Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function

1. (a) The point ( 1, 2 ) is on the graph of f , so f ( 1)= 2 .
(b) When x=2 , y is about 2.8 , so f (2) 2.8 .
(c) f (x)=2 is equivalent to y=2 . When y=2 , we have x= 3 and x=1 .
(d) Reasonable estimates for x when y=0 are x= 2.5 and x=0.3 .
(e) The domain of f consists of all x values on the graph of f . For this function, the domain is
3 x 3 , or 3,3 . The range of f consists of all y values on the graph of f . For this function, the range is 2 y 3 , or 2,3 .
(f) As x increases from 1 to 3 , y increases from 2 to 3 . Thus, f is increasing on the interval 1,3
.
2. (a) The point ( 4, 2 ) is on the graph of f , so f ( 4)= 2 . The point ( 3,4 ) is on the graph of g , so g(3)=4 .
(b) We are looking for the values of x for which the y values are equal. The y values for f and g are equal at the points ( 2,1 ) and ( 2,2 ) , so the desired values of x are 2 and 2 .
(c) f (x)= 1 is equivalent to y= 1 . When y= 1 , we have x= 3 and x=4 .
(d) As x increases from 0 to 4 , y decreases from 3 to 1 . Thus, f is decreasing on the interval 0,4
.
(e) The domain of f consists of all x values on the graph of f . For this function, the domain is
4 x 4 , or 4,4 . The range of f consists of all y values on the graph of f . For this function, the range is 2 y 3 , or 2,3 .
(f) The domain of g is 4,3 and the range is 0.5,4 .
3. From Figure 1 in the text, the lowest point occurs at about ( t,a ) = ( 12, 85) . The highest point occurs at about ( 17,115) . Thus, the range of the vertical ground acceleration is 85 a 115 . In
Figure 11, the range of the north south acceleration is approximately 325 a 485 . In Figure 12, the range of the east west acceleration is approximately 210 a 200 .
4. Example 1: A car is driven at 60 mi / h for 2 hours. The distance d traveled by the car is a function of the time t . The domain of the function is {t |0 t 2} , where t is measured in hours. The range of the function is { d |0 d 120} , where d is measured in miles.

Example 2: At a certain university, the number of students N on campus at any time on a particular day is a function of the time t after midnight. The domain of the function is {t |0 t 24} , where t is measured in hours. The range of the function is { N |0 N k} , where N is an integer and k 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

Example 3: A certain employee is paid $8.00 per hour and works a maximum of 30 hours per week.
The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s gross weekly pay P is a function of the number of hours worked h . The domain of the function is
0,30 and the range of the function is { 0,2.00,4.00,... ,238.00,240.00} .

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
[ 2,2] and the range is [ 1,2] .
7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is
[ 3,2] and the range is 3, 2 ) [ 1,3] .
8. No, the curve is not the graph of a function since for x=0 , points on the curve.

1 , and

2 , there are infinitely many

9. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years.
The person’s weight dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.
10. The salesman travels away from home from 8 to 9 A.M. and is then stationary until 10 : 00 . The salesman travels farther away from 10 until noon. There is no change in his distance from home until
1 : 00 , at which time the distance from home decreases until 3 : 00 . Then the distance starts increasing again, reaching the maximum distance away from home at 5 : 00 . There is no change from
5 until 6 , and then the distance decreases rapidly until 7 : 00 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

(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.
2

19. f (x)=3x x+2.
2

f (2)=3(2) 2+2=12 2+2=12.
2

f ( 2)=3( 2) ( 2)+2=12+2+2=16.
4

Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function

2

f (a)=3a a+2.
2

2

f ( a)=3( a) ( a)+2=3a +a+2.
2

2

2

2

f (a+1)=3(a+1) (a+1)+2=3(a +2a+1) a 1+2=3a +6a+3 a+1=3a +5a+4.
2

2

2 f (a)=2 f (a)=2(3a a+2)=6a 2a+4.
2

2

2

f (2a)=3(2a) (2a)+2=3(4a ) 2a+2=12a 2a+2.
2

22

2

4

2

4

2

f (a )=3(a ) (a )+2=3(a ) a +2=3a a +2. f (a)

