Mathematics 113Q, Section 2
Practice Exam 2, Fall 2004
Instructor: Adam Bowers
Name:
Answer the questions in the space provided. If you need more space, continue on the back
of the page. All answers must include supporting work or an explanation of your reasoning.
Without supporting work, even correct answers may be marked wrong. This exam has 14
questions, for a total of 165 points.
1. (10 points) The graph below is the graph of y = f 0 (x), the derivative of f .
y
0
0
x
(a) On which intervals is f concave up?
f is concave up when f 00 (x) > 0 which happens when f 0 (x) is increasing.
This happens whenever x ∈ (−2, −0.75) and x ∈ (2, ∞). Keep in mind
that the −0.75 is just an approximation, anything between −1 and
−0.5 would be acceptable.
(b) For which x values does f have local minimum values?
x = 3.
Use the first derivative test.
(c) For which x values does f have an inflection point?
The local max/min of f 0 :
x = −2, x = 2, x = −0.75
Math 113Q - 2
Exam 2 (practice exam)
2. (a) (5 points) Find the tangent line approximation to f (x) =
√
3
x + 1 at x=0.
1
f (x) ≈ 1 + x
3
(b) (5 points) Use the result above to estimate
√
3
1.02 = f (0.02) ≈ 1 +
√
3
1.02.
0.02
≈ 1.0066666666 . . .
3
3. (a) (5 points) Find the tangent line approximation to g(x) = ex at x=0.
g(x) ≈ 1 + x
(b) (5 points) Use the result above to estimate e−0.09 .
e−0.09 = g(−0.09) ≈ 1 − 0.09 = 0.91
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Math 113Q - 2
Exam 2 (practice exam)
4. (5 points) The function f (x) = x3 + P x2 + 13 (where P is constant) has a point of
inflection at x = 2. Determine the value of P .
Since f has a point of inflection at x = 2, f 00 (2) = 0.
Calculate:
f 00 (x) = 6x + 2P.
Then
f 00 (2) = 12 + 2P = 0,
so P = −6.
5. (15 points) Let f be a function whose derivative is f 0 (x) = (x − 2)e2x .
(a) Find the critical points of f and determine whether they are local maxima, local
minima, or a saddle point.
f 0 (x) = 0 only when x = 2, so this is t he only critical point.
To determine the type of critical point, I will use the first derivative test. When x < 2, f 0 (x) < 0 and when x > 2, f 0 (x) > 0. The
slopes change from negative to positive at x = 2, so f has a local
minimum at x = 2.
(b) Determine weather or not f has a global maximum.
No, because the global maximum must be at a local maximum and the
only critical point is a local minimum. (The interval is not closed.)
(c) Find the inflection points of f
The points of inflection occur when f 00 (x) = 0.
f 0 (x) using the product rule to get
Differentiate
f 00 (x) = e2x + (x − 2)(2e2x ) = e2x (2x − 3).
Since e2x 6= 0 for any x, we have that the only inflection point
is at x = 32 .
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Math 113Q - 2
Exam 2 (practice exam)
6. (10 points) Differentiate the following functions.
(a)
f (x) = x sinh(2x)
f 0 (x) = sinh(2x) + 2x cos(2x)
(b)
g(x) = cosh(x2 )2
g 0 (x) = 2 cosh(x2 ) · sinh(x2 ) · 2x
7. (15 points) Evaluate the following.
(a)
lim
x→0
=
cosh(2x)
x−1
cosh(0)
1
=
= −1
0−1
−1
(b)
sinh(3x)
x→0
4x
lim
Since sinh(3x) → sinh(0) = 0 and 4x → 0, we use l’H^
opital’s Rule:
= lim
x→0
3 cosh(3x)
3 cosh(0)
3
=
= .
4
4
4
(c)
lim
x→1
cosh(x) − 1
x−1
Since the numerator goes to cosh(1)−1, which is finite, and the
denominator goes to 0, the limit of the fraction is infinity (∞).
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Math 113Q - 2
Exam 2 (practice exam)
8. (15 points) A gardener is enclosing his garden in a rectangular fence. One side of the
garden is next to a barn, so fence will not be needed along that side.
(a) If the gardener has 120 feet of fencing, how big can his garden be?
Length: 2x + y = 120
Area: xy
First solve for y in terms of x using the equation for length.
Then you get an equation for area:
A(x) = x(120 − 2x) = 120x − 2x2 .
Then solving A0 (x) = 0 gives x = 30 and y = 60.
(b) Verify that your answer in part (a) is a local maximum.
A00 (x) = −4 < 0, so x = 30 is a local maximum, by the second
derivative test.
9. (15 points) Bob just opened a martial arts school. His profit equation looks like this:
P (x) = −x3 + 3x2 − 1,
where x is the number of students taught per day.
(a) Find the number of students taught per day which maximizes Bob’s profits.
Differentiate (and simplify):
P 0 (x) = −3x2 + 6x = −3x(x − 2).
