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CALCULUS Early Transcendentals Single Variable Tenth Edition Howard Anton Drexel University

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

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…………………………………………………………………………

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

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

Topics in Differentiation ……………………………

………………………………………

- ………

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

- …… 127

Applications of the Definite Integral in Geometry,

Science,

- and Engineering… 159

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

Chapter 8

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

Chapter 9

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

Conic Sections ………………………………………

- ………

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

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

- (a) −2
- (b) None
- (c) y = 0
- (d) −1
- 75 ≤ x ≤ 2
- x = −3,
- (e) ymax = 2
- 8 at x = −2
- ymin = −2

2 at x = 1

- (a) Yes
- (b) Yes
- (a) 1999,

$47,700

(c) No (vertical line test fails)

(d) No (vertical line test fails)

- (b) 1993,

$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

- √ 2 2 2 2 7

(a) √ f2(0) = 3(0) − 2 = −2

- f2 (2) = 3(2)2 − 2 = 10
- f (−2) = 3(−2) − 2 = 10
- f (3) = 3(3) − 2 = 25
- f ( 2) = 3( 2) − 2 = 4

f (3t) = 3(3t) − 2 = 27t − 2

- √ √ (b) f (0) = 2(0) = 0
- f (2) = 2(2) = 4
- f (−2) = 2(−2) = −4
- f (3) = 2(3) = 6
- f ( 2) = 2 2

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

- (a) Natural domain: x = 3

Range: y = 0

- (b) Natural domain: x = 0

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

(d) x2 − 2x + 5 = (x − 1)2 + 4 ≥ 4

- and is ≥ y ≥ 2
- √ 4 = 2

(e) Natural domain: sin x = 1,

- so x = (2n+ 12 )π,

- −1 ≤ sin x < 1,
- so 0 < 1−sin x ≤ 2,
- 1 1 1 and 1−sin x ≥ 2

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

- xx−2 = x + 2,

which is nonnegative for x ≥ −2

Natural √ √ domain: [−2,

2) ∪ (2,

The range of x + 2 is [0,

- for which x + 2 = 2

2) ∪ (2,

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

(b) C decreases for eight hours,

increases rapidly (but continuously),

- and then repeats

25 − x2

- √ 2 √25 − x ,
- − 25 − x2 ,
- −5 ≤ x ≤ 0 0 −1

Exercise Set 0

- 15 y 2 x –4
- (b) Domain: x = 0
- range: all y
- (a) Domain: x = 0
- range: all y
- (b) Domain: all x
- range: 0 < y ≤ 1

but the graph of y = x3 does not

- by deﬁnition
- 35 = (log 7
- 35)/(log 2) = (ln 7
- 35)/(ln 2) ≈ 2

6 = (log 0

- 6)/(log 5) = (ln 0
- 6)/(ln 5) ≈ −0
- –3 43
- x ≈ 1
- 47099 and x ≈ 7
- (a) No,

the curve passes through the origin

- √ (b) y = ( 4 2)x

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

- 75e−t/125 = 15,

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

- (c) y = 2−x = (1/2)x
- √ (d) y = ( 5)x

- (a) 140 dB
- (b) 120 dB
- (c) 80 dB
- no damage
- no damage

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

- respectively

IA = 107 I0 and IB = 109

- so IB /IA = 102

3 ≈ 200

- (a) log E = 4

2) = 16

- 7 ≈ 5 × 1016 J (b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E,
- respectively
- 5(M2 − M1 ) = log(10E) − log E = log 10 = 1,

- 5 = 2/3 ≈ 0

Chapter 0 Review Exercises y 5

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

- then V = 8 = x2 h,
- so h = 8/x2

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

- (b) The domain of C is (0,

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

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

- and the height is x,

so V = (6 − 2x)(5 − x)x ft3

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

- so 0 < x < 3

57 ft ×3

79 ft ×1

- 21 ft y 1 x –2

- −4 −3 −2 −1
- x f (x)
- (f ◦ g)(x)
- (g ◦ f )(x)
- −2 −3
- −3 −1 −4
- −3 −2 −1
- −1 −3
- −4 −2
- f (g(x)) = (3x + 2)2 + 1,
- g(f (x)) = 3(x2 + 1) + 2,
- so 9x2 + 12x + 5 = 3x2 + 5,
- 6x2 + 12x = 0,

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

- x = ±1

For f (g(h(x))) to be deﬁned,

- √ g(h(x)) = 1,
- x = ± 2

(f ◦ g ◦ h)(x) = 1/(2 − x2 )

- (a) even × odd = odd
- (b) odd × odd = even
- (c) even + odd is neither
- (d) odd × odd = even

