math 115 — practice for exam 1mconger/dhsp/115packetf18solns.pdf · math 115 — practice for...

Math 115 — Practice for Exam 1

Generated October 7, 2018

Name: SOLUTIONS

Instructor: Section Number:

1. This exam has 16 questions. Note that the problems are not of equal difficulty, so you may want toskip over and return to a problem on which you are stuck.

2. Do not separate the pages of the exam. If any pages do become separated, write your name on themand point them out to your instructor when you hand in the exam.

3. Please read the instructions for each individual exercise carefully. One of the skills being tested onthis exam is your ability to interpret questions, so instructors will not answer questions about examproblems during the exam.

4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that thegraders can see not only the answer but also how you obtained it. Include units in your answers whereappropriate.

5. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).However, you must show work for any calculation which we have learned how to do in this course. Youare also allowed two sides of a 3′′ × 5′′ note card.

6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of thegraph, and to write out the entries of the table that you use.

7. You must use the methods learned in this course to solve all problems.

Semester Exam Problem Name Points Score

Winter 2017 1 7 cats 10

Fall 2016 1 5 deep temp 10

Winter 2007 1 9 Big Bang 8

Fall 2015 1 6 6

Winter 2015 1 10 10

Fall 2017 1 2 globe 13

Fall 2017 1 7 peanut butter 13

Fall 2016 1 10 8

Fall 2010 1 8 apples 18

Fall 2016 1 4 10

Fall 2016 1 1 coasters 10

Fall 2010 1 7 refrigerator 14

Winter 2018 1 8 15

Fall 2015 1 8 12

Winter 2016 1 1 10

Winter 2002 1 5 10

Total 177

Recommended time (based on points): 159 minutes

Math 115 / Exam 1 (February 8, 2017) page 9

7. [10 points] Two housecats, Jasper and Zander, escape from their house at the same time andtravel along a straight line between their house and a tree. Let J(t) (respectively Z(t)) beJasper’s (respectively Zander’s) distance, in feet, from the tree t seconds after escaping. Thetable below shows some of the values of J(t) and Z(t). Assume that J(t) is invertible.

t 6 17 22 31 37

J(t) 41 33 21 14 2

Z(t) 39 32 31 36 43

a. [2 points] What is Jasper’s average velocity for 6 ≤ t ≤ 22? Be sure to include units.

Answer:21−41

22−6= −

20

16= −

5

4= −1.25 ft/sec.

b. [2 points] Estimate Z ′(31). Remember to show your work.

You can approximate it using either:

• Average rate of change in [31, 37]:43− 36

37− 31= 7/6.


31− 22= 5/9.

• Averaging these two: 1

2(7/6 + 5/9) = 0.861, or


37− 33= 0.8.

Answer: (picking one) 7/6

c. [3 points] Circle the one statement below that is best supported by the equation

Z(J−1(8)− 4) = 34.

i. 34 seconds after escaping, Zander is 4 feet closer to the tree than Jasper was 8seconds after escaping.

ii. Four seconds before Jasper is 8 feet from the tree, Zander is 34 feet from the tree.

iii. When Jasper is 4 feet further from the tree than he was 8 seconds after escaping,Zander is 34 feet from the tree.

iv. When Jasper is 4 feet closer to the tree than he was 8 seconds after escaping, Zanderis 34 feet from the tree.

v. Four seconds after Jasper is 8 feet from the tree, Zander is 34 feet from the tree.

d. [3 points] Circle the one statement below that is best supported by the equation

(J−1)′(3) = −0.2.

i. In the third second after leaving the house, Jasper travels about 0.2 feet.

ii. When Jasper is 3 feet from the tree, he is traveling about 0.2 feet/second slowerthan he was one foot earlier.

iii. Jasper gets about 1.5 feet closer to the tree during the third second after leavingthe house.

iv. It takes Jasper about one-tenth of a second to go from 3 feet to 2.5 feet from the tree.

v. One-half of a second before Jasper was 3 feet from the tree, he was about 2.9 feetfrom the tree.

