2020+Winter+Final+Practice
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School
Concordia University *
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Course
203
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
29
Uploaded by MegaSkunkPerson34
Jan Mikhail Alexei Ong
E
WeBWorK assignment due : 04/16/2020 at 11:59pm EDT.
1.
(1 point) Suppose
y
0
is the
y
-coordinate of the point of
intersection of the graphs below. Complete the statement below
in order to correctly describe what happens to
y
0
if the value of
a
(in the blue graph of
f
(
t
) =
a
(
1
+
r
)
t
below) is increased, and
all other quantities remain the same.
As
a
increases, the value of
y
0
•
A. increases
•
B. decreases
•
C. remains the same
(click on image to enlarge)
Solution:
SOLUTION
As
a
is increased, the
y
-intercept of the blue graph increases.
The entire blue graph will be shifted up, and the point of inter-
section shifts to the left and down, so the
y
-coordinate decreases.
Correct Answers:
•
B
2.
(1 point) The population of a colony of rabbits grows ex-
ponentially.
The colony begins with 15 rabbits; 5 years later
there are 330 rabbits.
(a) Express the population of the colony of rabbits,
P
, as a
function of time,
t
, in years.
P
(
t
) =
(b) Use the graph to estimate how long it takes for the popu-
lation of rabbits to reach 1000 rabbits.
It will take
years. (round to nearest 0.01 year)
Solution:
SOLUTION
a) Since the population grows exponentially, it can be de-
scribed by
P
=
ab
t
, where
P
is the number of rabbits and
t
is the
number of years which have passed. We know that
a
represents
the initial number of rabbits, so
a
=
15 and
P
=
15
(
b
)
t
. After 5
years, there are 330 rabbits so
330
=
15
(
b
)
5
22
=
b
5
(
b
5
)
1
/
5
=
22
1
/
5
b
≈
1
.
856
From this, we know that
P
(
t
) =
15
(
1
.
856
)
t
.
b) We want to find
t
when
P
=
1000.
Using a graph of
P
=
15
(
1
.
856
)
t
, we see (figure below) that the line
P
=
1000
and
P
=
15
(
1
.
856
)
t
intersect when
t
≈
6
.
791 years.
Correct Answers:
•
15*1.856ˆt
•
6.791
3.
(1 point)
Without using a calculator, match each exponential
function with its graph.
?
e
-
0
.
3
x
?
e
-
x
?
e
0
.
3
x
?
e
x
1
(Click on graph to enlarge)
Solution:
SOLUTION
The functions given in (b) and (a) represent exponential de-
cay while the functions given in (c) and (d) represent exponen-
tial growth.
Thus, (b) and (a) correspond to (I) and (II) (not
necessarily in that order) while (c) and (d) correspond to (III)
and (IV) (not necessarily in that order).
The function in (d) grows faster than the function in (c) since
it grows continuously at a rate of 1% while the function in (d)
grows continuously by 0.3%. Since (d) has a higher continuous
growth rate, its graph must decrease the fastest to 0 as
x
→ -
∞
,
thus it has graph IV and (c) has graph III.
Graphs (I) and (II) correspond to the exponential decay for-
mulas, with graph (I) decaying at a more rapid rate (since it
increases to
∞
quicker as
x
→ -
∞
it similarly decreases to 0
faster as
x
→
+
∞
). Thus formula I corresponds to graph (b)
and formula (a) corresponds to graph II. We have:
a) II
b) I
c) III
d) IV
Correct Answers:
•
II
•
I
•
III
•
IV
4.
(1 point) If
f
(
x
) =
8
x
and
g
(
x
) =
11
x
x
+
11
, find a simplified
formula for:
g
(
f
(
x
)) =
Solution:
SOLUTION
We solve by substituting the expression
f
(
x
) =
8
x
in for
x
in
g
(
x
) =
11
x
x
+
11
:
g
(
f
(
x
)) =
11
f
(
x
)
f
(
x
)+
11
=
11
·
8
x
8
x
+
11
.
Correct Answers:
•
11*8ˆx/(8ˆx+11)
5.