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

- therefore,

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) = ,

- or t = 374

which is the same date as t = 9

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

- (c) From t = 0 to t = 70
- 58 and from t = 313
- 92 to t = 365 (the same date as t = 0),
- for a total of about 122 days

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,

- cos(x/2) = q

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

- 4pq − 2p − 2q + 1 = 0,
- (2p − 1)(2q − 1) = 0
- thus whenever = 1/2,
- when x/2 = π/6,

B has coordinates (π/3,

1 + 3),

1 + 3),

and D'has coordinates (−2π/3,

- 1 − √ (5π/3,

1 − 3)

- (a) (b)

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

(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

- y = sin x
- (a) x = f (y) = 8y − 1
- (x) = y =
- 1 (x + 1)1/3

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

- for example f (0) = f (2) = 1

√ (c) x = f (y) = (ey )2 + 1

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

- ln(x − 1)
- x+2 y + 2 −1
- f (x) = y =
- y−1 x−1
- (e) x = f (y) = sin

1 − 2y y

- f −1 (x) = y =
- 2 + sin−1 x
- 1−x 2 −2 or > ,
- so this is also

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 >

- ,x< 3x 3π − 2 3π + 2
- (f ) x =
- y = tan 1 + 3 tan−1 y

Then sin[cos−1 (4/5)] = 3/5,

- sin[cos−1 (5/13)] = 12/13,
- cos[sin−1 (4/5)] = 3/5,

and cos[sin−1 (5/13)] = 12/13

- 4 12 (a) cos[cos−1 (4/5) + sin−1 (5/13)] = cos(cos−1 (4/5)) cos(sin−1 (5/13) − sin(cos−1 (4/5)) sin(sin−1 (5/13)) = − 5 13 3 5 33 =
- 5 13 65 4 5 + (b) sin[sin−1 (4/5) + cos−1 (5/13)] = sin(sin−1 (4/5)) cos(cos−1 (5/13)) + cos(sin−1 (4/5)) sin(cos−1 (5/13)) = 5 13 3 12 56 =

5 13 65 31

- y = 5 ft = 60 in,
- so 60 = log x,
- x = 1060 in ≈ 1
- 58 × 1055 mi
- 3 ln e2x (ex )3 + 2 exp(ln 1) = 3 ln e2x + 3 ln(ex )3 + 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 2

(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,

- (b) N = 80 when t = 9
- (c) 220 sheep

Chapter 0 Review Exercises

- (a) The function ln x − x0
- 2 is negative at x = 1 and positive at x = 4,

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

(This will be established later in this book

- ) (b) x = 3

654 and 3

- 32105 × 105

(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

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

- and it has an inverse
- y=f (x) y=x
- x /2 y=f(x)

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

- x = π/2

- y = π/2

Limits and Continuity Exercise Set 1

- (a) −1
- (c) does not exist (c) 0
- (a) −∞
- (b) −∞
- (c) −∞
- (a) +∞
- (b) +∞
- (d) 1 (e) −∞
- (f ) x = −2,

01 −0

001 −0

9950166 0

9995002 0

9999500 1

0000500 1

0005002 1

0050167

- (ii) 13

The limit appears to be 1

001 0 0

The limit is +∞

- 9999 −1 −1

7143 −7

0111 −67

001 −667

0 −6667

- (a) −0

25 −0

001 −0

- (b) 0 1

99 −0

999 −1

01 −1

2161 54

1415 −4

536 −53

19 −539

deﬁne f (x) = x for x = a and 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

- x2 − 1 = x − 1 which gets close to −2 as x gets close to −1,

thus y − 1 = −2(x + 1) or y = −2x − 1

- x4 − 1 = x3 + x2 + x + 1 which gets close to 4 as x gets close to 1,

thus y − 1 = 4(x − 1) or y = 4x − 3

(a) The length of the rod while at rest

- (b) The limit is zero

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

- (a) By Theorem 1

this limit is 2 + 2 · (−4) = −6

- (b) By Theorem 1

this limit is 0 − 3 · (−4) + 1 = 13

- (c) By Theorem 1

this limit is 2 · (−4) = −8

- (d) By Theorem 1
- this limit is (−4)2 = 16
- (e) By Theorem 1

this limit is (f ) By Theorem 1

- this limit is

6 + 2 = 2

1 2 =−

- (−4) 2

By Theorem 1

- this limit is 2 · 1 · 3 = 6

By Theorem 1

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

After simpliﬁcation,

- x4 − 1 = x3 + x2 + x + 1,

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

- x+5 x2 + 6x + 5 = ,

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

- x2 − 3x − 4 x−4

After simpliﬁcation,

- 2x2 + x − 1 = 2x − 1,

and the limit is 2 · (−1) − 1 = −3

- t2 + 5t − 2 t3 + 3t2 − 12t + 4 = ,

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

- 3 t − 4t t2 + 2t

The limit does not exist

The limit does not exist

The limit is +∞

- √ √ x−9 29

- √ = x + 3,
- and the limit is 9 + 3 = 6
- x−3 31
- by Theorem 1
- f (x) = 2x,
- g(x) = x,
- so lim f (x) = lim g(x) = 0,
- but lim f (x)/g(x) = 2