University of Michigan Department of Mathematics Winter, 2017 Math 115 Exam 1 Problem 7 (cats) Solution

Math 115 / Exam 1 (October 10, 2016) page 7

5. [10 points] Scientists bore a hole deep into the earth and lower an instrument to record thetemperature. As the instrument goes deeper, the temperature it records increases. Let T =g(w) be the temperature, in degrees Celsius, the instrument records when it is w hectometersbelow the surface of the earth. (Recall that 1 hectometer is 100 meters.) Assume that thefunction g is invertible and that the functions g and g−1 are continuous and differentiable.

a. [3 points] Using a complete sentence, give a practical interpretation of the equationg−1(68) = 49 in the context of this problem. Be sure to include units.

Solution: When the instrument records a temperature of 68◦C, the instrument is 4.9kilometers below the surface of the earth.

b. [4 points] Below is the first part of a sentence that will give a practical interpretation ofthe equation g′(13) = 0.6 in the context of this problem. Complete the sentence so thatthe practical interpretation can be understood by someone who knows no calculus. Besure to include units in your answer.When the instrument is lowered from 1300 meters to 1320 meters below the surface of the

earth, the temperature it records . . .

Solution: increases by approximately 0.12 Celsius degrees.

c. [3 points] Circle the one statement below that is best supported by the equation

(g−1)′(56) = 0.4.

i. The temperature recorded by the instrument is 56◦C when it is about 0.4 hectometersbelow the surface of the earth.

ii. The temperature recorded by the instrument increases from 56◦C to 56.4◦C when theinstrument is lowered approximately one more hectometer.

iii. When the instrument is lowered from 55.9 hectometers to 56 hectometers below thesurface of the earth, it detects an increase in temperature of about 0.04 Celsius degrees.

iv. The temperature recorded by the instrument increases from 56◦C and 57◦C when theinstrument is lowered about 1

0.4(= 2.5) hectometers further.

v.As the temperature recorded by the instrument increases from 55.9◦C to 56◦C, theinstrument is lowered about 4 meters further beneath the surface of the earth.

vi. When the instrument is 56 hectometers below the surface of the earth, the recordedtemperature is increasing at a rate of 0.04 Celsius degrees per meter.

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 1 Problem 5 (deep temp) Solution

9

9. (8 points) Cosmologists, through a technique best described as hocus pocus, measure a quantityT (t), the temperature of the universe in degrees Kelvin (K), where t is in gigayears (Gyr) afterthe Big Bang. Suppose that, currently, t = 13.6, T (13.6) = 2.4, and T ′(13.6) = −12.

[Note: A gigayear is 1 billion years, and the Kelvin temperature scale is an absolute temperaturescale where the lowest possible temperature is defined as being zero Kelvin.]

(a) For each of the following statements, state whether you agree or disagree with the conclu-sion and justify your reasoning.

(i) In the next billion years, the temperature of the universe will drop by approximately 12degrees Kelvin.

It is not possible for the temperature of the universe to drop by 12 Kelvin in the nextbillion years because the current temperature is 2.4 Kelvin and the Kelvin temperaturescale is absolute (the lowest possible temperature is 0 Kelvin).

(ii) In the next year, the temperature of the universe will drop by approximately 12

1,000,000,000

degrees Kelvin.

It is reasonable to say that the temperature of the universe will drop by approximately12

1,000,000,000Kelvin in the next year since T ′(13.6) = −12 so as time increases by one

billionth of a Gyr from the current time (t = 13.6 Gyr) the temperature should decreaseby twelve billionths Kelvin.

(b) Assume T (t) is decreasing and does not change concavity on the domain [13.6,∞). Do youexpect T (t) to be concave up or concave down on the domain [13.6,∞)? Justify your answerusing physical reasoning.

