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Test Bank For James stewart daniel k clegg saleem watson instructor s solutions manuals for calculus early transcendentals 9th edition All Chapters Included
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  • Student Solutions Manual for Stewart/Clegg/Watson\'s Calculus: Early Transcendentals, 9th (1-11)

Four Ways to Represent a Function 1. The functions () =  + √2 −  and () =  + √2 −  give exactly the same output values for every input value, so  and  are equal. 2. () = 2 −   − 1 = ( − 1)  − 1 =  for  − 1 6= 0, s...

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  • SOLUTION MANUALS FOR CALCULUS EARLY TRANSCENDENTALS 9th EDITION BY JAMES STEWART, DANIEL K. CLEGG & SALEEM WATSON
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1 FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function

√ √
1. The functions  () =  + 2 −  and () =  + 2 −  give exactly the same output values for every input value, so 
and  are equal.

2 −  ( − 1)
2.  () = = =  for  − 1 6= 0, so  and  [where () = ] are not equal because  (1) is undefined and
−1 −1
(1) = 1.

3. (a) The point (−2 2) lies on the graph of , so (−2) = 2. Similarly,  (0) = −2,  (2) = 1, and  (3)  25.

(b) Only the point (−4 3) on the graph has a ­value of 3, so the only value of  for which () = 3 is −4.

(c) The function outputs () are never greater than 3, so () ≤ 3 for the entire domain of the function. Thus, () ≤ 3 for
−4 ≤  ≤ 4 (or, equivalently, on the interval [−4 4]).

(d) The domain consists of all ­values on the graph of : { | −4 ≤  ≤ 4} = [−4 4]. The range of  consists of all the
­values on the graph of : { | −2 ≤  ≤ 3} = [−2 3].

(e) For any 1  2 in the interval [0 2], we have (1 )  (2 ). [The graph rises from (0 −2) to (2 1).] Thus, () is
increasing on [0 2].

4. (a) From the graph, we have  (−4) = −2 and (3) = 4.

(b) Since  (−3) = −1 and (−3) = 2, or by observing that the graph of  is above the graph of  at  = −3, (−3) is larger
than  (−3).

(c) The graphs of  and  intersect at  = −2 and  = 2, so  () = () at these two values of .

(d) The graph of  lies below or on the graph of  for −4 ≤  ≤ −2 and for 2 ≤  ≤ 3. Thus, the intervals on which
 () ≤ () are [−4 −2] and [2 3].

(e)  () = −1 is equivalent to  = −1, and the points on the graph of  with ­values of −1 are (−3 −1) and (4 −1), so
the solution of the equation  () = −1 is  = −3 or  = 4.

(f) For any 1  2 in the interval [−4 0], we have (1 )  (2 ). Thus, () is decreasing on [−4 0].

(g) The domain of  is { | −4 ≤  ≤ 4} = [−4 4]. The range of  is { | −2 ≤  ≤ 3} = [−2 3].

(h) The domain of  is { | −4 ≤  ≤ 3} = [−4 3]. Estimating the lowest point of the graph of  as having coordinates
(0 05), the range of  is approximately { | 05 ≤  ≤ 4} = [05 4].

5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 −85). The highest point occurs at about (17 115).

Thus, the range of the vertical ground acceleration is −85 ≤  ≤ 115. Written in interval notation, the range is [−85 115].


c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
° 9

,10 ¤ CHAPTER 1 FUNCTIONS AND MODELS

6. Example 1: A car is driven at 60 mih for 2 hours. The distance 
traveled by the car is a function of the time . The domain of the
function is { | 0 ≤  ≤ 2}, where  is measured in hours. The range
of the function is { | 0 ≤  ≤ 120}, 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 { | 0 ≤  ≤ 24}, where  is
measured in hours. The range of the function is { | 0 ≤  ≤ },
where  is an integer and  is the largest number of students on
campus at once.

Example 3: A certain employee is paid $800 per hour and works a pay

maximum of 30 hours per week. The number of hours worked is 240
238
rounded down to the nearest quarter of an hour. This employee’s 236

gross weekly pay  is a function of the number of hours worked .
4
The domain of the function is [0 30] and the range of the function is 2
{0 200 400     23800 24000}. 0 0.25 0.50 0.75 29.50 29.75 30 hours


7. We solve 3 − 5 = 7 for : 3 − 5 = 7 ⇔ −5 = −3 + 7 ⇔  = 35  − 75 . Since the equation determines exactly

one value of  for each value of , the equation defines  as a function of .