After simpliﬁcation,

- x+4−2 1 ,
- and the limit is 1/4
- =√ x x+4+2
- (a) After simpliﬁcation,
- x3 − 1 = x2 + x + 1,
- and the limit is 3
- (a) Theorem 1
- 2 doesn’t apply

moreover one cannot subtract inﬁnities

- (b) lim+ x→0
- 1 1 − x x2
- = lim+ x→0
- x−1 x2
- = −∞

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

- − = = 2 = and lim =

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

- or the limit could exist,

or equal any preassigned real number

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)

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

- lim g(x)

Exercise Set 1

- (a) −∞
- (b) +∞ (b) −1

(a) 3 + 3 · (−5) = −12 (e)

(b) 0 − 4 · (−5) + 1 = 21

- (f ) 3/(−5) = −3/5
- (c) 3 · (−5) = −15
- (d) (−5)2 = 25

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

- x f (x)

00001 0

000001 1

471128 1

560797 1

569796 1

570696 1

570786 1

- (b) The limit is π/2

The limit is −∞,

- by the highest degree term

- by the highest degree terms

- by the highest degree terms

- by the highest degree terms

The limit is −∞,

- by the highest degree terms

- by the highest degree terms
- √ 23

The limit is 3 −5/8 = − 3 5 /2,

- by the highest degree terms

√ 5 − x22 2 √ 5x − 2 = when x < 0

- 3 x+3 −1 − x − y2 + 1 √ 2−y when y < 0

- = 2 7 7 + 6y 2 + 6 y
- 3x4 + x = x2 − 8 1−

1 x3 8 x2

- when x < 0

- √ x2 + 3 + x 3 2 31

lim ( x + 3 − x) √ = lim √ = 0,

- by the highest degree terms

x→+∞ x2 + 3 + x x→+∞ x2 + 3 + x 33

- 1 − ex 1−0 = 1
- = x x→−∞ 1 + e 1+0 lim

- 1 + e−2x 1+0 = 1

= −2x x→+∞ 1 − e 1−0 lim

- (x + 1)x 1 x+1 = 1 + ,
- so lim = e from Figure 1

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,

- and lim e(t) = c
- t→+∞
- (b) −5
- (a) +∞ 49
- t→+∞
- lim p(x) = +∞

When n is even,

- lim p(x) = +∞
- when n is odd,
- lim p(x) = −∞
- x→−∞
- x→+∞
- x→+∞
- (b) Yes,

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,

- so does 2 2 et = ex

- x→+∞
- t→+∞
- (b) If f (t) → +∞ (resp

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

small) by taking t large enough

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

- limit −∞)

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→−∞

- without bound”

- x < c”,
- t = 1/x,
- lim f (t) = +∞
- t→+∞
- t = csc x,
- lim f (t) = +∞
- t→+∞

