This preview shows pages 1–3. Sign up to view the full content.
January 27, 2005 11:42
L24ch01
Sheet number 1 Page number 1
black
1
CHAPTER 1
Functions
EXERCISE SET 1.1
1. (a)
−
2
.
9
,
−
2
.
0
,
2
.
35
,
2
.
9
(b)
none
(c)
y
=0
(d)
−
1
.
75
≤
x
≤
2
.
15
(e)
y
max
=2
.
8 at
x
=
−
2
.
6;
y
min
=
−
2
.
2at
x
=1
.
2
2. (a)
x
=
−
1
,
4
(b)
none
(c)
y
=
−
1
(d)
x
,
3
,
5
(e)
y
max
=9at
x
=6;
y
min
=
−
x
3. (a)
yes
(b)
yes
(c)
no (vertical line test fails)
(d)
no (vertical line test fails)
4. (a)
The natural domain of
f
is
x
6
=
−
1, and for
g
it is the set of all
x
. Whenever
x
6
=
−
1,
f
(
x
)=
g
(
x
), but they have diFerent domains.
(b)
The domain of
f
is the set of all
x
≥
0; the domain of
g
is the same.
5. (a)
around 1943
(b)
1960; 4200
(c)
no; you need the year’s population
(d)
war; marketing techniques
(e)
news of health risk; social pressure, antismoking campaigns, increased taxation
6. (a)
around 1983
(b)
1966
(c)
the former
(d)
no, it appears to be levelling out
7. (a)
1999, $34,400
(b)
1985, $37,000
(c)
second year; graph has a larger (negative) slope
8. (a)
In thousands, approximately
43
.
2
−
37
.
8
6
=
5
.
4
6
per yr, or $900/yr
(b)
The median income during 1993 increased from $37.8K to $38K (K for ’kilodollars’; all ±gures
approximate). During 1996 it increased from $40K to $42K, and during 1999 it decreased
slightly from $43.2K to $43.1K. Thus the average rate of change measured on January 1 was
(40  37.8)/3 for the ±rst threeyr period and (43.2  40)/3 for the secondyear period, and
hence the median income as measured on January 1 increased more rapidly in the second
threeyear period. Measured on December 31, however, the numbers are (42  38)/3 and (43.1
 42)/3, and the former is the greater number. Thus the answer to the question depends on
where in the year the median income is measured.
(c)
1993
9. (a)
f
(0) = 3(0)
2
−
2=
−
2;
f
(2) = 3(2)
2
−
2 = 10;
f
(
−
2)=3(
−
2)
2
−
2 = 10;
f
(3) = 3(3)
2
−
2 = 25;
f
(
√
√
2)
2
−
2=4;
f
(3
t
) = 3(3
t
)
2
−
2=27
t
2
−
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;
f
(3
t
)=1
/
3
t
for
t>
1 and
f
(3
t
)=6
t
for
t
≤
1.
10. (a)
g
(3) =
3+1
3
−
1
;
g
(
−
1) =
−
1+1
−
1
−
1
;
g
(
π
π
+1
π
−
1
;
g
(
−
1
.
1) =
−
1
.
−
1
.
1
−
1
=
−
0
.
1
−
2
.
1
=
1
21
;
g
(
t
2
−
1) =
t
2
−
t
2
−
1
−
1
=
t
2
t
2
−
2
(b)
g
(3) =
√
3+1 = 2;
g
(
−
1
)=3
;
g
(
π
√
π
+1;
g
(
−
1
.
1
;
g
(
t
2
−
1
)=3i
f
t
2
<
2 and
g
(
t
2
−
1) =
√
t
2
−
1+1=

t

if
t
2
≥
2.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentJanuary 27, 2005 11:42
L24ch01
Sheet number 2 Page number 2
black
2
Chapter 1
11. (a)
x
6
=3
(b)
x
≤−
√
3or
x
≥
√
3
(c)
x
2
−
2
x
+ 5 = 0 has no real solutions so
x
2
−
2
x
+ 5 is always positive or always negative. If
x
= 0, then
x
2
−
2
x
+5=5
>
0; domain: (
−∞
,
+
∞
).
(d)
x
6
=0
(e)
sin
x
6
=1,so
x
6
=(2
n
+
1
2
)
π
,
n
,
±
1
,
±
2
,...
12. (a)
x
6
=
−
7
5
(b)
x
−
3
x
2
must be nonnegative;
y
=
x
−
3
x
2
is a parabola that crosses the
x
axis at
x
,
1
3
and opens downward, thus 0
≤
x
≤
1
3
(c)
x
2
−
4
x
−
4
>
0, so
x
2
−
4
>
0 and
x
−
4
>
0, thus
x>
4; or
x
2
−
4
<
0 and
x
−
4
<
0, thus
−
2
<x<
2
(d)
x
6
=
−
1
(e)
cos
x
≤
1
<
2, 2
−
cos
0, all
x
13. (a)
x
≤
3
(b)
−
2
≤
x
≤
2
(c)
x
≥
0
(d)
all
x
(e)
all
x
14. (a)
x
≥
2
3
(b)
−
3
2
≤
x
≤
3
2
(c)
x
≥
0
(d)
x
6
(e)
x
≥
0
15. (a)
Breaks could be caused by war, pestilence, ﬂood, earthquakes, for example.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 RichardWhite

Click to edit the document details