(1 point) If
f
(
x
) =
e
x
/
3
,
g
(
x
) =
7
x
+
7, and
h
(
x
) =
√
x
,
Find a simplified formula for:
f
(
g
(
x
))
h
(
x
) =
Solution:
SOLUTION
We start by computing
f
(
g
(
x
)) =
e
3
g
(
x
)
=
e
3
(
7
x
+
7
)
=
e
21
x
+
21
.
Next we compute
f
(
g
(
x
))
h
(
x
)
by multiplying the expression
above by
h
(
x
) =
√
x
:
f
(
g
(
x
))
h
(
x
) =
e
21
x
+
21
·
√
x
.
Correct Answers:
•
eˆ(7/3*x+7/3)*sqrt(x)
6.
(1 point) Let
f
(
x
) =
x
4
/
3
,
g
(
x
) =
(
5
x
-
2
)
3
8
, and
h
(
x
) =
tan
(
4
x
)
.
Find values for the constants
A
and
P
which result
in the simplified expression for
h
(
x
)
/
f
(
g
(
x
))
equal to the com-
bination of functions below:
h
(
x
)
f
(
g
(
x
))
=
A
tan
(
4
x
)
(
5
x
-
2
)
P
A
=
, and
P
=
Solution:
SOLUTION
We
start
by
computing
f
(
g
(
x
))
=
(
g
(
x
))
4
/
3
=
(
5
x
-
2
)
3
8
4
/
3
=
(
5
x
-
2
)
4
(
2
3
)
4
/
3
=
(
5
x
-
2
)
4
16
.
Thus
h
(
x
)
f
(
g
(
x
)
)
=
tan
(
4
x
)
(
5
x
-
2
)
4
/
16
=
16tan
(
4
x
)
(
5
x
-
2
)
4
.
Correct Answers:
•
16
•
4
7.
(1 point) The figures below shows the graph of
f
(
x
)
in
blue and the graph of
g
(
x
)
in red:
Based on the figure above, graph the function
f
(
g
(
x
)
)
.
Which of the graphs below labeled A-E most accurately
matches your graph?
(enter the corresponding letter)
2
None of these graphs
A
B
C
D
E
Solution:
SOLUTION
We can evaluate
f
((
g
(
x
))
at several points:
f
(
g
(
0
)) =
f
(
1
)
≈ -
0
.
5
f
(
g
(
1
)) =
f
(
0
.
5
)
≈ -
0
.
8
f
(
g
(
2
)) =
f
(
0
) =
-
1
.
Also observe
g
(
x
)
is linear and
f
(
x
)
is a quadratic function. Sub-
stituting the
x
’s in the formula for
f
(
x
)
with the linear equation
for
g
(
x
)
will result in a quadratic expression. Thus the graph of
f
(
g
(
x
))
must be parabola which goes through the points above.
Thus we can see that graph C is the correct graph.
Correct Answers:
•
C
8.
(1 point) Consider the graphs of
y
=
f
(
x
)
and
y
=
g
(
x
)
sketch in red and blue respectively on the graph below:
For each function, enter the letter of the graph A - F which rep-
resents it. Clearly not all of the graphs are matched with one of
the equations.
f
(
x
)+
g
(
x
)
f
(
x
)
-
g
(
x
)
g
(
x
)
-
f
(
x
)
A
B
C
D
E
F
(click on each individual graph to enlarge)
Solution:
SOLUTION
f
(
x
)+
g
(
x
)
C
f
(
x
)
-
g
(
x
)
D
g
(
x
)
-
f
(
x
)
F
Correct Answers:
•
C
•
D
•
F
9.
(1 point) Is the following statement true or false?
If
f
(
x
)
·
g
(
x
)
is an odd function, then both
f
(
x
)
and
g
(
x
)
must
be odd functions. Be sure you can explain your answer.
•
A. True
•
B. False
Solution:
SOLUTION
The statement is false.
For example, if
f
(
x
) =
x
and
g
(
x
) =
x
2
, then
f
(
x
)
·
g
(
x
) =
x
3
.
In this case,
f
(
x
)
·
g
(
x
)
is
an odd function, but
g
(
x
)
is an even function.
Correct Answers:
•
B
10.
