(d) Each positive integer has one and only one cube. This is a function.

(e) Some cars have several drivers. In a one-car family, everyone drives

the same car. So this is not a function.

(f) Each toe is attached to one and only one foot. This is a function.

(g) Each rational number succeeds one and only one integer. This is a

function.

(h) Each integer has one and only one successor. This is a function.

(i) Each real number has a well de¬ned square, and adding six is a well

de¬ned operation. This is a function.

11.

(a) (b)

273

Chapter 1

(c) (d)

(e) (f)

√

sin(8π/3) = sin(2π/3) = 3/2

12. (a)

√ √

tan(’5π/6) = [’1/2]/[’ 3/2] = 1/ 3

(b)

√

sec(7π/4) = 1/ cos(7π/4) = 2

(c)

√ √

csc(13π/4) = csc(5π/4) = 1/ sin(5π/4) = 1/[’ 2/2] = ’ 2

(d)

cot(’15π/4) = cot(’7π/4) = cot(π/4) = cos(π/4)/ sin(π/4) =

(e) √ √

[ 2/2]/[ 2/2] = 1

√

cos(’3π/4) = ’ 2/2

(f)

13. We check the ¬rst six identities.

√

cos π/3 = 1/2, sin π/3 = 3/2, cos2 π/3 + sin2 π/3 = [1/2]2 +

(a) √

[ 3/2]2 = 1/4 + 3/4 = 1.

√ √

cos π/3 = 1/2, sin π/3 = 3/2, ’1 ¤ 1/2 ¤ 1, ’1 ¤ 3/2 ¤ 1.

(b)

√ √2

tan π/3 = 3, sec π/3 = 2, tan π/3 + 1 = [ 3] + 1 = 3 + 1 =

2

(c)

22 = sec2 π/3.

√ √ √2

cot π/3 =√ 3, csc π/3 = 2/ 3, cot π/3 + 1 = [1/ 3] + 1 =

2

(d) 1/

4/3 = [2/ 3]2 = csc2 π/3.

274 Solutions to Exercises

sin(π/3 + (’π/6)) = sin(π/6) = 1/2, sin π/3 cos(’π/6) +

(e) √ √

cos π/3 sin(’π/6) = [ 3/2][ 3/2] + [1/2][’1/2] = 1/2.

√

cos(π/3 + (’π/6)) = cos(π/6) = √ cos π/3 cos(’π/6) ’

(f) 3/2,

√ √

sin π/3 sin(’π/6) = [1/2][ 3/2] ’ [ 3/2][’1/2] = 3/2.

14. We shall do (a), (c), (e).

F/2

Fig. S1.14(a)

Fig. S1.14(c)

Fig. S1.14(e)

= (15/2)—¦

15. (a) θ

= ’60—¦

(b) θ

= 405—¦

(c) θ

= (405/4)—¦

(d) θ

= (540/π )—¦

(e) θ

= (’900/π )—¦

(f) θ

θ = 13π/36 radians

16. (a)

θ = π/18 radians

(b)

θ = ’5π/12 radians

(c)

275

Chapter 1

θ = ’2π/3 radians

(d)

θ = π 2 /180 radians

(e)

θ = 157π/9000 radians

(f)

f —¦ g(x) = [(x ’ 1)2 ]2 + 2[(x ’ 1)2 ] + 3; g —¦ f (x) = ([x 2 + 2x +

17. (a)

3] ’ 1)2 .

3√

32

f —¦ g(x) = x ’ 2 + 1; g —¦ f (x) = [ x + 1]2 ’ 2.

(b)

f —¦ g(x) = sin([cos(x 2 ’ x)] + 3[cos(x 2 ’ x)]2 ); g —¦ f (x) =

(c)

cos([sin(x + 3x 2 )]2 ’ [sin(x + 3x 2 )]).

f —¦ g(x) = eln(x’5)+2 ; g —¦ f (x) = ln(ex+2 ’ 5).

