Math 220 - Fall 2013 Test 1 Solutions - Prof. Timothy John Pilachowski, Exams of Calculus

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 xxxy 3
3
123 += is increasing or
decreasing at x = –1.
( ) ( ) ( )
432131211,32 2
2==+=
+=
yxxy , 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
follows the formula
( )
2
2
15
2
π
++= t
ttm 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
15
2+=
tm feet per hour.
Note that
π
/2 is a constant, so its derivative equals 0.
1.c. (14 points) Given 2
1
32
x
xxy += , find y
.
Rewrite: 2
2
2
1
32
1
32
+=+= xxx
x
xxy
. Then 3
2
2
3
22
1
==
xxy .
2.a. (20 points) Given the function
( )
x
xxh 9
16 += , with a domain limited to x > 0,
i. Find both coordinates of any x-intercepts. 22
16
9
9160
9
160 xx
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.
4
3
9160
9
16916,916 2
2
21 ±=====
+= xx
x
xyxxy
The negative value is not in the domain and is
therefore extraneous.
interval
4
3
0<< x
4
3
=x x<
4
3
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
32 18
29,916
x
xyxy ==
=
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 016,
+
xyx , slant asymptote is y = 16x
241212
3
4
9
4
3
16 =+=
+
=y
At
24,
4
3 the graph of h(x) has a minimum.
pf2

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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 + xy ′− = − + − − = − − =− , 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

follows the formula ( )

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 = xx + , find y ′.

Rewrite:

2

2

2

1 2 3

− = − + = xx + x

x

y x x. Then

3 2 2

− (^1) − y ′ == − xx.

2.a. (20 points) Given the function ( )

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

2.b. (16 points) A widget company sets the price ( p ) using the equation ( ) 2 30

p x = − xx + , where x

is the number of widgets sold. Find the number of widgets which will generate maximum revenue.

Revenue = number sold times price: R ( ) x x x x x 2 x 30 x

To find a maximum use first derivative:

2 2 Rx =− xx + =− x + x − =− x + x − ⇒ x =−

The negative value is extraneous.

Selling 3 widgets generates maximum revenue.

3.a. (14 points) Given ( ) ( )

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

2

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 =