(1 point) Let
f
(
x
) =
x
+
1
x
2
+
14
x
+
49
. Use interval notation to
indicate the domain of
f
(
x
)
.
Note:
You should enter your answer in
interval notation
.
If the set is empty, enter ”” without the quotation marks.
Domain =
Correct Answers:
•
(-infinity,-7) U (-7,infinity)
11.
(1 point) Find an expression for the function
f
(
x
)
whose
graph is given by the bottom half of the parabola
x
+(
y
-
7
)
2
=
0
.
f
(
x
)
=
Correct Answers:
•
-sqrt(-x) + 7
12.
(1 point) A Norman window has the shape of a rectangle
surmounted by a semicircle. If the perimeter of the window is
24 feet, express the area
A
of the window as a function of the
width
x
(across the base) of the window.
A
(
x
)
=
Correct Answers:
•
x*(24-x-pi*x/2)/2 + pi*xˆ2/8
3
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13.
(1 point) The function
f
(
x
) =
√
2
x
-
x
2
is given graphed
below:
Note: Click on graph for larger version in new browser
window.
(A) Starting with the formula for
f
(
x
)
, find a formula for
g
(
x
)
, which is graphed below:
Note: Click on graph for larger version in new browser
window.
g
(
x
)
=
(B) Starting with the formula for
f
(
x
)
, find a formula for
h
(
x
)
, which is graphed below:
Note: Click on graph for larger version in new browser
window.
h
(
x
)
=
Correct Answers:
•
3*(2*(x-2)-(x-2)**2)**(1/2)
•
2+-1*(2*(x--3)-(x--3)**2)**(1/2)
14.
(1 point) Suppose that
f
(
x
) =
√
8
x
-
8
and
g
(
x
) =
2
x
2
-
7
.
For each function
h
given below, find a formula for
h
(
x
)
and the
domain of
h
. Enter the domains using
interval notation
.
(A)
h
(
x
) = (
f
◦
g
)(
x
)
.
h
(
x
)
=
Domain =
(B)
h
(
x
) = (
g
◦
f
)(
x
)
.
h
(
x
)
=
Domain =
(C)
h
(
x
) = (
f
◦
f
)(
x
)
.
h
(
x
)
=
Domain =
(D)
h
(
x
) = (
g
◦
g
)(
x
)
.
h
(
x
)
=
Domain =
Correct Answers:
•
sqrt(8*2*xˆ2+8*(-7)+(-8))
•
(-infinity,-2] U [2,infinity)
4
•
2*(8*x+(-8))+(-7)
•
[1,infinity)
•
sqrt(8*sqrt(8*x+(-8))+(-8))
•
[1.125,infinity)
•
2*(2*xˆ2+(-7))ˆ2+(-7)
•
(-infinity,infinity)
15.
(1 point) The expression
(
5
-
7
)
4
x
can be written as 5
f
(
x
)
,
where
f
(
x
)
is a function of
x
. Find
f
(
x
)
.
f
(
x
)
=
Correct Answers:
•
(-7)*(4)*x
16.
(1 point)
Consider the angle
θ
, which is labeled as
Q
in blue
on the graph, with corresponding point
P
on the circle.
Sketch each of the angles given below, then select the
point on the circle that best corresponds to the angle.
Angle
Point
π
2
+
θ
?
π
2
-
θ
?
π
+
θ
?
π
-
θ
?
(Click on graph to enlarge)
Solution:
SOLUTION
(a) F
(b) A
(c) D
(d) G
Correct Answers:
•
F
•
A
•
D
•
G
17.
(1 point)
Part 1 of 3:
In this multi-part problem, we will use algebra to verify the
identity
sin
(
t
)
1
-
cos
(
t
)
=
1
+
cos
(
t
)
sin
(
t
)
.
First, using algebra we may rewrite the equation above as
sin
(
t
) =
1
+
cos
(
t
)
sin
(
t
)
·
Solution:
SOLUTION
Alternately, we can directly derive this by multiplying the
denominator by 1
+
cos
(
t
)
to get sin
2
(
t
)
:
sin
(
t
)
1
-
cos
(
t
)
=
sin
(
t
)
1
-
cos
(
t
)
(
1
+
cos
(
t
))
(
1
+
cos
t
)
=
sin
(
t
)(
1
+
cos
(
t
))
1
-
cos
2
t
=
sin
(
t
)(
1
+
cos
(
t
))
sin
2
(
t
)
=
1
+
cos
(
t
)
sin
(
t
)
.
