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首页James Stewart 的 Calculus 第五版答案
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1, 2
(
)
f f( 1)= 2
x=2 y 2.8 f(2) 2.8
f(x)=2 y=2 y=2 x= 3 x=1
x y=0 x= 2.5 x=0.3
f x f
3 x 3 3,3 f y f
2 y 3 2,3
x 1 3 y 2 3 f 1,3
4, 2
(
)
f f( 4)= 2 3,4
(
)
g
g(3)=4
x y y f g
2,1
(
)
2,2
(
)
x 2 2
f(x)= 1 y= 1 y= 1 x= 3 x=4
x 0 4 y 3 1 f 0,4
f x f
4 x 4 4,4 f y f
2 y 3 2,3
g 4,3 0.5,4
t,a
(
)
= 12, 85
(
)
17,115
(
)
85 a 115
325 a 485
210 a 200
60 / 2 d
t t|0 t 2
{
}
t
d|0 d 120
{
}
d
N
t t|0 t 24
{
}
t
N|0 N k
{
}
N k
1. (a) The point is on the graph of , so .
(b) When , is about , so .
(c)
is equivalent to . When , we have and .
(d) Reasonable estimates for when are and .
(e) The domain of
consists of all values on the graph of . For this function, the domain is
, or . The range of consists of all values on the graph of . For this function,
the range is , or .
(f) As
increases from to , increases from to . Thus, is increasing on the interval
.
2. (a) The point
is on the graph of , so . The point is on the graph of , so
.
(b) We are looking for the values of
for which the values are equal. The values for and
are equal at the points and , so the desired values of are and .
(c) is equivalent to . When , we have and .
(d) As increases from to , decreases from to . Thus, is decreasing on the interval
.
(e) The domain of
consists of all values on the graph of . For this function, the domain is
, or . The range of consists of all values on the graph of . For this function,
the range is
, or .
(f) The domain of is and the range is .
3. From Figure 1 in the text, the lowest point occurs at about
. The highest point
occurs at about
. Thus, the range of the vertical ground acceleration is . In
Figure 11, the range of the north
south acceleration is approximately . In Figure 12,
the range of the east
west acceleration is approximately .
4. Example 1: A car is driven at
mi h for hours. The distance traveled by the car is a function
of the time . The domain of the function is , where is measured in hours. The range of
the function is , where is measured in miles.
Example 2: At a certain university, the number of students on campus at any time on a particular
day is a function of the time
after midnight. The domain of the function is , where is
measured in hours. The range of the function is , where is an integer and is the
largest number of students on campus at once.
1
Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function
$8.00 30
P h
0,30 0,2.00,4.00,... ,238.00,240.00
{
}
[ 2,2] [ 1,2]
[ 3,2] 3, 2
)
[ 1,3]
x=0 1 2
160 20 10
120 5
170 30 190
30
8 9 10 00
10
1 00 3 00
5 00
5 6 7 00
Example 3: A certain employee is paid per hour and works a maximum of hours per week.
The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s
gross weekly pay is a function of the number of hours worked . The domain of the function is
and the range of the function is .
5. No, the curve is not the graph of a function because a vertical line intersects the curve more than
once. Hence, the curve fails the Vertical Line Test.
6. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is
and the range is .
7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is
and the range is .
8. No, the curve is not the graph of a function since for
, , and , there are infinitely many
points on the curve.
9. The person’s weight increased to about pounds at age and stayed fairly steady for years.
The person’s weight dropped to about
pounds for the next years, then increased rapidly to
about pounds. The next years saw a gradual increase to pounds. Possible reasons for the
drop in weight at years of age: diet, exercise, health problems.
10. The salesman travels away from home from to A.M. and is then stationary until : . The
salesman travels farther away from
until noon. There is no change in his distance from home until
: , at which time the distance from home decreases until : . Then the distance starts
increasing again, reaching the maximum distance away from home at : . There is no change from
until , and then the distance decreases rapidly until : P.M., at which time the salesman
reaches home.
11. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the
water will slowly warm up to room temperature.
2
Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function
12. The summer solstice (the longest day of the year) is around June 21, and the winter solstice (the
shortest day) is around December 22.
13. Of course, this graph depends strongly on the geographical location!
14. The temperature of the pie would increase rapidly, level off to oven temperature, decrease rapidly,
and then level off to room temperature.
15.
16. (a)
(b)
(c)
3
Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function
f(x)=3x
2
x+2.
f(2)=3(2)
2
2+2=12 2+2=12.
f( 2)=3( 2)
2
( 2)+2=12+2+2=16.
(d)
17. (a)
(b) From the graph, we estimate the number of cell phone subscribers in Malaysia to be about 540 in
1994 and 1450 in 1996.
18. (a)
(b) From the graph in part (a), we estimate the temperature at 11:00 A.M. to be about 84.5 C.
19.
4
Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function
f(a)=3a
2
a+2.
f( a)=3( a)
2
( a)+2=3a
2
+a+2.
f(a+1)=3(a+1)
2
(a+1)+2=3(a
2
+2a+1) a 1+2=3a
2
+6a+3 a+1=3a
2
+5a+4.
2f(a)=2 f(a)=2(3a
2
a+2)=6a
2
2a+4.
f(2a)=3(2a)
2
(2a)+2=3(4a
2
) 2a+2=12a
2
2a+2.
f(a
2
)=3(a
2
)
2
(a
2
)+2=3(a
4
) a
2
+2=3a
4
a
2
+2.
f(a)
2
= 3a
2
a+2
2
= 3a
2
a+2
(
)
3a
2
a+2
(
)
=9a
4
3a
3
+6a
2
3a
3
+a
2
2a+6a
2
2a+4=9a
4
6a
3
+13a
2
4a+4.
f(a+h)=3(a+h)
2
(a+h)+2=3(a
2
+2ah+h
2
) a h+2=3a
2
+6ah+3h
2
a h+2.
r+1 V r+1
(
)
=
4
3
r+1
(
)
3
=
4
3
r
3
+3r
2
+3r+1
(
)
r r+1
V r+1
(
)
V r
(
)
=
4
3
r
3
+3r
2
+3r+1
(
)
4
3
r
3
=
4
3
3r
2
+3r+1
(
)
f(x)=x x
2
f(2+h)=2+h (2+h)
2
=2+h (4+4h+h
2
)=2+h 4 4h h
2
= h
2
+3h+2
(
)
f(x+h)=x+h x+h
(
)
2
=x+h (x
2
+2xh+h
2
)=x+h x
2
2xh h
2
f(x+h) f(x)
h
=
x+h x
2
2xh h
2
x+x
2
h
=
h 2xh h
2
h
=
h(1 2x h)
h
=1 2x h
f(x)=
x
x+1
f(2+h)=
2+h
2+h+1
=
2+h
3+h
f(x+h)=
x+h
x+h+1
f(x+h) f(x)
h
=
x+h
x+h+1
x
x+1
h
=
x+h
(
)
x+1
(
)
x x+h+1
(
)
h x+h+1
(
)
x+1
(
)
=
1
x+h+1
(
)
x+1
(
)
f(x)=x/(3x 1) x 0=3x 1 x=
1
3
x R|x
1
3
{
}
= ,
1
3
1
3
,
f(x)=(5x+4) x
2
+3x+2
(
)
/
x
0=x
2
+3x+2
0=(x+2)(x+1) x= 2
1 x R|x 2, 1
{
}
=( , 2) ( 2, 1) ( 1, )
20. A spherical balloon with radius has volume . We
wish to find the amount of air needed to inflate the balloon from a radius of to . Hence, we need
to find the difference .
21. , so ,
, and
.
22. , so , , and
.
23. is defined for all except when , so the domain is
.
24. is defined for all except when or
, so the domain is .
25.
5
Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function
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