Then t also tends to +∞,

- and 61

Set t = −x,

- then get lim
- t→−∞
- t = e by Figure 1

From the hint,

lim bx = lim e(ln b)x x→+∞

- x→+∞
- ln 2x t + ln 2 = ,
- so the limit is 1
- ln 3x t + ln 3

- = ⎪ ⎪ ⎩ +∞ if b > 1
- v 200 160 120 80 40 t

(b) lim v = 190 1 − lim e−0

- 168t = 190,

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

- t→∞
- t→∞

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

- n 2 3 4 5 6 7 1 + 10−n 1

00001 1

000001 1

- 0000001 1 + 10n 101 1001 10001 100001 1000001 10000001 n (1 + 10−n )1+10 2

71828 2

- 718282 The limit appears to be e

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

(c) The exponents are being multiplied by a,

- so the result is ea

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

After a long division,

- f (x) = x + 2 +

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

- x→±∞
- y 15 9 y=x+2 3 –12
- –6 –3
- –9 –15

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

- f (x) = −x2 +1+
- y 12 x=3
- y = –x 2 + 1
- –6 –12

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

The only vertical asymptote is at x = 1

- x→±∞
- y 5 3 y = sin x x –4
- x=1 –4

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

- 1 if and only if |x| < 0

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

- 1 if and only if |x − 3| < (0

1)/4 = 0

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

- 001 at x = 4

000124998,

- which corresponds to δ = 0

000124998

- and f (x) = 16 − = 15
- 999 at x = 3

999874998,

- for which δ = 0

000125002

- 000125 (to six decimals)
- (a) x0 = (1

95)2 = 3

x1 = (2

05)2 = 4

|4 − 4

- 2025| ) = 0
- |(x3 −4x+5)−2| < 0
- 05 is equivalent to −0
- 05 < (x3 −4x+5)−2 < 0
- which means 1
- 95 < x3 −4x+5 < 2

Now x3 −4x+5 = 1

- 95 at x = 1
- and x3 −4x+5 = 2
- 05 at x = 0

So δ = min (1

- 0616 − 1,

1 − 0

9558) = 0

- 80274) and (1
- 19301) belong to the graph

Set x0 = 0

- 87 and x1 = 1

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

we see that if x0 < x < x1 then 1

- 80274 < f (x) < 2
- and therefore 1
- 8 < f (x) < 2

- |2x − 8| = 2|x − 4| < 0
- 1 when |x − 4| < 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 when |x − 3| < 0

05 = δ

- x−3 x−3 13

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

- so we can choose δ = 0

1 1 x − 5 |x − 5| < ,

- so we can choose 15

Assume δ ≤ 1

- 05 · 20 = 1

and hence any δ > 0 will work

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

2 2x + x − 1 = |2x| < if |x| < /2,

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

- then take δ =
- if |x − 0| = |x| < δ,
- then |x − 0| = |x| <

- let > 0

(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,

- since clearly x < 11
- (c) δ = min(1,
- |3x + 32| · |x − 10| < 65 · |x − 10| < 65 · /65 =

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

- so δ = min(1,

- x + 1 < −1/2,
- |x + 1| > 1/2

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

- so δ = min(1/2,

Exercise Set 1

√ √ √ x + 2 x − 4 1 √ 35

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

- so δ = min(2 ,
- x + 2 x + 2 2 37

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

and |f (x) − 0| = |x| < δ =

- if x is irrational

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

1 here,

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

- so N = 99

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

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

- and N = −10

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

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

- x21 1−
- = 1 − ,
- x1 = − = 1 − ,
- x2 = 1 + x21

1 + x22

- (b) N =
- (c) N = −
- 01 if |x| > 10,

x2 x 1 < 0

- 001 if |x + 1| > 1000,
- x > 999,

N = 999

- − 1 = x+1 x + 1
- 1 47
- − 0 < 0
- 005 if |x + 2| > 200,
- −x − 2 > 200,
- x < −202,

x+2 4x − 1 11 < 0

- 1 if |2x + 5| > 110,
- −2x − 5 > 110,
- 2x < −115,
- x < −57

N = −57

− 2 = 2x + 5 2x + 5 1 1 1 51

- 2 < if |x| > √ ,
- so N = √

11 4x − 1 < if |2x+5| > 11 ,

- when −2x−5 > 11 ,
- which means 2x < − 11 −5,
- or x < − 11 − 5 ,
- − 2 = 2x + 5 2x + 5
- 2 2 5 11 so N = − −

√ 2 2 2 x 2 √ √ 2 2 2 2 55

- √ ,
- so N = 1 +
- < if x − 1 > ,
- when x > 1 + ,
- or x > 1 + − 2 = √
- x−1 x − 1 57
- 1 1 > 100 if |x| < 2 x 10
- 1 1 > 1000 if |x − 1| < |x − 1| 1000

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

- (d) −

1 1 1 ,

|x| < < −10000 if x4 < 4 x 10000 10

1 1 1 1 ,

- or 0 < |x − 3| < √ ,
- so δ = √

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

- 1 1 1 > M when 0 < |x| < ,
- so δ =
- |x| M M

If M < 0 then −

- 1 1 1 1 < M when 0 < x4 < − ,
- or |x| < ,
- so δ =
- 1/4 x4 M (−M ) (−M )1/4

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

- so δ =

If x > 4 then

√ x − 4 < if x − 4 < 2 ,

- or 4 < x < 4 + 2 ,
- so δ = 2

- so δ =

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

- 1 − x > ,
- or x < 1 − ,
- so we can choose δ = −

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 ,

- 1 − x < ,
- or x > 1 − ,
- so we can choose δ =
- 1−x M M M 73
- (a) Given any M > 0,