Correct Answers:
•
1-cos(t)
18.
(1 point) Find the inverse function (if it exists) of
h
(
x
) =
4
x
4
-
5. If the function is not invertible, enter
NONE
.
h
-
1
(
x
) =
(Write your inverse function in terms of the independent vari-
able x.)
Solution:
SOLUTION
This function is not invertible. The graph does not pass the
horizontal line test. If we try to solve for
x
in
y
=
4
x
4
-
5, we
get the following:
y
=
4
x
4
-
5
y
+
5
=
4
x
4
x
4
=
y
+
5
4
x
=
±
4
r
y
+
5
4
Since this is not one to one, we say the inverse
h
-
1
(
x
)
does not
exist.
Correct Answers:
•
NONE
5
19.
(1 point) If
t
=
g
(
v
)
represents the time in hours it takes
to drive to the next town at velocity
v
mph.
Which if the following statement(s) correctly explain the mean-
ing of
g
-
1
(
t
)
? Check all that may apply.
•
A. The velocity in mph of the car if it takes
t
minutes to
drive to the next town.
•
B. How many hours it takes to reach a velocity of
t
mph.
•
C. The velocity in mph of the car if it takes
t
hours to
drive to the next town.
•
D. The number of hours it takes to drive
t
miles.
•
E. The velocity in mph of the car after you have driven
for
t
miles.
•
F. None of the above
Solution:
SOLUTION
The inverse function
g
-
1
(
t
)
represents the velocity needed
for a trip of
t
hours. Its units are mph.
Correct Answers:
•
C
20.
(1 point)
If you are given the graph of
f
, how do you find the graph of
f
-
1
?
(a) Reflect it over the x-axis.
(b) Reflect it over the y-axis.
(c) Reflect it over
y
=
x
.
(d) Reflect it over
y
=
-
x
.
Correct Answers:
•
c
21.
(1 point)
A function is given by a table of values, a graph, a formula,
or a verbal description. Determine whether it is one-to-one. If
it is one-to-one, enter ”y” below. If not, enter ”n” below.
Correct Answers:
•
y
22.
(1 point)
A function is given by a table of values, a graph, a formula,
or a verbal description. Determine whether it is one-to-one. If
it is one-to-one, enter ”y” below. If not, enter ”n” below.
Correct Answers:
•
n
23.
(1 point)
Find a formula for the inverse of the function.
f
(
x
) =
4
x
-
1
2
x
+
3
.
f
-
1
(
x
) =
Correct Answers:
•
(3x+1)/(4-2x)
24.
(1 point)
Find a formula for the inverse of the function.
f
(
x
) =
e
x
3
.
f
-
1
(
x
) =
Correct Answers:
•
(ln(x))ˆ(1/3)
25.
(1 point)
Find a formula for the inverse of the function.
f
(
x
) =
ln
(
x
+
8
)
.
f
-
1
(
x
) =
Correct Answers:
•
eˆx-(8)
26.
(1 point)
Express the given quantity as a single logarithm.
ln
x
+
2ln
y
-
7ln
z
Correct Answers:
•
ln(x*yˆ{2}/ zˆ{7})
27.
(1 point)
Simplify cos
(
sin
-
1
x
)
.
Correct Answers:
•
sqrt(1-xˆ2)
6
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28.
(1 point)
Let
f
(
x
) =
x
3
-
14
x
. Calculate the difference quotient
f
(
2
+
h
)
-
f
(
2
)
h
for
h
=
.
1
h
=
.
01
h
=
-
.
01
h
=
-
.
1
If someone now told you that the derivative (slope of the tangent
line to the graph) of
f
(
x
)
at
x
=
2 was an integer, what would
you expect it to be?
Correct Answers:
•
-1.39000000000003
•
-1.93989999999999
•
-2.05990000000007
•
-2.59
•
-2
29.
