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Instructions and 10 practice problems for Math 115 Exam 1. The problems cover topics such as functions, derivatives, and integrals. guidelines for answering the questions, including showing work and not using a calculator. The problems vary in difficulty and require interpretation of questions. The document also includes a table with past exam problems and scores.
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Semester Exam Problem Name Points Score Fall 2015 1 4 scones 13 Fall 2017 1 2 globe 13 Fall 2017 1 3 5 Winter 2017 1 4 pests 10 Fall 2017 1 10 6 Winter 2020 1 10 9 Winter 2017 1 2 5 Winter 2020 1 8 11 Fall 2019 1 9 10 Winter 2017 1 1 19 Total 101
Recommended time (based on points): 91 minutes
Math 115 / Exam 1 (October 13, 2015) page 6
Answer:
h
b. [5 points] Below is the first part of a sentence that will give a practical interpretation of the equation h′(6) = 15 in the context of this problem. Complete the sentence so that the practical interpretation can be understood by someone who knows no calculus. Be sure to include units in your answer. If Algie decreases the amount of baking soda per scone from 6 grams to 5.8 grams, then...
Solution: If Algie decreases the amount of baking soda per scone from 6 grams to 5. grams, then the height of each scone will decrease by approximately 3 millimeters.
c. [3 points] Algie makes a batch of scones, with each scone containing k grams of baking soda (for some constant k). When the scones come out of the oven, he decides they are each 10 millimeters shorter than he would like. Write a mathematical expression involving k, h, and h−^1 for the number of grams of baking soda per scone he should use to get scones of the desired height.
Answer: h−^1 (h(k) + 10)
d. [3 points] Algie does some calculations and determines that
h−^1 (30)
Based on this information, which of the following statements must be true? Circle all of the statements that must be true or circle none of these. A. If Algie makes 40 scones, each with 30 grams of baking soda, then the scones will rise to a height of 60 millimeters.
B. If Algie wants to make 40 scones, then he must use 60 grams of baking soda.
If Algie wants to make scones of height 30 millimeters and he has 60 grams of baking soda, then the maximum number of scones he can make is 40.
D. A scone containing 1.5 grams of baking soda rises to a height of 30 millimeters.
E. A scone containing 30 grams of baking soda rises to a height of 1.5 millimeters.
F. none of these
University of Michigan Department of Mathematics Fall, 2015 Math 115 Exam 1 Problem 4 (scones) Solution
b. [5 points] After climbing the globe, Gabe jumps onto a small ferris wheel. Let H(t) be his height, 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)
cos
( (^) π 75
(t − 120)
π 75 (t − 120) = cos−^1
t 0 = 120 +
π cos−^1
(smallest positive) tans = t 0 − P =
π cos−^1
where the period of H(t) is P = (^2) ππ 75
University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 2 (globe) Solution
B(k) = e−^4 k 2 tan(k + 3). Use the limit definition of the derivative to write an explicit expression for B′(5). Your answer should not involve the letter B. Do not attempt to evaluate or simplify the limit. Please write your final answer in the answer box provided below. Solution: B′(5) = lim h→ 0
e−4(5+h) 2 tan(h + 8) − e−^100 tan(8) h
University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 3 Solution
In each of the following parts, the corresponding portion of the graph of a function obtained from k by one or more transformations is shown, to- gether with a list of possible formulas for that function. In each case, circle all possible formu- las for the function shown. Note that the graphs are not all drawn at the same scale. − 5 − 4 − 3 − 2 − 1 1 2 3 4 5
y = k(x)
x
y
a. [3 points]
y = m(x)
x
y
k(x) − 2
k(x) + 2
k(−x) − 2
k(−x) + 2
k(x) − 2
F. 2k(x) − 2
G. 2k(x) + 2
H. − 2 k(−x) − 2
I. − 2 k(−x) − 2
J. − 2 k(x) + 2
K. none of these
b. [3 points]
y = q(x)
x
y
A. k(2x + 2)
B. k(− 2 x − 2)
C. −k(2x + 2)
D. k(− 2 x + 2)
E. −k(0.5(x + 2))
F. −k(0. 5 x − 2)
G. k(0. 5 x + 2)
H. k(0.5(x − 2))
I. k(2(x + 1))
J. −k(0.5(x − 2))
K. none of these
University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 1 Problem 10 Solution
Math 115 / Exam 1 (February 11, 2020) page 11
[9 points] Let P (t) be a town’s population, in thousands of people, t years after the beginning of
Some values of P ′(t), the derivative of P (t), are given in the table below.
t − 8 − 3 0 3 6 8 12 15 P ′(t) 2 2 0 0 3 0 − 6 − 2
Assume that between each pair of consecutive values of t given in the table, P ′(t) is either always increasing, always decreasing, or always constant.
a. [1 point] Let y = P ′(t). What are the units of y?