Solving P 0 (x) = 0 for x gives x = 0 or x = 2. Plugging these
values into the profit equation gives:
P (0) = −1, P (2) = 3.
Thus the maximum profit occurs when x = 2. [Since $3 isn’t very
much money, we’ll assume the profit is given in hundreds of dollars.]
Notice that limx→∞ P (x) = −∞.
(b) Is your answer in part (a) a local maximum.
P 00 (x) = −6x + 6, so P 00 (2) = −6 < 0. Thus x = 2 is a local
maximum by the second derivative test.
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Math 113Q - 2
Exam 2 (practice exam)
10. (15 points) A square-bottomed box with no top has a volume of 4 cubic inches.
(a) What are the dimensions of the box that minimizes the surface area?
Volume: x2 h = 4
Surface Area: x2 + 4xy
First solve for h in terms of x using the equation for volume.
Then you get an equation for surface area:
A(x) = x2 + 4x(
4
16
) = x2 + .
2
x
x
Then
16
x2
and solving A0 (x) = 0 gives x3 = 8 or x = 2 and h = 1.
A0 (x) = 2x −
(b) Verify that your answer in part (a) is a local minimum.
A00 (x) = 2 + 2·16
, so A00 (2) = 6 > 0.
x3
by the second derivative test.
NOTE: Question 11 is on the next page.
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Thus x = 2 is a local minimum
Math 113Q - 2
Exam 2 (practice exam)
11. (15 points) A small home is to occupy a lot with area 1500 square feet. In addition to
this, the owner wishes to have a lawn surrounding the house. In front and back of the
house, the lawn will be 5 feet wide. On both sides, the lawn will extend 3 feet.
(a) What is the smallest property on which this home (with lawn) can fit?
Area of building:
Area of Property:
xy = 1500
(x + 6)(y + 10)
First solve for y in terms of x using the equation for the area
of the building. Then you get an equation for the area of the
property:
A(x) = (x + 6)(
1500
6 · 1500
+ 10) = 10x +
+ 1560.
x
x
Then
9000
x2
and solving A0 (x) = 0 gives x2 = 900 or x = 30 and y = 50.
A0 (x) = 10 −
(b) Verify that your answer in part (a) is a local minimum.
A00 (x) = 2·9000
, so A00 (30) = 32 > 0.
x3
by the second derivative test.
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Thus x = 30 is a local minimum
Math 113Q - 2
Exam 2 (practice exam)
12. (15 points) Let f (x) = (x − 4)e−x for x ≥ 0.
(a) Does f have a horizontal asymptote? Justify your answer by computing the relevant
limit.
Use l’H^
opital’s rule to find the limit of f (x) as x → ∞:
y = x→∞
lim
x−4
= 0.
ex
Thus the horizontal asymptote is the line y = 0, or the x-axis.
(b) Find the best possible bounds for f ; that is, find the largest A and smallest B such
that
A ≤ f (x) ≤ B for all x ≥ 0.
Differentiate f :
f 0 (x) = e−x − (x − 4)e−x = e−x (5 − x).
Thus the only critical point is at x = 5. Evaluating f at the
critical points and endpoints (in this case 0 and ∞), we have:
f (0) = −4, f (5) = e−5 , x→∞
lim f (x) = 0.
Thus our bounds are A = −4 and B = e−5 .
13. (a) (10 points) Use the Racetrack Principle to show that ln x ≤ x − 1 for all x ≥ 1.
Let g(x) = ln x and h(x) = x−1. The interval is [1, ∞). We have
g(1) = ln(1) = 0 and h(1) = 1 − 1 = 0. Also we have g 0 (x) = x1 and
h0 (x) = 1. But x ≥ 1, so x1 ≤ 1. Thus
g(1) = h(1) and
1
g 0 (x) = ≤ 1 = h0 (x).
x
Therefore, by the Racetrack Principle, g(x) ≤ h(x) or ln x ≤ x − 1
[substituting back in for g and h.]
NOTE: Part (b) is on the next page.
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Math 113Q - 2
Exam 2 (practice exam)
(b) Is ln x ≤ x − 1 for all 0 < x ≤ 1?
This is true. Use the Racetrack Principle just as before, but
this time you will have
g 0 (x) =
1
≥ 1 = h0 (x)
x
on the interval (0, 1], where g and h are equal at the right endpoint,
not the left.
14. (5 points) Let f (x) = x cos x. Explain why there must be a critical point of f on the
interval [0, π/2].
We wish to show there is a point c in the interval [0, π/2] where f 0 (c) =
0. We use the Mean Value Theorem. First we state that f satisfies
the necessary conditions for the Mean Value Theorem: f is continuous
and differentiable for all x in [0, π/2]. Then it follows from MVT
that there is some c in the open interval (0, π/2) such that
f 0 (c) =
f (π/2) − f (0)
0−0
=
= 0.
π/2 − 0
π/2
Since f 0 (c) = 0, we know that f has a critical point at x = c.
concludes the problem.
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