(1 point) Let
F
be the function whose graph is shown
below. Evaluate each of the following expressions.
(If a limit does not exist or is undefined, enter “DNE”.)
1.
lim
x
→-
1
-
F
(
x
)
=
2.
lim
x
→-
1
+
F
(
x
)
=
3.
lim
x
→-
1
F
(
x
)
=
4.
F
(
-
1
)
=
5.
lim
x
→
1
-
F
(
x
)
=
6.
lim
x
→
1
+
F
(
x
)
=
7.
lim
x
→
1
F
(
x
)
=
8.
lim
x
→
3
F
(
x
)
=
9.
F
(
3
)
=
The graph of
y
=
F
(
x
)
.
Correct Answers:
•
0
•
0
•
0
•
1
•
2
•
3
•
DNE
•
0
•
DNE
30.
(1 point) Let
f
(
x
) =
13
if
x
<
-
9
-
x
+
4
if
-
9
≤
x
<
2
-
2
if
x
=
2
4
if
x
>
2
.
Sketch the graph of this function and find the following limits,
if they exist.
(
If a limit does not exist, enter
DNE
.)
1.
lim
x
→-
9
-
f
(
x
)
=
2.
lim
x
→-
9
+
f
(
x
)
=
3.
lim
x
→-
9
f
(
x
)
=
4.
lim
x
→
2
-
f
(
x
)
=
5.
lim
x
→
2
+
f
(
x
)
=
6.
lim
x
→
2
f
(
x
)
=
Correct Answers:
•
13
•
13
•
13
•
2
•
4
•
DNE
31.
(1 point) Evaluate the limit
lim
x
→-
6
x
2
+
7
x
+
6
x
+
6
Correct Answers:
•
-5
32.
(1 point) Evaluate the limit
lim
a
→
1
a
3
-
1
a
2
-
1
Correct Answers:
•
1.5
33.
(1 point) Evaluate the limit
lim
x
→
∞
4
+
5
x
5
-
6
x
Enter
I
for
∞
,
-I
for
-
∞
, and
DNE
if the limit does not exist.
Limit =
Correct Answers:
•
-0.833333333333333
7
34.
(1 point) Evaluate the limit
lim
x
→
∞
10
x
+
9
5
x
2
-
5
x
+
11
Enter
I
for
∞
,
-I
for
-
∞
, and
DNE
if the limit does not exist.
Limit =
Correct Answers:
•
0
35.
(1 point) The horizontal asymptotes of the curve
y
=
4
x
(
x
4
+
1
)
1
4
are given by
y
1
=
and
y
2
=
where
y
1
>
y
2
.
Correct Answers:
•
4
•
-4
36.
(1 point)
Evaluate the following limits.
(a) lim
x
→
∞
8
e
x
+
6
=
(b)
lim
x
→-
∞
8
e
x
+
6
=
[NOTE: If needed, enter INF or infinity for
∞
and -INF or
-infinity for
-
∞
.]
Correct Answers:
•
0
•
1.33333
37.
(1 point)
Evaluate the following limits. If needed, enter INF for
∞
and
MINF for
-
∞
.
(a)
lim
x
→
∞
p
x
2
+
9
x
+
1
-
x
=
(b)
lim
x
→-
∞
p
x
2
+
9
x
+
1
-
x
=
Correct Answers:
•
4.5
•
INF
38.
(1 point)
For the function
g
whose graph is given, state the value of the
given quantity, if it exists. If it does not exist, enter ”n” below.
(a) lim
t
→
0
-
g
(
t
)
(b) lim
t
→
0
+
g
(
t
)
(c) lim
t
→
0
g
(
t
)
(d) lim
t
→
2
-
g
(
t
)
(e) lim
t
→
2
+
g
(
t
)
(f) lim
t
→
2
g
(
t
)
(g)
g
(
2
)
(h) lim
t
→
4
g
(
t
)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Correct Answers:
•
-1
•
-2
•
n
•
2
•
0
•
n
•
1
•
3
39.
(1 point)
Guess the value of the limit (if it exists) by evaluating the
function at values close to where the limit is to be done. If it
does not exist, enter ”n” below. If the answer is infinite, use ”i”
to represent infinity.
lim
x
→
0
cos
x
-
1
x
Correct Answers:
•
0
40.