The function T (t) should be concave up since as the universe expands, the temperature ofthe universe decreases and T (t) should asymptotically approach zero.

University of Michigan Department of Mathematics Winter, 2007 Math 115 Exam 1 Problem 9 (Big Bang) Solution


5. [8 points] Remember to show your work carefully throughout this problem.

Algie and Cal go on a picnic, arriving at 12:00 noon.

a. [5 points] Five minutes after they arrive, they notice that 5 ants have joined their picnic.More ants soon appear, and after careful study, they determine that the number of antsappears to be increasing by 20% every minute. Find a formula for a function A(t) modelingthe number of ants present at the picnic t minutes past noon for t ≥ 5.

Solution: Since this is an exponential function, there are constants c and b such thatA(t) = cbt. We can see immediately that b = 1.2. We can then use the fact that weknow that A(5) = 5 to find c: c(1.2)5 = 5, so c = 5/(1.2)5, which is approximately 2.01.Alternatively, we can use a horizontal shift to say that this is 5(1.2)t−5.

Answer: A(t) =5(1.2)(t−5) = 5

1.25(1.2)t

b. [3 points] Algie and Cal notice that their food is, unfortunately, also attracting flies.The number of flies at their picnic t minutes after noon can be modeled by the functiong(t) = 1.8(1.25)t. Algie and Cal decide they will end their picnic when there are at least1000 flies. How long will their picnic last? Include units.

Solution: We wish to find t such that 1.8(1.25)t = 1000. Then

1.8(1.25)t = 1000

ln(1.8(1.25)t) = ln(1000)

ln(1.8) + t ln(1.25) = ln(1000)

t ln(1.25) = ln(1000)− ln(1.8)

t = ln(1000/1.8)/ ln(1.25) ≈ 28.3.

So they end their picnic about 28.3 minutes after noon (when it started).

Answer: About 28.3 minutes

6. [6 points] Consider the function

R(w) = 2 + (ln(w))cos(w).

Use the limit definition of the derivative to write an explicit expression for R′(π).Your answer should not involve the letter R. Do not attempt to evaluate or simplify the limit.

Please write your final answer in the answer box provided below.

Answer: R′(π) = limh→0

(

2 + (ln(π + h))cos(π+h))

−(

2 + (ln(π))cos(π))

h

University of Michigan Department of Mathematics Fall, 2015 Math 115 Exam 1 Problem 6 Solution


10. [10 points] On the axes provided below, sketch the graph of a single function y = h(x)satisfying all of the following:

• h(x) is defined for all x in the interval −6 < x < 6.

• h′(x) < 0 for all x < −3.

• limx→−2+

h(x) = −1.

• h′(0) = 0.

• The average rate of change of h(x) between x = −1 and x = 2 is 1.

• h(x) is not continuous at x = 3.

• h(x) > 0 for all x > 3.

• h′(x) > 0 for all x > 4.

Make sure that your sketch is large and unambiguous.

1

2

3

4

5

−1

−2

−3

−4

−5

1 2 3 4 5 6−1−2−3−4−5−6

y

x

bc

b

b

b

b

University of Michigan Department of Mathematics Winter, 2015 Math 115 Exam 1 Problem 10 Solution

2. [13 points] After Blizzard left Arizona, Gabe the mouse found a large globe (a sphere) to climb.The globe has a diameter of 40 inches and it is attached to a 12-inch-long pole. Gabe starts at thebase of the pole at point P . He climbs up to the bottom of the globe at point Q. He then climbs theglobe along a semicircle until he stops at the top of the globe at point R (see the diagram below).Note that the diagram is not drawn to scale.

Complete path:

P

Q

R

12 in•

•

•

P

Q

Gabe

•

•

•

G(t)

a. [8 points] Assume that Gabe walks through the path at a velocity of 3 inches per second. LetG(t) be Gabe’s height above the ground (in inches) t seconds after he started his climb at pointP . Find a piecewise-defined formula for G(t). Be sure to include the domain for each piece.