(d)

f —¦g(x) = sin([ln(x 2 ’x)]2 +[ln(x 2 ’x)]); g—¦f (x) = ln([sin(x 2 +

(e)

x)]2 ’ [sin(x 2 + x)]).

’x 2 2 x2 2

f —¦ g(x) = e[e ] ; g —¦ f (x) = e’[e ] .

(f)

f —¦ g(x) = [(2x ’ 3)(x + 4)] · [(2x ’ 3)(x + 4) + 1] · [(2x ’ 3)(x +

(g)

4)+2]; g—¦f (x) = (2[(x(x +1)(x +2)]’3)([(x(x +1)(x +2)]+4).

f is invertible, with f ’1 (t) = (t ’ 5)1/3 .

18. (a)

g is not invertible since g(0) = g(1) = 0.

(b)

h is invertible, with h’1 (t) = sgn t · t 2 .

(c)

f is invertible, with f ’1 (t) = (t ’ 8)1/5 .

(d)

g is invertible, with g ’1 (t) = ’[ln t]/3.

(e) √

h is not invertible, since sin π/4 = sin 9π/4 = 2/2.

(f)

f is not invertible, since tan π/4 = tan 9π/4 = 1.

(g) √

g is invertible, with g ’1 (x) = sgn x · |x|.

(h)

19. We will do (a), (c), (e), and (g).

Fig. S1.19(a)

276 Solutions to Exercises

Fig. S1.19(c)

Fig. S1.19(e)

(g) Not invertible.

√

’1

Invertible, f (t) = t.

20. (a)

Invertible, g ’1 (t) = et .

(b)

Invertible, h’1 (t) = Sin’1 t.

(c)

Invertible, f ’1 (t) = Cos’1 t.

(d)

Invertible, g ’1 (t) = Tan’1 t.

(e)

Not invertible because h(’1) = h(1) = 1.

(f)

√

Invertible, f ’1 (t) = [3 + 9 + 4t]/2.

(g)

Chapter 2

lim x · ex = 0 because x tends to 0 and ex tends to 1.

1. (a)

x’0

x2 ’ 1

= lim x + 1 = 2.

(b) lim

x’1 x ’ 1 x’1

lim (x ’ 2) · cot(x ’ 2) = lim [(x ’ 2)/ sin(x ’ 2)] · cos(x ’ 2) =

(c)

x’2 x’2

1 · 1 = 1. [Here we use the non-trivial fact, explored in Chapter 5,

that lim (sin h/ h) = 1.]

h’0

lim x · ln x = lim ln x x = ln 1 = 0. [Here we use the non-trivial

(d)

x’0 x’0

fact, explored in Chapter 5, that lim x x = 1.]

x’0

’ 7t + 12

t2

= lim (t ’ 4) = ’1.

(e) lim

t ’3

t’3 t’3

277

Chapter 2

s 2 ’ 3s ’ 4

= lim (s + 1) = 5.

(f) lim

s’4

s’4 s’4

ln x

= lim ln[x 1/(x’1) ] = lim ln(1 + h)1/ h = ln e = 1.

(g) lim

x’1 x ’ 1 x’1 h’0

[Here we use the non-trivial fact, explored in Chapters 5 and 6, that

lim (1 + h)1/ h = e, where e is Euler™s number.]

h’0

x2 ’ 9

= lim x ’ 3 = ’6.

(h) lim

x’’3 x + 3 x’’3

2. (a) lim f (x) does not exist, so f is not continuous.

x’’1

lim f (x) = 1/2 and f (3) = 1/2 so f is continuous at c = 3.

(b)

x’3

lim f (x) = 0. If we de¬ne f (0) = 0, which is plausible from the

(c)

x’0

graph, then f is continuous at 0.

lim f (x) = 0. If we de¬ne f (0) = 0, which is plausible from the

(d)

x’0

graph, then f is continuous at 0.

lim f (x) = 1 and f (1) = 1 so f is continuous at c = 1.

(e)

x’1

lim f (x) does not exist so f is not continuous at c = 1.

(f)

x’1

lim f (x) = 0 and f (0) = 0 so f is continuous at c = π .

