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Math 160 - Exam 3 Solutions - Summer 2012 - Jaimos F Skriletz
Answer the following problems to the best of your ability. To receive full credit, answers must be supported by a sufficient
amount of work using the methods presented in class.
1. (15 pts) Find all local extrema of the polynomial function f (x) = x3 − 3x2 − 24x + 21 (be sure to show how you used
the first and/or second derivative to find the local extrema).
The critical points are found by solving
f (x) = 3x2 − 6x − 24 = 0
3(x2 − 2x − 8) = 0
3(x + 2)(x − 4) = 0
x = −2
or x = 4
The sign table for f 0 (x) is
−2
(+)
f 0 (x)
4
(−)
(+)
Thus (−2, f (−2)) = (−2, 49) is a local maximum.
And (4, f (4)) = (4, −59) is a local minimum.
2. (10 pts) Find the intervals for which the function f (x) = 9x2 − x3 − 30x + 17 is concave up and concave down. What
are the inflection points of f (x)?
f 0 (x) = 18x − 3x2 − 30
f 00 (x) = 18 − 6x = −6(x − 3)
The only zero for f 00 (x) is x = 3 and the sign table is
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f 00 (x)
f (x) is concave up on (−∞, 3).
f (x) is concave down on (3, ∞).
(3, f (3)) = (3, −19) is an inflection point.
(+)
(−)
2
Math 160 - Exam 3 Solutions - Summer 2012 - Jaimos F Skriletz
3. (15 pts) Use the following information to sketch the function f (x). Assume that f (x) is continuous on all real numbers.
lim f (x) = 0
x→−∞
lim f (x) = +∞
x→+∞
f (−1) = −1
f (0) = −4
f (1) = −5
f 0 (x)
f (x)
f (4) = 0
f 00 (x)
4. (10 pts) What is the absolute maximum and minimum of f (x) = 2x3 + 15x2 + 24x − 15 on the domain [−3, 1].
f 0 (x) = 6x2 + 30x + 24 = 6(x2 + 5x + 4) + 6(x + 4)(x + 1) = 0
The two critical points are x = −4 and x = −1. The only one in [−3, 1] is x = −1 so the extrema is at one of the
following three points
x
−3
−1
1
f (x)
−6
−26
26
Thus the absolute maximum is 26 and the absolute minimum is −26.
Math 160 - Exam 3 Solutions - Summer 2012 - Jaimos F Skriletz
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5. (10 pts) Consider the price-demand relation
x + p2 = 1500
(a) Find the elasticity of demand for the current market price p = 20 and state if the demand is inelastic or elastic
−pf 0 (p)
(recall E(p) =
)
f (p)
f (p) = x = 1500 − p2
f 0 (p) = −2p
2p2
−p(−2p)
=
2
1500 − p
1500 − p2
2(20)2
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E(20) =
<1
=
2
1500 − (20)
11
E(p) =
Thus the demand is inelastic.
(b) In order to increase revenue would you need to raise or lower the current market price?
Since the demand is inelastic, increasing the price will result in increasing the revenue.
6. (10 pts) Evaluate the following limit: lim
x→1
ln(x)
ln(1)
0
=
=
3 − 3x
3 − 3(1)
0
So by L’Hopital’s
1
1
ln(x)
−1
= lim x = 1 =
x→1 3 − 3x
x→1 −3
−3
3
lim
Math 160 - Exam 3 Solutions - Summer 2012 - Jaimos F Skriletz
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7. (10 pts)
A rancher builds a rectangular fence. He uses brick for one
side of the fence, and wood for the other three sides (as
shown). It costs $14 per foot for the brick fence and $6 per
foot for the wood fence.
Let x be the width and y be the height as shown.
(a) Suppose the rancher spent $1200 on the fence, find the enclosed area as a function of the length of one of the sides.
The cost of the fence is
C = 14x + 6(2y + x) = 14x + 12y + 6x = 20x + 12y
If the cost is $1200 then
20x + 12y = 1200
12y = −20x + 1200
1200
−5
−20
x+
=
x + 100
y=
12
12
3
5 2
Thus the area is A = xy = x( −5
3 x + 100) = − 3 x + 100x
(b) What are the dimensions of the fence that maximize the area?
The area function is a parabola facing downward since a = −5/3 < 0 thus the max is at the vertex
A0 = −
10
x + 100 = 0
3
10
− x = −100
3
3
x = 100( ) = 30
10
If x = 30 then y = − 35 (30) + 100 = 50.
Thus x = 30 feet and y = 50 feet gives maximum area.
Math 160 - Exam 3 Solutions - Summer 2012 - Jaimos F Skriletz
8. (20 pts)
(a) Evaluate the following indefinite integral:
Z
(6x2 − 10x + 5)dx = 2x3 − 5x2 + 5x + C
(b) Find f (x) given that f 0 (x) = 3ex − 10x + 2 and f (0) = 5
Z
f (x) =
(3ex − 10x + 2)dx = 3ex − 5x2 + 2x + C
Since f (0) = 3e0 − 5(0)2 + 2(0) + C = 3 + C = 5 then C = 5 − 3 = 2. Thus
f (x) = 3ex − 5x2 + 2x + 2
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