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The solutions to test 1 of math 220, taught by t. Pilachowski in fall 2013. It includes step-by-step calculations for determining the increasing/decreasing behavior of a function, finding the rate of change of water level in a pond, finding the derivative of a given function, finding x-intercepts, relative extrema, points of inflection, and equations of asymptotes.
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Math 220, T Pilachowski, Fall 2013
Fall 2013 MATH 220 – TEST 1(B) (0.3 – 2.7) [Pilachowski] ANSWER KEY
1.a. (14 points) Use calculus to determine whether the function y x x 3 x
3
= + − is increasing or
decreasing at x = –1.
2 2 y ′ = x + x − y ′− = − + − − = − − =− , so the function is decreasing.
1.b. (12 points) During a thunderstorm, the rise and fall of the water level (in feet) in a drainage pond
t m t t where t is time in hours. Find an equation to express how
quickly the water level is changing. Don’t forget to put the correct label on your answer.
“How quickly” refers to a rate of change: 2
m ′^ =− 2 t + feet per hour.
Note that π/2 is a constant, so its derivative equals 0.
1.c. (14 points) Given 2
x
y = x − x + , find y ′.
Rewrite:
2
2
2
1 2 3
− = − + = x − x + x
x
y x x. Then
3 2 2
− (^1) − y ′ == − x − x.
x
h x x
= 16 + , with a domain limited to x > 0,
i. Find both coordinates of any x -intercepts.
2 2
0 16 x x x
= x + ⇒ = + ⇒ − = which
has no real solution, so there are no x -intercepts.
ii. Find the x -coordinate of any relative extrema (within the limited domain x > 0), stating whether each
is a maximum or minimum. Use calculus tools, show your steps, and verify your conclusion.
2
2
1 2 = + ′= − = − = ⇒ = ⇒ =±
− − x x
x
y x x y x
The negative value is not in the domain and is
therefore extraneous.
interval
0 < x < 4
x = < x 4
sign of h ′^ ( – ) 0 ( + )
You could also have shown that the second
derivative > 0 at this x -value.
iii. Find the x -coordinate of any points of inflection (within the limited domain x > 0). Use calculus
tools, show your steps, and verify your conclusion.
3
x
y ′ = − x y ′′= − ∗− x =
− − which never = 0, so there are no points of inflection.
iv. Find the equation of any asymptotes. Show how you found your answers.
x in denominator ⇒ vertical asymptote is x = 0; as x → ∞, y → 16 x + 0 , slant asymptote is y = 16 x
y =
At (^)
the graph of h ( x ) has a minimum.
Math 220, T Pilachowski, Fall 2013
p x = − x − x + , where x
is the number of widgets sold. Find the number of widgets which will generate maximum revenue.
To find a maximum use first derivative:
2 2 R ′ x =− x − x + =− x + x − =− x + x − ⇒ x =−
The negative value is extraneous.
Selling 3 widgets generates maximum revenue.
5 g x = 2 x − 3 , use calculus to determine where the curve has a point of
inflection.
4 4 3 3 g ′ = x − ∗ = x − ⇒ g ′′= ∗ x − ∗ = x − = at 2
x =
interval
x < 2
x = < x 2
sign of g ′′^ ( – ) 0 ( + )
3.b. (4 extra credit points) Determine the equations of all asymptotes for
2 7
x
y.
(2 x – 7) in the denominator means there is a vertical asymptote at
2
x =.
As x →∞, denominator → ∞, and y → 0 so there is a horizontal asymptote at y = 0.
3.c. (10 points) Frankincents is putting up a storage building shaped like a box with a square base. Let
x = length of a side of the base and let y = height of the shed. The cement for the base costs $10 per
2 ft. The material for the four sides and the top costs $7 per
2 ft. She wants to maximize the volume
while keeping her cost to $1000. Find the objective equation and the constraint for this scenario, and
label which is which. DO NOT SOLVE; JUST THE WRITE THE OBJECTIVE AND
CONSTRAINT.
goal is to find maximum volume: objective function is V x y
area of base = area of top =
2 x ; cost of base =
2 10 x ; cost of top =
2 7 x
area of side = xy ; cost of one side = 7 xy ; cost of four sides = 4(7 xy ) = 28 xy
constraint is cost: 10 7 28 17 28 1000
2 2 2 x + x + xy = x + xy =
Concavity changes so the graph has a point of
inflection at 2
x =