2

2

2

(

2

= 3a a+2 = 3a a+2
4

3

2

3

) ( 3a2 a+2)

2

2

4

3

2

=9a 3a +6a 3a +a 2a+6a 2a+4=9a 6a +13a 4a+4.
2

2

2

2

2

f (a+h)=3(a+h) (a+h)+2=3(a +2ah+h ) a h+2=3a +6ah+3h a h+2.
4
3
2
3 4 r +3r +3r+1 . We
( r+1 ) =
3
3 wish to find the amount of air needed to inflate the balloon from a radius of r to r+1 . Hence, we need
4
4 3 4
3
2
2
r +3r +3r+1 r =
3r +3r+1 . to find the difference V ( r+1 ) V ( r ) =
3
3
3

(

20. A spherical balloon with radius r+1 has volume V ( r+1 ) =

(

2

)

2

(

2

2

21. f (x)=x x , so f (2+h)=2+h (2+h) =2+h (4+4h+h )=2+h 4 4h h =
2

2

2

2

)

)

( h2+3h+2) ,

2

f (x+h)=x+h ( x+h ) =x+h (x +2xh+h )=x+h x 2xh h , and
2

2

2

2

h 2xh h h(1 2x h) f (x+h) f (x) x+h x 2xh h x+x
=
=
=
=1 2x h . h h h h x 2+h
2+h
x+h
=
, so f (2+h)=
, f (x+h)=
, and x+1 2+h+1 3+h x+h+1 x x+h f (x+h) f (x) x+h+1 x+1
1
( x+h ) ( x+1 ) x ( x+h+1 )
=
=
=
. h h h ( x+h+1 ) ( x+1 )
( x+h+1 ) ( x+1 )

22. f (x)=

23. f (x)=x/(3x 1) is defined for all x except when 0=3x 1

{

x R| x

1
3

}

=

/(

,
2

1
,
3

1
3

)

x=

1
, so the domain is
3

.
2

24. f (x)=(5x+4) x +3x+2 is defined for all x except when 0=x +3x+2
, 2) ( 2, 1) ( 1, ) .
1 , so the domain is { x R| x 2, 1} =(

0=(x+2)(x+1)

x= 2 or

25.
5

Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function

3

f (t)= t + t is defined when t

0 . These values of t give real number results for t , whereas any

value of t gives a real number result for

3

t . The domain is 0,

26. g(u)= u + 4 u is defined when u 0 and 4 u 0

/

4

2

).

u 4 . Thus, the domain is 0 u 4= 0,4 .

2

2

x 5x is defined when x 5x>0 x(x 5)>0 . Note that x 5x 0 since that would
27. h(x)=1 result in division by zero. The expression x(x 5) is positive if x5 . (See Appendix A for
,0) (5, ) . methods for solving inequalities.) Thus, the domain is (
2

2

28. h(x)= 4 x . Now y= 4 x

2

2

y =4 x

2

2

x +y =4 , so the graph is the top half of a circle of

radius 2 with center at the origin. The domain is
From the graph, the range is 0 y 2 , or 0,2 .

{ x|4 x2 0} = { x|4

29. f (x)=5 is defined for all real numbers, so the domain is R , or ( horizontal line with y intercept 5 .

,

1
(x+3) is defined for all real numbers, so the domain is R , or (
2
3 a line with x intercept 3 and y intercept
.
2

2

x } =[ 2,2] .

) . The graph of f is a

30. F(x)=

31. f (t)=t 6t is defined for all real numbers, so the domain is R , or (

} ={ x|2

2

x

,

,

) . The graph of F is

) . The graph of f is a

2

parabola opening upward since the coefficient of t is positive. To find the t intercepts, let y=0 and
2

solve for t . 0=t 6t=t(t 6)

t=0 and t=6 . The t coordinate of the vertex is halfway between the t
6

Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function

2

intercepts, that is, at t=3 . Since f (3)=3 6 3= 9 , the vertex is (3, 9) .