Solution: From point P to Q: It takes the ant 4 seconds to climb 12 inches at a velocity of 3inches per second. During that time, the ant climbs at a constant rate of 3 inches per secondsstarting at the floor, hence G(t) = 3t for 0 ≤ t ≤ 4.

From point Q to R: The distance L along the semicircle traveled by the ant is L = 1

2(2πR),

where R is the radius of the circle. In this case R = 20 inches, then L = 20π. Hence it takes theant T = L

3= 20π

3seconds to go from point Q to R a t a velocity of 3 inches per second. Its height

is given by a sinusoidal function with midline at k = 12+ 20 = 32, amplitude A = 1

2(40) = 20,

period P = 2T = 40π

3and a minimum at (4, 12). Hence G(t) = 32 − 20 cos(B(t − 4)). The

constant B = 2π

P= 2π

40π

3

= 3

20for 4 ≤ t ≤ 4 + T . Hence

G(t) =

3t for 0 ≤ t ≤ 4.

32− 20 cos(

3

20(t− 4)

)

for 4 ≤ t ≤ 4 + 20π

3

University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 2 (globe) Solution

b. [5 points] After climbing the globe, Gabe jumps onto a small ferris wheel. Let H(t) be hisheight, in inches, above the ground t seconds after Gabe jumped, where

H(t) = 12 + 9 cos( π

75(t− 120)

)

.

Find the the smallest positive value of t at which Gabe’s height above the ground is 10.5 inches.Clearly show each step of your algebraic work. Give your answer in exact form.

Solution:

12 + 9 cos( π

75(t− 120)

)

= 10.5

cos( π

75(t− 120)

)

= −1

6π

75(t− 120) = cos−1

(

−1

6

)

t0 = 120 +75

πcos−1

(

−1

6

)

(smallest positive) tans = t0 − P =75

πcos−1

(

−1

6

)

− 30.

where the period of H(t) is P = 2ππ

75

= 150.

University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 2 (globe) Solution

7. [13 points] After testing different ingredients in their parents’ garages, Imran and Nicole have recentlyopened new organic peanut butter companies.

a. [3 points] Two months after opening, Imran’s company, Chunky Munky, has produced a totalof 256 pounds of peanut butter. Imran thinks Chunky Munky produces peanut butter at aconstant rate of 690 pounds every 6 months. Assuming Imran is correct, write a formula forP (m), the total amount of peanut butter, in pounds, that Chunky Munky will have producedm months after opening.

Solution: Since the production increases at a constant rate, then P (m) must be a linear

function. The slope of P (m) is690

6= 115 pounds per month. Since P (2) = 256, then using

the point slope formula for linear functions we get

P (m) = 256 + 115(m− 2).

b. [4 points] Nicole’s company, Lots O’ Crunch, has produced a total of 182 pounds of peanutbutter two months after opening and a total 454 pounds of peanut butter five months afteropening. Nicole thinks that Lots O’ Crunch produces peanut butter exponentially. AssumingNicole is correct, write a formula for Q(x), the total amount of peanut butter, in pounds, LotsO’ Crunch will have produced x months after opening. Decimal approximations must be rounded

to at least three decimal places.

Solution: We know that Q(2) = 182 and Q(5) = 454 where Q(x) = abx. Then

ab5 = 454

ab2 = 182

b3 =454

182

b =

(

454

182

)1

3

≈ 1.356 and a =182

b2=

182(

454

182

)2

3

≈ 98.95.

Then Q(x) = 182

( 454

182)23

(

(

454

182

)1

3

)

x

≈ 98.95(1.356)x.

University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 7 (peanut butter) Solution

Ann Arbor’s leading local peanut butter company is Sticky PB Company. The total amount ofpeanut butter produced by Sticky PB Company m months after Chunky Munky opens is given by

S(m) = 1500e0.32m.

c. [2 points] By what percent is Sticky PB Company’s production growing every month? Roundyour answer to two decimal places.