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

- x + 1 > M ,
- or x > M − 1,
- so N = M − 1
- (b) Given any M < 0,

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

- x + 1 < M ,
- or x < M − 1,
- so N = M − 1
- 4 (amperes) 7

5 + δ 7

5 − δ

- (d) 0
- (e) It approaches inﬁnity

Exercise Set 1

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

- (d) Yes

No: f (3) is not deﬁned

- (c) No: f (1) is not deﬁned
- (f ) Yes
- (d) Yes
- (e) Yes
- x→2−
- (f ) Yes

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

- (d) Yes
- (c) No: lim f (x) = f (2)

- (g) Yes

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,

- x = 1 and x = −1

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

- x→4−
- by Theorem 1
- f (x) = g(x) = 2 if x = 3,
- f (3) = 1,
- g(3) = 3
- use Theorem 1
- 3 with g(x) =

(a) f is continuous for x < 1,

- and for x > 1
- lim− f (x) = 5,
- lim+ f (x) = k,

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

(b) f is continuous for x < 2,

- and for x > 2
- lim f (x) = 4k,
- lim f (x) = 4 + k,
- so if 4k = 4 + k,
- k = 4/3 then f − + x→2
- is continuous for all x
- f is continuous for x < −1,
- −1 < x < 2 and x > 2
- lim f (x) = 3m + k = 3m + 4,
- x→2−
- lim f (x) = 4,
- lim f (x) = k,
- so k = 4 is required
- x→−1−
- x→−1
- lim f (x) = 9,
- so 3m + 4 = 9,

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

- and m = 5/3
- (a) x = 0,

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

- (b) x = −3

deﬁne f (−3) = −3 = lim f (x),

then the discontinuity is removable

- x→−3

(c) f is undeﬁned at x = ±2

- at x = 2,
- lim f (x) = 1,

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

- at x→2
- x = −2,
- lim f (x) does not exist,

so the discontinuity is not removable

- x→−2

Discontinuity at x = 1/2,

- not removable
- at x = −3,
- removable

(b) 2x2 + 5x − 3 = (2x − 1)(x + 3) 39

- 6) of the two continuous functions g(x) = x3 and h(x) = x1/5
- it is thus continuous

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

- (a) Let h = x − c,
- x = h + c

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

- (b) With g(h) = f (c + h),

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

so g(h) is continuous at h = 0

That is,

- f (c + h) h→0
- is continuous at h = 0,
- so f is continuous at x = c

and f (x) = −1 on 1 ≤ x ≤ 2

If f (x) = x3 + x2 − 2x − 1,

- then f (−1) = 1,
- f (1) = −1

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

- since f (−1
- 3) < 0 and f (−1

2) > 0,

- the midpoint x = −1
- 25 of [−1
- 2] is the required approximation of the root

- since f (0
- 7) < 0 and f (0

8) > 0,

- the midpoint x = 0

75 of [0

- 8] is the required approximation

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

- since f (2
- 2) < 0 and f (2

3) > 0,

- use the interval [2

Since f (2

- 23) < 0 and f (2
- 24) > 0 the midpoint x = 2

235 of [2

- 24] is the required approximation of the root

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)

- then the statement follows

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

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

Then for large z,

- f (z) = area of ellipse,
- and for small z,
- f (z) = 0

so it is continuous everywhere

Discontinuities at x =

- and x = + 2nπ,

sin−1 u is continuous for −1 ≤ u ≤ 1,

- so −1 ≤ 2x ≤ 1,
- or −1/2 ≤ x ≤ 1/2

3) ∪ (3,

- (−∞,
- −1] ∪ [1,
- (a) f (x) = sin x,
- g(x) = x3 + 7x + 1
- (b) f (x) = |x|,
- g(x) = sin x
- (c) f (x) = x3 ,
- g(x) = cos(x + 1)
- 1 1 = cos lim = cos 0 = 1

lim cos x→+∞ x→+∞ x x 19

- lim sin−1
- x→+∞
- x 1 − 2x
- sin x
- lim e x→0
- = sin−1
- x x→+∞ 1 − 2x
- 1 π = sin−1 − =−
- lim sin x
- = e0 = 1
- sin 3θ sin 3θ = 3 lim = 3

- lim+ θ→0
- sin θ = θ2

- sin θ = +∞

tan 7x tan 7x 7 sin 7x 3x 7 7 = · · ,

- so lim = ·1·1=

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

√ sin x sin x 1 √ = lim x lim = 0

- 5 x→0+ 5 x x→0+ x

sin x2 sin x2 = lim x lim = 0

- x→0 x→0 x→0 x2 x

t2 = 33

- 1 − cos2 t
- t sin t
- t→0 1 − cos2 t
- 1 + cos θ θ2 (1 + cos θ) θ2 · = = 35
- 1 − cos θ 1 + cos θ 1 − cos2 θ