2

4 t
(2+t)(2 t)
=
32. H(t)=
, so for t 2 , H(t)=2+t . The domain is {t |t 2} . So the graph of H is
2 t
2 t the same as the graph of the function f (t)=t+2 (a line) except for the hole at ( 2,4 ) .

33. g(x)= x 5 is defined when x 5 0 or x 5 , so the domain is 5,
2

y =x 5

) . Since y= x 5

2

x=y +5 , we see that g is the top half of a parabola.

34.
F(x)= 2x+1 =

=

{
{

2x+1
(2x+1)

if 2x+1 0 if 2x+12x+ x x<
30 x
. Hence, the domain of A is 00 20 2x>0 x0 12 2x>0 x>0 . Combining these restrictions gives us the domain 00 .

31. The function f is one to one, so its inverse exists and the graph of its inverse can be obtained by reflecting the graph of f about the line y=x .

4

Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms

32. The function f is one to one, so its inverse exists and the graph of its inverse can be obtained by reflecting the graph of f about the line y=x . For the graph of 1/ f , the y coordinates are simply the
1
reciprocals of f . For example, if f (5)=9, then 1/ f (5)= . If we draw the horizontal line y=1 , we see
9
that the only place where the graphs intersect is on that line.

33. (a) It is defined as the inverse of the exponential function with base a , that is, log x=y a y

a =x .

(b) ( 0, )
(c)
(d) See Figure .
34. (a) The natural logarithm is the logarithm with base e , denoted ln x .
(b) The common logarithm is the logarithm with base 10 , denoted log x .
(c) See Figure .
6

35. (a) log 64=6 since 2 =64 .
2

1
2 1
(b) log
.
= 2 since 6 =
6 36
36
36. (a) log 2=
8

(b) ln e

2

1
1/3
since 8 =2 .
3

= 2

37. (a) log 1.25+log 80=log
10

10

2

( 1.25 80 ) =log 10100=log 1010 =2
10
3

(b) log 10+log 20 3log 2=log ( 10 20 ) log 2 =log
5
5
5
5
5
5

200
2
=log 25=log 5 =2
5
5
8
5

Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms

( log 23+log 25) =2log 215=15 [ Or: 2 ( log 23+log 25) =2log 23 2log 25=3 5=15 ]

38. (a) 2
(b) e

3ln 2

3

=e

ln (2 )

=e

ln 8

=8 [ Or: e

3ln 2

( ln 2) 3=23=8 ]

= e

2

39. 2ln 4 ln 2=ln 4 ln 2=ln 16 ln 2=ln a 16
=ln 8
2

b

a

b

a b

40. ln x+aln y bln z=ln x+ln y ln z =ln (x y ) ln z =ln (xy /z )
2

(1+x ) x
1
2
1/2
2
41. ln (1+x )+ ln x ln sin x=ln (1+x )+ln x ln sin x=ln [(1+x ) x ] ln sin x=ln
2
sin x
2

ln 10
0.926628
12 ln 12 ln 8.4
(b) log 8.4=
3.070389
2 ln 2
42. (a) log 10=

43. To graph these functions, we use log

1.5

x=

ln x ln x and log x=
. These graphs all approach
50
ln 1.5 ln 50

+

as x 0 , and they all pass through the point ( 1,0 ) . Also, they are all increasing, and all approach as x
. The functions with larger bases increase extremely slowly, and the ones with smaller bases do so somewhat more quickly. The functions with large bases approach the y axis more closely as x

+

0 .

x

44. We see that the graph of ln x is the reflection of the graph of e about the line y=x , and that the graph of log

x

10

x

x is the reflection of the graph of 10 about the same line. The graph of 10 increases x more quickly than that of e . Also note that log

10

x

as x

more slowly than ln x .