8. We solve 32 − 2 = 5 for : 32 − 2 = 5 ⇔ −2 = −32 + 5 ⇔  = 32 2 − 52 . Since the equation determines

exactly one value of  for each value of , the equation defines  as a function of .
√
9. We solve 2 + ( − 3)2 = 5 for : 2 + ( − 3)2 = 5 ⇔ ( − 3)2 = 5 − 2 ⇔ ⇔  − 3 = ± 5 − 2
√
 = 3 ± 5 − 2 . Some input values  correspond to more than one output . (For instance,  = 1 corresponds to  = 1 and
to  = 5.) Thus, the equation does not define  as a function of .

10. We solve 2 + 5 2 = 4 for : 2 + 5 2 = 4 ⇔ 5 2 + (2)  − 4 = 0 ⇔

√ √
−2 ± (2)2 − 4(5)(−4) −2 ± 42 + 80 − ± 2 + 20
= = = (using the quadratic formula). Some input
2(5) 10 5

values  correspond to more than one output . (For instance,  = 4 corresponds to  = −2 and to  = 25.) Thus, the

equation does not define  as a function of .
√
11. We solve ( + 3)3 + 1 = 2 for : ( + 3)3 + 1 = 2 ⇔ ( + 3)3 = 2 − 1 ⇔  + 3 = 3
2 − 1 ⇔
√
 = −3 + 3
2 − 1. Since the equation determines exactly one value of  for each value of , the equation defines  as a
function of .



c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°

, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11

12. We solve 2 − || = 0 for : 2 − || = 0 ⇔ || = 2 ⇔  = ±2. Some input values  correspond to more than

one output . (For instance,  = 1 corresponds to  = −2 and to  = 2.) Thus, the equation does not define  as a function
of .

13. The height 60 in ( = 60) corresponds to shoe sizes 7 and 8 ( = 7 and  = 8). Since an input value  corresponds to more

than output value , the table does not define  as a function of .

14. Each year  corresponds to exactly one tuition cost . Thus, the table defines  as a function of .

15. 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.

16. 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].

17. 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].

18. No, the curve is not the graph of a function since for  = 0, ±1, and ±2, there are infinitely many points on the curve.

19. (a) When  = 1950,  ≈ 138◦ C, so the global average temperature in 1950 was about 138◦ C.

(b) When  = 142◦ C,  ≈ 1990.

(c) The global average temperature was smallest in 1910 (the year corresponding to the lowest point on the graph) and largest
in 2000 (the year corresponding to the highest point on the graph).

(d) When  = 1910,  ≈ 135◦ C, and when  = 2000,  ≈ 144◦ C. Thus, the range of  is about [135, 144].

20. (a) The ring width varies from near 0 mm to about 16 mm, so the range of the ring width function is approximately [0 16].

(b) According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the
late 1800s, and has been steadily warming since then. In the mid­19th century, there was variation that could have been
associated with volcanic eruptions.

21. 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.




22. The temperature of the pie would increase rapidly, level off to oven

temperature, decrease rapidly, and then level off to room temperature.




c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°

, 12 ¤ CHAPTER 1 FUNCTIONS AND MODELS

23. (a) The power consumption at 6 AM is 500 MW which is obtained by reading the value of power  when  = 6 from the

graph. At 6 PM we read the value of  when  = 18 obtaining approximately 730 MW

(b) The minimum power consumption is determined by finding the time for the lowest point on the graph,  = 4 or 4 AM. The
maximum power consumption corresponds to the highest point on the graph, which occurs just before  = 12 or right
before noon. These times are reasonable, considering the power consumption schedules of most individuals and
businesses.

24. Runner A won the race, reaching the finish line at 100 meters in about 15 seconds, followed by runner B with a time of about

19 seconds, and then by runner C who finished in around 23 seconds. B initially led the race, followed by C, and then A.
C then passed B to lead for a while. Then A passed first B, and then passed C to take the lead and finish first. Finally,
B passed C to finish in second place. All three runners completed the race.


25. Of course, this graph depends strongly on the 26. The summer solstice (the longest day of the year) is
geographical location! around June 21, and the winter solstice (the shortest day)
is around December 22. (Exchange the dates for the
southern hemisphere.)




27. As the price increases, the amount sold decreases. 28. The value of the car decreases fairly rapidly initially, then
somewhat less rapidly.




29.




c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°

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