θ sin θ

θ→0 1 − cos θ

- (1 + cos θ),
- 1 = lim sin t,
- so the limit does not exist
- lim+ sin t→+∞ x x→0 39
- 2 − cos 3x − cos 4x 1 − cos 3x 1 − cos 4x 1 − cos 3x 1 − cos 3x 1 + cos 3x sin2 3x = +

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

Thus x 1 + cos 3x lim

- 2 − cos 3x − cos 4x sin 3x sin 4x sin 3x sin 4x = lim · + lim · = 3 · 0 + 4 · 0 = 0

x→0 x→0 x x 1 + cos 3x x 1 + cos 4x

093497 0

100932 0

100842 0

098845 0

091319 0

- (b) Let t = x − 5

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

- consider f (x) = tan−1 x

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

- sin x◦ πx◦

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

t→0 (180t/π) 180 sin kx 1 = k,

- lim f (x) = 2k 2 ,
- so k = 2k 2 ,

and the nonzero solution is k =

- + kx cos kx 2 x→0
- lim− f (x) = k lim
- (a) lim
- sin t = 1
- (b) lim− t→0
- 1 − cos t = 0 (Theorem 1
- (c) sin(π − t) = sin t,

- t→0 sin t sin x
- t = x − 1

sin(πx) = sin(πt + π) = − sin πt

- and lim

sin(πx) sin πt = − lim = −π

- t→0 x−1 t
- √ √ cos x − sin x = 55
- t = x − π/4,

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

- lim = − 2 lim = − 2

− t→0 t t x − π/4 x→π/4 57

- sin x = 1
- sin−1 5x 5x = 5 lim = 5
- x→0 x→0 sin 5x 5x
- −|x| ≤ x cos
- ≤ |x|,

which gives the desired result

- no conclusions can be drawn

lim f (x) = 0 by the Squeezing Theorem

- x→+∞

- (a) Let f (x) = x − cos x
- f (0) = −1,
- f (π/2) = π/2

5 y=x 1 0

- y = cos x x

(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

- 8 and g(39) > 9

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

8 exactly

- (a) 1 (h) 2 3
- x f (x)
- (b) Does not exist
- (i) 1/2

The limit is

- (c) Does not exist
- (−1)3 − (−1)2 = 1
- −1 − 1

- x2 + 4x + 3 x+1 2

By the highest degree terms,

- the limit is 11
- (a) y = 0
- (b) None

32 25 =

- (c) y = 2

- sin 3x = cos 3x,
- and the limit is 1

- 3x − sin(kx) sin(kx) =3−k ,
- so the limit is 3 − k

- tan t → −∞,
- so the limit in question is 0
- x/3 (−3) ,
- so the limit is e−3
- (a) f (x) = 2x/(x − 1)
- y 10 x 10
- (a) lim f (x) = 5
- (b) δ = (3/4) · (0

048/8) = 0

- (a) |4x − 7 − 1| < 0
- 01 means 4|x − 2| < 0
- or |x − 2| < 0
- so δ = 0

2 4x − 9 − 6 < 0

- 05 means |2x + 3 − 6| < 0
- or |x − 1
- so δ = 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

- 001 provided |x − 4| < 0
- 001/9 = 1/9000,
- take δ = 1/9000,

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

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

- − < f (x) − f (x0 ) < ,

f (x) > f (x0 ) − = f (x0 )/2 > 0,

- for x0 − δ < x < x0 + δ

(a) f is not deﬁned at x = ±1,

- continuous elsewhere
- (b) None
- continuous everywhere

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

- continuous elsewhere

for x > 2 f is a polynomial and is continuous

At x = 2,

f (2) = −13 = 13 = lim+ f (x),

- so f is not continuous there

- f (x) = −1 for a ≤ x 0),
- minimum: t = 3 (slope < 0)

(d) (3 − 18)/(4 − 2) = −7

- 5 cm/s (slope of estimated tangent line to curve at t = 3)

It is a straight line with slope equal to the velocity

- (a) msec =
- 2 f (1) − f (0) = =2 1−0 1 f (x1 ) − f (0) 2x21 − 0 = lim = lim 2x1 = 0 x1 →0 x1 →0 x1 − 0 x1 →0 x1 − 0
- (b) mtan = lim