Solution: Since b = e.32, then r = b−1 = e.32−1 ≈ 0.38. Hence it grows by 38% every month.

d. [4 points] After a lot of analysis, Imran determines that Chunky Munky’s total peanut butterproduction m months after opening is best modeled by the exponential function

C(m) = 100(1.6)m.

According to this model, when will Chunky Munky and Sticky PB Company have produced thesame amount of peanut butter? Show all your work and leave your answer in exact form.

Solution:

Method 1:

1500e0.32m = 100(1.6)m

ln(

1500e0.32m)

= ln (100(1.6)m)

ln(1500) + ln(e.32m) = ln(100) + ln((1.6)m)

ln(1500) + 0.32m = ln(100) +m ln(1.6)

0.32m−m ln(1.6) = ln(100)− ln(1500)

m(0.32− ln(1.6)) = ln(100)− ln(1500)

m =ln(100)− ln(1500)

0.32− ln(1.6)

Method 2:

1500e0.32m = 100(1.6)m

e.32m

(1.6)m=

1

15

(

e.32

1.6

)

m

=1

15

ln

((

e.32

1.6

)

m)

= ln

(

1

15

)

m ln

(

e.32

1.6

)

= ln

(

1

15

)

then m =ln(

1

15

)

ln(

e.32

1.6

)

University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 7 (peanut butter) Solution


10. [8 points]The entire graph of a function m is shown on theright. Use this graph to answer the questions in partsa. and b. below.

Note that the scales on the axes of the graphs on thispage are not all the same.

−3 −2 −1 1 2 3

−2

−1

1

2

y = m(t)

t

y

a. [4 points]The graph of a function h is shown on the right.It is a transformation of the graph of m. Write aformula for h(t) in terms of m and t.

−6 −4 −2 2 4 6−2

2

4

6y = h(t)

t

y

Answer: h(t) =−2m(1

2t) + 2

b. [4 points] Determine which one of the graphs A–F below is the graph of y = −m(−2t +1)− 3. Then find the values of p and q shown on the graph you chose.

To recieve credit, you must circle an option (A–F) next to the word “Answer” below andwrite your values of p and q in the spaces provided.

A.

−7 −5 −3 −1 1 3 5

p

q

ty

B.

−1 −0.5 0.5 1 1.5 2

p

q

ty

C.

−8 −6 −4 −2 2 4

p

q

ty

D.

−2 −1.5 −1 −0.5 0.5 1

p

q

ty

E.

−5 −3 −1 1 3 5 7

p

q

ty

F.

−4 −2 2 4 6 8

p

q

ty

Remember: to receive credit on this problem, you must circle one option below and write yourvalues of p and q in the spaces provided.

Answer: A B C D E F

p = −2 and q = −4



8. [18 points] The figure below gives the graph of a function A = b(m). The function is periodicand a full period is shown on the graph.

0 2 4 6 8 10 12m0

2000

4000

6000

8000

10 000

12 000A

a. [8 points] For each of the following graphs, give an expression for the function depictedin terms of the function b.

f(m)

0 2 4 6 8 10 12m0

2000

4000

6000

8000

10 000

12 000A

f(m) = b(m + 6)

g(m)

0 2 4 6 8 10 12m0

5000

10 000

15 000

20 000

25 000

30 000

35 000A

g(m) = 3b(m + 1)

h(m)

0 2 4 6 8 10 12m0

5000

10 000

15 000

20 000

25 000

30 000

35 000A

h(m) = 3b(m − 1)

j(m)

0 2 4 6 8 10 12m0

2000

4000

6000

8000

10 000

12 000A

j(m) = −b(m) + 11000

University of Michigan Department of Mathematics Fall, 2010 Math 115 Exam 1 Problem 8 (apples) Solution


b. [4 points] The function b from the previous page represents the number of bushels ofMichigan-grown organic apples, A, available in Michigan grocery stores as a function ofthe number of months, m, after January 1. The function A = b(m) is repeated below.