(1 point)
Evaluate the following limit. If the answer is positive infi-
nite, type ”I”; if negative infinite, type ”N”; and if it does not
8
exist, type ”D”.
lim
x
→
∞
s
12
x
3
-
5
x
+
4
1
+
9
x
2
+
3
x
3
Correct Answers:
•
2
41.
(1 point) Let
f
(
x
) =
11
x
2
-
7
x
.
(a) Use the limit process to find the slope of the line tangent
to the graph of
f
at
x
=
3.
Slope at
x
=
3:
(b) Find an equation of the line tangent to the graph of
f
at
x
=
3.
Tangent line:
y
=
Correct Answers:
•
59
•
59*(x-3)+78
42.
(1 point) Let
f
(
x
) =
5
x
4
-
2.
(a) Use the limit process to find the slope of the line tangent
to the graph of
f
at
x
=
2.
Slope at
x
=
2:
(b) Find an equation of the line tangent to the graph of
f
at
x
=
2.
Tangent line:
y
=
Correct Answers:
•
160
•
160*(x-2)+78
43.
(1 point) Let
f
and
g
be functions that satisfy
f
0
(
2
) =
-
8
and
g
0
(
2
) =
1. Find
h
0
(
2
)
for each function
h
given below:
(A)
h
(
x
) =
10
f
(
x
)
.
h
0
(
2
)
=
(B)
h
(
x
) =
-
5
g
(
x
)
.
h
0
(
2
)
=
(C)
h
(
x
) =
6
f
(
x
)+
13
g
(
x
)
.
h
0
(
2
)
=
(D)
h
(
x
) =
10
g
(
x
)
-
7
f
(
x
)
.
h
0
(
2
)
=
(E)
h
(
x
) =
12
f
(
x
)+
7
g
(
x
)
-
4.
h
0
(
2
)
=
(F)
h
(
x
) =
-
3
g
(
x
)
-
5
f
(
x
)+
4
x
.
h
0
(
2
)
=
Correct Answers:
•
-80
•
-5
•
-35
•
66
•
-89
•
41
44.
(1 point)
Identify the graphs A (blue), B(red) and C (green) as the
graphs of a function
f
(
x
)
and its derivatives
f
0
(
x
)
and
f
00
(
x
)
.
(Clicking on the sketch will give you a version of the picture in
a separate window.)
is the graph of the function,
f
(
x
)
.
is the graph of the function’s first derivative,
f
0
(
x
)
.
is the graph of the function’s second derivative,
f
00
(
x
)
.
Hint:
Remember that
f
0
(
x
)
is itself a function, and we can
find the derivative of the function
f
0
(
x
)
, which is called the sec-
ond derivative of the function
f
(
x
)
and denoted by
f
00
(
x
)
.
Correct Answers:
•
C
•
A
•
B
45.
(1 point) Let
f
(
x
) =
√
2
+
2
x
f
0
(
3
) =
Correct Answers:
•
0.353553390593274
9
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46.
(1 point) Let
f
(
x
) =
√
29
-
x
The slope of the tangent line to the graph of
f
(
x
)
at the point
(
4
,
5
)
is
.
The equation of the tangent line to the graph of
f
(
x
)
at
(
4
,
5
)
is
y
=
mx
+
b
for
m
=
and
b
=
.
Hint: the slope at
x
=
4 is given by
m
=
lim
h
→
0
f
(
4
+
h
)
-
f
(
4
)
h
Correct Answers:
•
-0.1
•
-0.1
•
5.4
47.
(1 point) A function
f
(
x
)
is said to have a
removable
discontinuity at
x
=
a
if:
1.
f
is either not defined or not continuous at
x
=
a
.
2.
f
(
a
)
could either be defined or redefined so that the new
function is continuous at
x
=
a
.
Let
f
(
x
) =
(
9
x
+
-
8
x
+
18
x
(
x
-
2
)
,
if
x
6
=
0
,
2
2
,
if
x
=
0
Show that
f
(
x
)
has a removable discontinuity at
x
=
0 and deter-
mine what value for
f
(
0
)
would make
f
(
x
)
continuous at
x
=
0.