Answer: = thousands of people per year
For each of the following, circle all correct answers.
b. [2 points] At which of the following time(s) is the town’s population increasing?
t = − 6 t = 2 t = 7 t = 13 none of these
c. [2 points] On which of the following interval(s) is the town’s population constant?
(− 7 , −5) (1, 2) (7, 10) none of these
d. [2 points] On which of the following interval(s) is P (t) linear?
(− 7 , −5) (1, 2) (7, 10) none of these
e. [2 points] At which of the following time(s) is the town’s population the largest?
t = 3 t = 6 t = 8 t = 15
© 2020 Univ. of Michigan Dept. of Mathematics (Creative Commons BY−NC−SA license) Winter, 2020 Math 115 Exam 1 Problem 10 Solution
Math 115 / Exam 1 (February 11, 2020) page 9
y = j(x)
x
y
a. [7 points] On the axes below, carefully sketch the graph of j′(x), the derivative of j(x), on the interval − 2 < x < 8. Be sure that your graph carefully indicates where j′(x) is zero, positive, and negative, and where j′(x) is increasing, decreasing, and constant.
y = j′(x)
x
y
b. [4 points] Shown below is a portion of the graph of a function k(x) which can be obtained from j(x) through one or more graph transformations. Find a formula for k(x) in terms of j(x).
y = k(x)
x
y
Answer: k(x) = j(−(x^ −^ 6))^ −^1
© 2020 Univ. of Michigan Dept. of Mathematics (Creative Commons BY−NC−SA license) Winter, 2020 Math 115 Exam 1 Problem 8 Solution
Math 115 / Exam 1 (October 7, 2019) page 9
a. [4 points] A portion of the graph of a polynomial function q(x) is shown below. Find a possible formula for q(x) of the smallest possible degree. Assume that all of the key features of the graph are shown.
y = q(x)
x
y
Solution: We see that the degree of q(x) must be even. The zeros of q(x) are −1, 2, and 4. Note that 2 is a double zero. We use the point (0, 4) to find the leading coefficient.
q(x) = C(x + 1)(x − 2)^2 (x − 4) 4 = C(0 + 1)(0 − 2)^2 (0 − 4)
−
Answer: q(x) =
(x + 1)(x − 2)^2 (x − 4)
b. [3 points] Find the formula for a rational function r(x) that has a hole with an x-value of 5, a vertical asymptote at x = 1, and a horizontal asymptote at y = −2. Solution: Note that the factor of x in the numerator could be replaced by any degree 1 factor (x + B) with B a constant.
Answer: r(x) =
−2(x − 5)x (x − 5)(x − 1)
c. [3 points] Consider the function z(x) = 4 −x^ − 2 x^2 15 x + 3x^2
Find lim x→∞ z(x) and lim x→−∞ z(x). If the value does not represent a real number (including the case of limits that diverge to ∞ or −∞), write “dne” or “does not exist.”
Solution: Note that 4−x^ =
4
)x , so (^) xlim→∞ 4 −x^ = 0 and (^) x→−∞lim 4 −x^ does not exist (as 4−x diverges to ∞ as x → −∞).
Answer: lim x→∞ z(x) =
(^3) and lim x→−∞ z(x) = DNE
University of Michigan Department of Mathematics Fall, 2019 Math 115 Exam 1 Problem 9 Solution
Math 115 / Exam 1 (February 8, 2017) page 3
The graphs of the functions f (x) and g(x) are included here for your convenience.
y = f (x)
x
y
y = g(x)
x
y
g. [3 points] Find all the values of x with − 5 < x < 4 at which the function f (x) is not continuous.
Answer: −^2 ,^0 ,^2
h. [2 points] What is the range of y = g(x)?
Answer: [0,5]
i. [2 points] For which of the following values of x is f ′(x) > 0? Circle all that apply.
x = − 5 x = − 1 x = 1. 5 x = e none of these
University of Michigan Department of Mathematics Winter, 2017 Math 115 Exam 1 Problem 1 Solution