Which of the graphs on the preceding page could best correspond to the statement:

“In Washington, the apple growing season starts a month earlier, and the peak grocerystore supply is three times as much as in Michigan.” Explain your answer.

Solution: The graph of g(m) best corresponds to the statement above. The graph ofg(m) has a peak which is three times higher than that of b(m) and the graph has beenshifted one unit to the left to signify the growing season beginning one month earlier.

0 2 4 6 8 10 12m0

2000

4000

6000

8000

10 000

12 000A

c. [6 points] Using the graph of b(m), repeated above, sketch a well-labeled graph of b′(m).

2 4 6 8 10 12m

b¢

Note, graphs may differ–answer is not unique.

University of Michigan Department of Mathematics Fall, 2010 Math 115 Exam 1 Problem 8 (apples) Solution


4. [10 points] Consider the function f defined by f(x) =(x+ 1.8)(x+ 2.1)

(2x+ 1.8)(3x− 6.9)(x+ 2.1).

You do not have to show your work/reasoning on this problem. However, any work that youdo show may be considered for partial credit.

a. [3 points] What is the domain of f?

Answer: all real numbers except −0.9, 1.3, and −2.1

b. [2 points] Find the equations of all vertical asymptotes of the graph of y = f(x).If there are none, write none.

Answer: x = −0.9 and x = 1.3

c. [2 points] Let g(x) = e−0.4x.

Find the equations of all horizontal asymptotes of the graph of y =g(x)

f(x).

If there are none, write none.

Solution: g(x) is a positive exponential decay function and dominates any rational

function as x → ∞. In particular, limx→∞

g(x)

f(x)= 0 and lim

x→−∞

g(x)

f(x)= ∞ (DNE), so the

only horizontal asymptote of the graph of y =g(x)

f(x)is y = 0.

Answer: y = 0

d. [3 points] Find a formula for a rational function h(x) such that limx→∞

f(x)

h(x)= 8.

Solution: There are many possible answers. Some examples include:

• h(x) =1

8 · 6 · x=

1

48x, and

• h(x) =1

8f(x) =

(x+ 1.8)(x+ 2.1)

8(2x+ 1.8)(3x− 6.9)(x+ 2.1).

Answer: h(x) =

1

48x



1. [10 points] Laquita decides to visit an amusement park during Fall Break and rides severalroller coasters, including the Classic Amazing Looping Coaster and the Ultra Mountain.Let R(t) be the distance, in feet, that the CAL Coaster has moved along the track t secondsafter the ride begins. The ride lasts a total of 60 seconds. Several values of R(t) are shown inthe following table. t 0 10 25 30 40 45 55 60

R(t) 0 496 1103 1327 1817 2136 2718 3141

For parts a.– c., remember to show your work and reasoning clearly.

a. [2 points] Find the average velocity of the CAL Coaster during the last 15 seconds of theride, i.e. for 45 ≤ t ≤ 60. Include units.

Solution:R(60)−R(45)

60− 45=

3141− 2136

60− 45=

1005

15= 67.

Answer:67 ft/sec

b. [2 points] Estimate the instantaneous velocity of the CAL Coaster 30 seconds after theride begins. Include units.

Solution: We estimate using average velocity based on nearby measurements.

Average velocity for 25 ≤ t ≤ 30:R(30)−R(25)

30− 25=

1327− 1103

5=

224

5= 44.8

(Note that other answers, such as those incorporating (40, 1817), were accepted if workand appropriate units were shown.)

Answer: About 44.8 ft/sec

c. [2 points] Estimate R′(55).

Solution: We estimate the derivative using the average rate of change of R(t) for

55 ≤ t ≤ 60:R(60)−R(55)

60− 55=

3141− 2718

5= 84.6

(Again, answers incorporating (45, 2136) were accepted with appropriate work.)

