January 27, 2005 11:42
L24ch01
Sheet number 1 Page number 1
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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.