Must redefine
f
(
0
) =
.
Hint: Try combining the fractions and simplifying.
The discontinuity at
x
=
2 is not a removable discontinuity, just
in case you were wondering.
Correct Answers:
•
-0.5
48.
(1 point)
A rock is thrown off of a 100 foot cliff with an upward velocity
of 50 m/s. As a result its height after t seconds is given by the
formula:
h
(
t
) =
100
+
50
t
-
5
t
2
What is its height after 5 seconds?
What is its velocity after 5 seconds?
(Positive velocity means it is on the way up, negative velocity
means it is on the way down.)
Correct Answers:
•
225
•
0
49.
(1 point)
(a) From the graph of
f
, state whether or not
f
is continuous
(enter ”y” for yes, ”n” for no below) at:
Part (a):
(1)
(2)
(3)
(4)
(5)
(b) For each of the numbers in part (a), determine if
f
is
continuous from the right, or from the left, or neither (enter ”r”,
”l”, or ”n”, respectively below. In the case of continuity at a
point from both sides, enter ”b”.)
Part (b):
(1)
(2)
(3)
(4)
(5)
Correct Answers:
•
n
•
n
•
y
•
n
•
n
•
n
•
l
•
b
•
r
•
r
50.
(1 point)
If
f
and
g
are continuous functions with
f
(
0
) =
3 and
lim
x
→
0
f
(
x
)
g
(
x
) =
6, find
g
(
0
)
.
g
(
0
) =
Correct Answers:
•
2
51.
(1 point)
(a) Find the slope of the tangent line to the curve
y
=
2
x
+
3
at
the point where
x
=
a
.
(b-d) Find the slopes of the tangent lines at the points whose
x-coordinates are: (b) -1, (c) 0, and (d) 1.
(a)
(b)
(c)
(d)
Correct Answers:
•
-2/(a+3)ˆ2
•
-0.5
10
•
-0.222222222222222
•
-0.125
52.
(1 point)
If
f
(
x
) =
3
x
2
-
5
x
, find
f
0
(
2
)
f
0
(
2
) =
Use it to find an equation of the tangent line to the parabola
y
=
3
x
2
-
5
x
at the point (2,2).
y
=
Correct Answers:
•
7
•
7 x - 12
53.
(1 point)
The following limit represents the derivative of some func-
tion
f
at some number
a
.
lim
h
→
0
(
1
+
h
)
10
-
1
h
What are
f
and
a
?
f
(
x
)
=
a
=
Correct Answers:
•
xˆ10
•
1
54.
(1 point) If
f
(
x
) =
4
x
1
+
x
2
find
f
0
(
4
)
.
f
0
(
4
)
=
Use this to find the equation of the tangent line to the curve
y
=
4
x
1
+
x
2
at the point
(
4
,
16
17
)
. The equation of this tangent line
can be written in the form
y
=
mx
+
b
.
The equation of the tangent line is
y
=
.
Correct Answers:
•
4*(1-4ˆ2)/[(1+4ˆ2)ˆ2]
•
0.941176--0.207612*4
55.
(1 point) Let
f
(
t
) =
√
2
t
7
. Find
f
0
(
t
)
.
f
0
(
t
)
=
Find
f
0
(
3
)
.
f
0
(
3
)
=
Correct Answers:
•
-(1.41421*7*tˆ6/[(tˆ7)ˆ2])
•
-0.00150884
56.
(1 point) Let
f
(
x
) =
6
x
5
√
x
+
7
x
3
√
x
.
f
0
(
x
) =
[NOTE: Your answer should be a function in terms of the vari-
able ’
x
’ and not a number!]
Correct Answers:
•
6*(5+1/2)*xˆ(5-1/2)-7*(3+1/2)/[xˆ(3+3/2)]
57.
(1 point) If
f
(
x
) =
3
-
x
2
2
+
x
2
find
f
0
(
x
)
.
f
0
(
x
)
=
Find
f
0
(
1
)
.
f
0
(
1
)
=
Correct Answers:
•
[-2*x*(2+xˆ2)-(3-xˆ2)*2*x]/[(2+xