Solution manual for calculus 5th edition by james stewart kokoska|ISBN:9780357632499|All answers included

SOLUTION AND ANSWER GUIDE
CALCULUS 5TH EDITION JAMES STEWART, KOKOSKA
Chapter 1-13
AP
CHAPTER 1: SECTION 1.1

TABLE OF CONTENTS
EX
End of Section Exercise Solutions .............................................................................................................. 1




END OF SECTION EXERCISE SOLUTIONS
SO

1.1.1
(a) f (1) = 3
(b) f (−1)  −0.2
(c) f (x) = 1 when x = 0 and x = 3.
LU
(d) f (x) = 0 when x ≈ –0.8.
(e) The domain of f is −2  x  4. The range of f is −1  y  3.
(f) f is increasing on the interval−2  x  1.
TI
1.1.2
(a) f (−4) = −2; g(3) = 4
(b) f (x) = g(x) when x = –2 and x = 2.
O
(c) f (x) = −1 when x ≈ –3.4.

(d) f is decreasing on the interval 0  x  4.
N

(e) The domain of f is −4  x  4. The range of f is −2  y  3.
S
(f) The domain of g is −4  x  4. The range of g is 0.5  y  4.



1.1.3

, (a) f (2) = 12 (b) f (2) = 16 (c) f (a) = 3a2 − a + 2
(d) f (−a) = 3a2 + a + 2 (e) f (a +1) = 3a2 + 5a + 4 (f) 2 f (x) = 6a2 − 2a + 4
(g) f (2a) = 12a2 − 2a + 2 (h) f (a2) = 3a4 − a2 + 2
 f (a)2 = (3a2 − a + 2)
2
(i) = 9a4 − 6a3 +13a2 − 4a + 4
(j) f (a + h) = 3 ( a + h) − (a + h) + 2 = 3a2 + 3h2 + 6ah − a − h + 2
2
AP

1.1.4

f (3 + h) − f (3) (4 + 3(3 + h) − (3 + h)2 ) − 4 9 + 3h − 9 − 6h − h2) −3h − h2
= = = = − (3 + h)
h h h h
EX

1.1.5
f (a + h) − f (a) a3 + 3a2h + 3ah2 + h3 − a3 h(3a2 + 3ah + h2 )
= = 3a2 + 3ah + h2
=
SO
h h h


1.1.6
1 1 a x 1
f (x) − f (a) x−a ax − ax a−x =−
LU
  




=  = =
x−a x−a x−a ax(x − a) ax
TI
1.1.7

x + 3 1+ 3 x + 3 x + 3 − 2x − 2 −x +1 x −1
− −2
f (x) − f (1) x +1 1+1 = x +1
= = x +1 x +1 = − x +1 = − 1
=
x −1 x −1
O
x −1 x −1 x −1 x −1 x +1
N
1.1.8
x+4
The domain of f (x) = is x 
| x  −3,3.
S
x2 − 9

1.1.9

2x3 − 5
The domain of f (x) = is x  | x  −3, 2.
x2 + x − 6

,1.1.10

The domain of f (t) = 3
2t −1 is all real numbers.
AP
1.1.11

g (t ) = − is defined when 3 − t  0  t  3 and 2 − t  0  t  2. Thus, the domain is t  2,
or (−, 2.

1.1.12
EX

1
The domain of h(x) =− is (−, 0) (5, ).


1.1.13
SO
The domain of F( p) = 2 − p is0  p  4.

1.1.14

The domain of f (u) = u +1 is u  | u  −2, −1.
1
1+
LU
u +1
1.1.15
(a) This function shifts the graph of y = |x| down two units and to the left one unit.
(b) This function shifts the graph of y = |x| down two units
TI
(c) This function reflects the graph of y = |x| about the x-axis, shifts it up 3 units and then to the left 2
units.
(d) This function reflects the graph of y = |x| about the x-axis and then shifts it up 4 units.
O
(e) This function reflects the graph of y = |x| about the x-axis, shifts it up 2 units then four units to the
left.
(f) This function is a parabola that opens up with vertex at (0, 5). It is not a transformation of y = |x|.
N

1.1.16
S
(a) g ( f (x)) = g (x2 +1) =10(x2 +1)

(b) f (g (4)) = f (10(4)) = 402 +1 = 1601

(c) g (g (−1)) = g (10(−1)) = 10(−10) = −100

, (d) ( ) ( (
f g ( f (2)) = f g 22 +1 )) = f (10(5)) = f (50) = 502 +1 = 2501
(e) 1 1 1 1
=  = =
f (g ( x)) f (10x) (10x)
2
+1 100x2 + 1

1.1.17

4 − x2 is −2  x  2, and the range is
AP
The domain of h(x) =
0  y  2. The graph is the top half of a circle of radius 2 with center at
the origin.
EX
1.1.18

The domain of f (x) = 1.6x − 2.4 is all real numbers.
SO

1.1.19

t2 −1
LU
The domain of g(t) = ist  | t  −1.
t +1
TI
O

1.1.20
x −1
N
f (x) =
The domain of x2 −1 isx  | x  −1,1.
S