<< . .

. 31
( : 41)



. . >>

This is not a function.
(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) .

<< . .

. 31
( : 41)



. . >>