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

- (c) mtan = lim

x1 →x0

- y Secant 2 x

(d) The tangent line is the x-axis

- (a) msec =
- 1/3 − 1/2 1 f (3) − f (2) = =− 3−2 1 6
- (b) mtan = lim

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

- (c) mtan = lim

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

- (a) mtan = lim

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

- (b) mtan = 2(−1) = −2 17
- (a) mtan

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

- x1 ) − (x0 + x1 − x0
- = lim

x1 →x0

- 1 1+ √ √ x1 + x0
- 1 =1+ √ 2 x0
- 1 3 (b) mtan = 1 + √ = 2 2 1 19

Velocity represents the rate at which position changes

- (a) 72◦ F at about 4:30 P

(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)

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

- about 10 cm/year

Growth rate (cm/year)

- 30 20 10 t (yrs)
- 3 · 403 = 19,200 ft

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

- (c) Solve s'= 0

3t3 = 1000

- t ≈ 14
- 938 so vave ≈ 1000/14

938 ≈ 66

943 ft/s

- 3(40 + h)3 − 0

3 · 403 0

- 3(4800h + 120h2 + h3 ) = lim = lim 0
- 3(4800 + 120h + h2 ) = 1440 ft/s h→0 h→0 h→0 h h
- (d) vinst = lim 29
- (a) vave =
- 6(4)4 − 6(2)4 = 720 ft/min 4−2 6t41 − 6(2)4 6(t41 − 16) 6(t21 + 4)(t21 − 4) = lim = lim = lim 6(t21 + 4)(t1 + 2) = 192 ft/min t1 →2 t1 →2 t1 →2 t1 →2 t1 − 2 t1 − 2 t1 − 2
- (b) vinst = lim

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

- f (1) = 2
- f (3) = 0,
- f (5) = −2
- f (6) = −1

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

- (b) f (2) = m = 3
- (c) The same,
- f (2) = 3
- y − (−1) = 5(x − 3),

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),

- y = 4x − 2
- f (x) = lim

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) = lim

√ √ √ √ √ √ 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)

- f (x) = lim

- 1 5 1 (x − 8),
- y = x +

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

- f (x) = lim
- f (x) = lim

Δx→0

- 1 1 √ √ −√ x − x + Δx x − (x + Δx) x x + Δx √ = lim = lim = √ √ √ √ √ Δx→0 Δx x x + Δx Δx→0 Δx x x + Δx( x + Δx x + Δx)
- −1 1 √ = − 3/2

= 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

- h→0 h→0 h
- f (t) = lim
- (f ) E y

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

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

- (a) f (x) =
- (b) f (x) = x2 and a = 3
- x and a = 1

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

- and = −2

h→0 h→0 dx h→0 h h dx x=1 5

- y = −2x + 1 37
- w f (w) − f (1) w−1

- f (w) − f (1) w−1 39

2 − 2

- 12 f (2) − f (1) 2

34 − 2

- 12 f (2) − f (0) 2

34 − 0

- 58 f (3) − f (1) = = 0
- 3−1 2 2−1 1 2−0 2

(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

- (a) F ≈ 200 lb,

dF/dθ ≈ 50

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

- so the extra 301 − 300

(b) μ = (dF/dθ)/F ≈ 50/200 = 0

- (a) T ≈ 115◦ F,

- 35◦ F/min

(b) k = (dT /dt)/(T − T0 ) ≈ (−3

- 35)/(115 − 75) = −0

lim− f (x) = lim+ f (x) = f (1),

- so f is continuous at x = 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

- lim = lim = lim 2 = 2,
- so f (1) = 2

− + + h h h→0 h→0 h→0 h→0+ y 5

- 4) that lim x sin(1/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 − f (x0 ) <

Since x − x0

- f (x) − f (x0 ) > > 0

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

- (a) Let = |m|/2

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

- (b) For 0 < |x| < δ,
- |f (x)| < 12 |mx|

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

- |f (x) − mx| > |f (x)|

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

- then b = 0 since f (0) = 0

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

Exercise Set 2

- by Theorems 2

2 and 2

24x7 + 2,

- by Theorems 2
- by Theorem 2
- − (7x6 + 2),
- by Theorems 2
- −3x−4 − 7x−8 ,
- by Theorems 2

3 and 2

- 24x−9 + 1/ x,
- by Theorems 2
- f (x) = exe−1 −
- 10 x−1−
- by Theorems 2

3 and 2

- (3x2 + 1)2 = 9x4 + 6x2 + 1,
- so f (x) = 36x3 + 12x,
- by Theorems 2
- y = 10x − 3,
- y (1) = 7