Answer: R′(55) ≈ 84.6

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 1 Problem 1 (coasters) Solution


d. [4 points] Let h(t) be Laquita’s height, in feet, above the ground, t seconds after her rideon the Ultra Mountain begins. A graph of h(t) is shown below.

14 32 42 67

248

109

180

23

y = h(t)

t (sec)

y (ft)

Let the quantities I–V be defined as follows:

I. The number 0.

II. Laquita’s instantaneous vertical velocity, in ft/sec, at t = 14.

III. h′(32)

IV. Laquita’s average vertical velocity, in ft/sec, between t = 14 and t = 42.

V. Laquita’s instantaneous vertical velocity, in ft/sec, at t = 67.

Rank the quantities in order from least to greatest by filling in the blanks below with theoptions I–V. You do not need to show your work.

III < V < I (0) < IV < II

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 1 Problem 1 (coasters) Solution


7. [14 points] David is living on the 37th floor of a fancy building. He wants to get rid of anancient (very energy inefficient) refrigerator that was in the building before alterations weremade to the apartment. The refrigerator will not fit through the new doors of the apartment,so it must be pushed down a rather rickety ramp out the window. The ramp is 350 feet long.Below is a table showing the distance from the window along the ramp at given times:

time (seconds) 0 2 4 6 8 10 12 14 16

distance from window (feet) 0 3.9 18.6 43.1 79.4 122.5 174.6 240.1 313.6

Suppose s(t) = d is the distance from the window, in feet, as a function of time, t, in seconds.

a. [3 points] Compute the average velocity of the refrigerator over the interval 4 ≤ t ≤ 12.

Solution: The average velocity over the time interval 4 ≤ t ≤ 12 is

174.6 − 18.6

12 − 4= 19.5 ft / sec.

b. [4 points] Approximate the instantaneous velocity of the refrigerator when t = 8 seconds.

Solution: Avg. vel. over 6 ≤ t ≤ 8 is79.4 − 43.1

2= 18.15 ft / sec..

Avg. vel. over 8 ≤ t ≤ 10 is122.5 − 79.4

2= 21.55 ft / sec.

Or, we could average those velocities, giving the approximate instantaneous velocityat t = 8 as 19.85 ft / sec. [Note: any of the three approximations were accepted.]

c. [3 points] Approximately where will the refrigerator be after 18 seconds? Justify youranswer.Solution: The refrigerator will be on the sidewalk.

This is because the approximation of the velocity at 16 seconds is

313.6 − 240.1

2= 36.75 ft / sec.

which would mean that in 2 more seconds it would travel about 73.5 ft,but it is only 36.4 ft to the ground.

d. [4 points] Based upon the information in the table, does s appear to be concave up orconcave down at t = 8? Justify your answer.

Solution: The velocity appears to be increasing at t = 8, because the average velocityfor 6 ≤ t ≤ 8 is less than the average velocity for 8 ≤ t ≤ 10. Thus, if s

′ is increasing, s′′

is positive, and the graph would be concave up.

University of Michigan Department of Mathematics Fall, 2010 Math 115 Exam 1 Problem 7 (refrigerator) Solution


8. [15 points] Consider the functions f(x) and g(x) given by the formula and graph below.

f(x) =

2x3 − 2x2 for x ≤ 1,

x3 + 1 for x > 1.

−4 −3 −2 −1 1 2 3 4

−6

−5

−4

−3

−2

−1

1

2

3

x

g(x)

a. [5 points] Circle the correct answer(s) in each of the following questions.

Solution:

i) At which of the following values of x is the function g(x) not continuous?

x = −3 x = −1 x = 0 x = 2 x = 3.5

ii) At which of the following values of x is the function f(x) + g(x) continuous?

x = −2 x = −1 x = 0 x = 1 x = 2

Note that g(x) is linear on the interval (−4,−2),(1, 2) and (2, 3). All your answers below shouldbe exact. If any of the quantities do not exist, write dne.

b. [2 points] Find limx→2

(2f(x) + g(x)).

Solution: 2(23 + 1)− 2.5 = 15.5 Answer: 15.5

c. [2 points] Find limx→∞

f(2x)

x3.

Solution: limx→∞

f(2x)

x3= lim

x→∞

8x3 + 1

x3= 8 Answer: 8

d. [2 points] Find limx→∞

g(

x2e−x + 3)

.

Solution: limx→∞

g(

x2e−x + 3)

= limu→3+

g (u) = 1 Answer: 1

e. [2 points] For which value(s) of p does limx→p+

g(x) = 1?

Solution: Answer: p = −3.5, 3

f. [2 points] Find limx→−1−

f(−x).

Solution: Answer: 2



8. [12 points] A portion of the graph of a function f is shown below.

1 2 3 4 5 6 7 8 9 10

−3

−2

−1

1

2

3

0x

yy = f(x)

a. [2 points] Give all values c in the interval 0 < c < 10 for which limx→c

f(x) does not exist. If

there are none, write NONE.

Answer: c = 1, 8

b. [2 points] Give all values c in the interval 0 < c < 10 for which limx→c+

f(x) does not exist.

If there are none, write NONE.

Answer: c = NONE

c. [2 points] Give all values c in the interval 0 < c < 10 for which f(x) is not continuous atc. If there are none, write NONE.

Answer: c = 1, 3, 8

d. [6 points] With f as shown in the graph above, define a function g by the formula

g(x) =

B + 2x2 + 3x3 +Ax5

12 + 6x3 + 4x5if x ≤ 0

f(x) if 0 < x < 10

where A and B are nonzero constants.Find values of A and B so that both of the following conditions hold.

• g(x) is continuous at x = 0.

• limx→−∞

g(x) =1

2.

If no such values exist, write none in the answer blanks.Be sure to show your work or explain your reasoning.

Solution: To satisfy the first condition, we first compute g(0) by plugging in x = 0 to

the rational functionB + 2x2 + 3x3 +Ax5

12 + 6x3 + 4x5to find lim

x→0−

g(x) = g(0) =B

12. In order for

g(x) to be continuous at x = 0, we must also have limx→0+

g(x) =B

12.

Now limx→0+

g(x) = limx→0+

f(x) = −1 (from the graph), so B

12= −1, and B = −12.

To satisfy the second condition, we compute that

limx→−∞

g(x) = limx→−∞

B + 2x2 + 3x3 +Ax5

12 + 6x3 + 4x5=

A

4.

In order for this limit to equal 1

2, we must have A/4 = 1/2, so A = 2.

Answer: A = 2 and B = −12University of Michigan Department of Mathematics Fall, 2015 Math 115 Exam 1 Problem 8 Solution


1. [10 points] A portion of the graph of a function f is shown below.

1

2

3

4

5

6

1 2 3 4 5 6x

y

y = f(x)

Note: You may assume that pieces of the function that appear linear are indeed linear.

Use the graph above to evaluate each of the expressions below, and write your answer on theanswer blank provided. If any of the quantities do not exist (including the case of limits thatdiverge to ∞ or −∞), write dne.

a. [1 point] f(1)

Answer: 3

b. [1 point] limx→5

f(x)

Answer: 0

c. [1 point] limq→3

f(q)

Answer: DNE

d. [1 point] limz→2

f(2)

Answer: 5

e. [1 point] limr→6−

f(r)

Answer: 1

f. [1 point] limh→0

f(4.25 + h)− f(4.25)

h

Answer: −4

g. [1 point] limp→0.5

f(p)

p

Answer: 4

h. [1 point] limt→3

f(t)f(t+ 2)

Answer: 0

i. [1 point] limx→3+

f(f(x))

Answer: 3

j. [1 point] lims→1

f(f(s))

Answer: 5


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