(g)

x’π

lim f (x) = 2 · e2 and f (2) = 2 · e2 so f is continuous at c = 2.

(h)

x’2

3. (a) We calculate

f (2 + h) ’ f (2)

f (2) = lim

h

h’0

[(2 + h)2 + 4(2 + h)] ’ [22 + 4 · 2]

= lim

h

h’0

[4 + 4h + h2 + 8 + 4h] ’ [4 + 8]

= lim

h

h’0

h2 + 8h

= lim

h

h’0

= lim h + 8

h’0

= 8.

The derivative is therefore equal to 8.

(b) We calculate

f (1 + h) ’ f (h)

f (1) = lim

h

h’0

278 Solutions to Exercises

[’1/(1 + h)2 ] ’ [’1/12 ]

= lim

h

h’0

’1 ’ [’(1 + h)2 ]

= lim

h(1 + h)2

h’0

2h + h2

= lim

h’0 h + 2h2 + h3

2+h

= lim

h’0 1 + 2h + h2

=2

The derivative is therefore equal to 2.

(x 2 + 1) · 1 ’ x · 2x 1 ’ x2

d x

= =2

4. (a) .

dx x 2 + 1 (x 2 + 1)2 (x + 1)2

d d d2

sin(x 2 ) = sin (x 2 ) · x = [cos(x 2 )] · 2x.

(b)

Y

dx dx dx

d d d

FL

tan(t 3 ’t 2 ) = tan (t 3 ’t 2 )· (t 3 ’t 2 ) = sec2 (t 3 ’ t 2 ) ·

(c)

dt dt dt

(3t ’ 2t).

2

AM

d x2 ’ 1 (x 2 + 1) · (2x) ’ (x 2 ’ 1) · (2x) 4x

= =2

(d) .

dx x 2 + 1 (x 2 + 1)2 (x + 1)2

x · cos x

cos x

d

[x · ln(sin x)] = 1·ln(sin x)+x · = ln(sin x)+

(e) .

TE

sin x sin x

dx

d s(s+2)

= es(s+2) · [1 · (s + 2) + s · 1] = es(s+2) · [2s + 2].

(f) e

ds

d sin(x 2 ) d

2 2

= esin(x ) · [sin(x 2 )] = esin(x ) · cos(x 2 ) · 2x.

(g) e

dx dx

ex + 1

1

ln(e + x) = x · (e + 1) = x

x x

(h) .

e +x e +x

Since the ball is dropped, v0 = 0. The initial height is h0 = 100.

5. (a)

Therefore the position of the body at time t is given by

p(t) = ’16t 2 + 0 · t + 100.

The body hits the ground when

0 = p(t) = ’16t 2 + 100

or t = 2.5 seconds.

(b) Since the ball has initial velocity 10 feet/second straight down, we

know that v0 = ’10. The initial height is h0 = 100. Therefore the

279

Chapter 2

position of the body at time t is given by

p(t) = ’16t 2 ’ 10 · t + 100.

The body hits the ground when

0 = p(t) = ’16t 2 ’ 10t + 100

or t ≈ 2.207 seconds.

(c) Since the ball has initial velocity 10 feet/second straight up, we

know that v0 = 10. The initial height is h0 = 100. Therefore the

position of the body at time t is given by

p(t) = ’16t 2 + 10 · t + 100.

The body hits the ground when

0 = p(t) = ’16t 2 + 10t + 100

or t ≈ 2.832 seconds.

1

d

sin(ln(cos x)) = cos(ln(cos x)) · · (’ sin x)

6. (a)

cos x

dx

’ sin x

= cos(ln(cos x)) · .

cos x

d sin(cos x)

= esin(cos x) · cos(cos x) · (’ sin x).

(b) e

dx

1

d

ln(esin x + x) = sin x · (cos x + 1).

(c)

+x

dx e

1

d

arcsin(x 2 + tan x) = · [2x + sec2 x].

(d)

1 ’ [x 2 + tan x]2

dx

’1 1 ex

d

arccos(ln x ’ e /5) = · ’

x

(e) .