6

Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms

45. 3 ft =36 in, so we need x such that log x=36
2

68 , 719 , 476 , 736 in

1ft
1mi
12in 5280ft

36

x=2 =68 , 719 , 476 , 736 . In miles, this is

1 , 084 , 587.7 mi.

46.

0.1

From the graphs, we see that f (x)=x >g(x)=ln x for approximately 0a and a a

2k 2

2k

. Then 1+a

>1+a

2k 2

2k

2n

1
1+a

2k 2

1+

1
1+a

1+

2k 2

1+

2k 1

1
1+a

a

2n 1

<

1
1+a

2k

1
1+a

2k+1

. We have thus shown, by induction, that the odd terms are

2k+1

increasing and the even terms are decreasing. Also all terms lie between 1 and 2 , so both

{b } n 2n+1

{a } n and

are bounded monotonic sequences and are therefore convergent by Theorem 11. Let

4+3a
1
1 n lim a =L . Then lim a
=L also. We have a =1+
=1+
=
,
2n
2n+2
n+2
1+1+1/(1+a )
(3+2a )/(1+a ) 3+2a n n n n n n

so
14

Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences

4+3a a 2n

=

2n+2

4+3L
3+2L

. Taking limits of both sides, we get L=

3+2a

2n

2

(since L>0 ). Thus, lim a = 2 . Similarly we find that lim a
2n

n

{p }

73. (a) Suppose

n

p

=

n+1

lim p

n

n

n

=

n+1

n

L= 2

= 2 . So, by part (a), lim a = 2 .

2n+1

n

blim p

a+ p

L =2

n

n

converges to p . Then

bp

n

2

3L+2L =4+3L

p=

a+lim p

n

n

bp a+ p

2

p +ap=bp

p( p+a b)=0

b p p a n b n n
=
< p since 1+
>1 .
(b) p = n+1 a+ p a n a p n n
1+
a b b b 2 p ,p< p< p ,p<
(c) By part (b), p <
1
0
2
1
0
3 a a a n n b b p< p , so lim p lim p =0 since bb a and p is decreasing. In either case the sequence
0

{p } n n

{ } n is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem. It then

follows from part (a) that lim p =b a . n n

15

Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series

n

1. Using Theorem 5 with

b (x 5) , b =

n=0 n

f

n

( n)

( 8) f (5)
(a)
, so b =
.
8 n! 8!
/

2. (a) Using Formula 6, a power series expansion of f at 1 must have the form f (1)+ f (1)(x 1)+
/

. Comparing to the given series, 1.6 0.8(x 1)+

, we must have f (1)= 0.8 . But from the graph,

/

f (1) is positive. Hence, the given series is not the Taylor series of f centered at 1 .
1 / /
/
2
(b) A power series expansion of f at 2 must have the form f (2)+ f (2)(x 2)+ f (2)(x 2) +
.
2
1 / /
2
3
Comparing to the given series, 2.8+0.5(x 2)+1.5(x 2) 0.1(x 2) +
, we must have f (2)=1.5
2
/ /

; that is, f (2) is positive. But from the graph, f is concave downward near x=2 , so f be negative. Hence, the given series is not the Taylor series of f centered at 2 .

/ /

(2) must

3. n 0
1
2
3
4

( n) f (x) cos x sin x cos x sin x cos x

f

( n)

( 0)

1
0
1
0
1

We use Equation 7 with f (x)=cos x . cos x = f (0)+ f /(0)x+ f
2

( 3)
( 4)
(0) 2 f (0) 3 f ( 0 ) 4 x+ x+ x+ 2!
3!
4!

/ /

4

x x =1
+
2! 4!

n 2n

=

( 1) x n=0 (2n)!

n 2n

( 1) x
If a = n ( 2n )! a lim n , then

2n+2

n+1

a

n

=lim n x
(2n)!
2n
(2n+2)!
x

2

=x lim n 1

( 2n+2 ) ( 2n+1 )

=0