2t − 1,

- by Theorems 2

2 and 2

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

- dy/dx|x=1 = 15

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

- f (1) ≈
- dy = −8x7 ,
- dy/dx|x=1 = −8
- 999699 − (−1) f (1
- 01) − f (1) = = 0
- and by diﬀerentiation,
- f (1) = 3(1)2 − 3 = 0

- by Theorems 2

2 and 2

3πr2 ,

- by Theorems 2

2 and 2

- the exact x2

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

- dx x=2 dV = 4πr2 37

dV (b) = 4π(5)2 = 100π dr r=5

- y − 2 = 5(x + 3),
- y = 5x + 17
- (a) dy/dx = 21x2 − 10x + 1,
- d2 y/dx2 = 42x − 10 (c) dy/dx = −1/x2 ,
- d2 y/dx2 = 2/x3
- dy/dx = 24x − 2,
- d2 y/dx2 = 24

(d) dy/dx = 175x4 − 48x2 − 3,

- d2 y/dx2 = 700x3 − 96x
- (a) y = −5x−6 + 5x4 ,
- y = 30x−7 + 20x3 ,

y = −210x−8 + 60x2 (b) y = x−1 ,

- y = −x−2 ,
- y = 2x−3 ,

y = −6x−4 (c) y = 3ax2 + b,

- y = 6ax,
- y = 6a 45
- (a) f (x) = 6x,
- f (x) = 6,
- f (x) = 0,
- f (2) = 0 (b)

dy d2 y d2 y 3 = 30x4 − 8x,

- = 120x − 8,
- = 112 dx dx2 dx2 x=1

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

- y = 3x2 + 3,
- y = 6x,

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

- dy dy = 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,

- 5/6) and (2,

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

- −2) is on the graph
- −2 = a(0)2 + b(0) + c,
- c = −2

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

- 0 = a + b − 2

The slope is dy/dx = 2ax + b

at x = 0 the slope is b so b = −1,

- thus a = 3

1) and (2,

- 4) are on the secant line so its slope is (4 − 1)/(2 + 1) = 1

- x = 1/2
- y = −2x,
- so at any point (x0 ,
- y0 ) on y = 1 − x2 the tangent line is y − y0 = −2x0 (x − x0 ),
- or y = −2x0 x + x20 + 1

The point√(2,

- 0) is to be on the line,
- so 0 = −4x0 + x20 + 1,
- x20 − 4x0 + 1 = 0

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

The points are (2 + 3,

- −6 − 4 3) and (2 − 3,
- −6 + 4 3)

- y = 3ax2 + b

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,

- so x = −2x0

1 1 x 2

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

- or y = − 2 +

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

- the y-axis at 2/x0 ,

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

- y = −

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,

- and the derivative equals 3

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

- namely 2x|x=1 = 2,

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)

- lim − x→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

- f is continuous at 2/3,

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

- x→2/3−
- x→2/3+
- x→2/3+
- diﬀerentiable at 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

- diﬀerentiable at x = 2

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)

- (a) f (x) = nxn−1 ,
- f (x) = n(n − 1)xn−2 ,

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

- (c) From parts (a) and (b),

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

Let g(x) = xn ,

- f (x) = (mx + b)n

Use Exercise 52 in Section 2

- but with f and g permuted

- the result follows

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 ,

- as predicted by Exercise 75

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

- f (x) =
- 2x2 + 4x + 2 + 1 = 2 + (x + 1)−2 ,

so f (x) = −2(x + 1)−3 = −2/(x + 1)3

- (x + 1)2

Exercise Set 2

- (a) f (x) = 2x2 + x − 1,
- f (x) = 4x + 1 3
- (a) f (x) = x4 − 1,
- f (x) = 4x3 5
- f (x) = (3x2 + 6)

(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) = (x3 + 7x2 − 8)

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

- f (x) =

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

- f (x) =

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

- f (x) = f (x) = =
- 2x3/2 + x − 2x1/2 − 1 ,

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)

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

- d 3x + 2

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

- dx dx x x2 dx x x
- 3x + 2 dy 2 −5x−6 + x−5 + 1 − 2

so = 5(−5) + 2(−2) = −29

- x x dx x=1 25
- f (x) =

(x2 + 1) · 1 − x · 2x 1 − x2 = ,

- so f (1) = 0
- (x2 + 1)2 (x2 + 1)2
- (a) g (x) = (b) g (x) =
- 1 1 xf (x) + √ f (x),

g (4) = (2)(−5) + (3) = −37/4

xf (x) − f (x) (4)(−5) − 3 = −23/16

- g (4) = x2 16

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

(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