  (x) = 3g(x)2 g  (x)

In general,

Letting g(x) = x7 + 2x − 3,

we have f  (x) = 3(x7 + 2x − 3)2 (7x6 + 2)

 (x + 3) · 2 − (2x − 1) · 1 7 dy  7 dy =

so = dx (x + 3)2 (x + 3)2 dx x=1 16

x(3) − (3x + 2)(1) 3x + 2 d'−5 3x + 2 dy −6 = x +1 + x +1 −5x + x +1 = = 23

so = 5(−5) + 2(−2) = −29

(x2 + 1) · 1 − x · 2x 1 − x2 = ,

g  (4) = (2)(−5) + (3) = −37/4

xf  (x) − f (x)  (4)(−5) − 3 = −23/16

(a) F  (x) = 5f  (x) + 2g  (x),

F  (2) = 5(4) + 2(−5) = 10

(b) F  (x) = f  (x) − 3g  (x),

F  (2) = 4 − 3(−5) = 19

(c) F  (x) = f (x)g  (x) + g(x)f  (x),

F  (2) = (−1)(−5) + (1)(4) = 9

(d) F  (x) = [g(x)f  (x) − f (x)g  (x)]/g 2 (x),

F  (2) = [(1)(4) − (−1)(−5)]/(1)2