Math 151 - Practice Exam 2 - solutions
Problem 1 Find
dy
dx
where y = x2 + 10x + x7/4 .
7
Solution. None of the big Rules needed. f 0 (x) = 2x + 10 + x3/4 .
4
Problem 2 Consider the function
f (x) =
x−2
.
3x − 1
a) Find f 0 (2).
b) What is the slope-intercept equation of the tangent line to the graph of
f (x) at x = 2?
Solution. a) We need the Quotient Rule with u = x − 2 and v = 3x − 1.
d x−2
(3x − 1) − 3(x − 2)
5
=
=
.
2
dx 3x − 1
(3x − 1)
(3x − 1)2
Then we substitute x = 2 to get f 0 (2) = 5/25 = 1/5.
b) Since f (2) = 0, we have
1
y = (x − 2) + 0
5
which expands to the slope-intercept form y =
x
5
− 25 ..
Problem 3 Consider the function
√
f (x) = x5 (2x3 + 7 x).
a) Find f 0 (x) using the Product Rule (do not simplify).
b) Expand the formula for f (x) and compute f 0 (x) using the Rule for powers
of x.
√
Solution. a) With u = x5 and v = 2x3 + 7 x, we first compute u0 = 5x4
√
and v 0 = 6x2 + 7/(2 x). Then
√
√
f 0 (x) = x5 (6x2 + 7/(2 x)) + 5x4 (2x3 + 7 x).
b) We expand f (x) to
f (x) = 2x8 + 7x5.5 .
Then we can apply the basic rules of 4.1 to get the same answer in a different
form,
f 0 (x) = 16x7 + 38.5x4.5 .
Yes, they agree.
Problem 4 Suppose f (x) and g(x) are functions with values f (a) = 2,
g(a) = 3, f 0 (a) = 5 and g 0 (a) = −1. Find the derivative of f (x)(g(x) − 4) at
x = a.
Solution. Use the product rule,
d
[f (x)(g(x) − 4)] = f (x)g 0 (x) + f 0 (x)(g(x) − 4).
dx
Now we substitute x = a and the given information to get the value
2(−1) + 5(3 − 4) = −7.
Problem 5 Let f (x) = x3 − 9x2 + 24x − 1.
Find all the points at which the graph of f has a horizontal tangent line.
Solution.
f 0 (x) = 3x2 − 18x + 24 = 3(x2 − 6x + 8) = 3(x − 2)(x − 4).
This is zero for x = 2 and x = 4, and these points are exactly where the
graph of f has a horizontal tangent line.
Problem 6 The graph of a function f (x) is shown below. All answers in
this question are integers.
a) Indicate all x-values where f (x) has a relative maximum.
b) Indicate all x-values where f (x) has a relative minimum.
c) List all relative maximum values of f (x).
d) List all relative minimum values of f (x).
e) List all intervals where this function is increasing.
Solution. a) x = −2, 0, 2. b) x = −3, −1, 1, 3. c) y = 6, 5. d) y = 4, 3
(if a value happens more than once, no need to list it again).
e) The function is increasing in [−3, −2], [−1, 0], and [1, 2].
Problem 7 Consider the function
f (x) = 4x4 − 2x2 + 1
a) Determine all critical numbers for this function.
b) Determine the intervals where this function is increasing.
c) Determine the intervals where this function is decreasing.
d) Find all numbers where this function has a local maximum.
e) Find all numbers where this function has a local minimum.
Solution. First, we compute
f 0 (x) = 16x3 − 4x = 4x(4x2 − 1).
a) The derivative equals zero for x = 0 and x = ±1/2. These are the critical
numbers.
b) Sampling gives f 0 (x) > 0 in (−1/2, 0) and in (1/2, ∞). These are the
intervals where f (x) is increasing.
c) Sampling gives f 0 (x) < 0 in (−∞, −1/2) and in (0, 1/2). These are the
intervals where f (x) is decreasing.
d) From part b) and c), x = 0 is the only number where f (x) has a local
maximum.
e) From part b) and c), x = ±1/2 are the only numbers where f (x) has a
local minimum.
Problem 8 Assume that the unit price of manufacturing x batches of paper
towels is
1
p(x) = 412 − x2
3
(a) What is the revenue R(x) from selling these x batches?
(b) Find the marginal revenue R0 (x) at x = 20.
(c) Use your answer in part (b) to approximate the additional revenue from
selling 2 extra batches in addition to the 20 batches from part (b). The
exact answer will not give credit.
(d) Find an equation for the tangent line to the graph of R(x) at x = 20. No
need to simplify!
Solution. (a) R = xp = 412x − x3 /3.
(b) R0 (x) = 412 − x2 , so R0 (20) = 12.
(c) R(22) ≈ R(20) + 2R0 (20) = R(20) + 24, so 24 dollars additional revenue.
(d) y = 12(x − 20) + 8240 − 8000/3.
Problem 9 Assume that the unit price of manufacturing x batches of
flibbertigibbets is
p(x) = 20.1 − 0.02x
and the total cost is
C(x) = 150 + 0.1x.
(a) What is the total profit P (x) from selling these x batches, and the average
profit P̄ (x)?
(b) Find the marginal average profit at x = 20.
(c) For what value of x is the average profit P̄ (x) maximal? Round to one
digit after the decimal point.
Solution. (a) P (x) = xp(x) − C(x) = 20x − 0.02x2 − 150. So
P̄ (x) =
150
20x − 0.02x2 − 150
= 20 − 0.02x −
x
x
(b) P̄ 0 (x) = −0.02 + 150/x2 , so P̄ 0 (20) = 0.355.
(c) We need to solve P̄ 0 (x) = 0,√so 0.02x2 = 150. After discarding the
negative answer, this gives x = 50 3 ≈ 86.6.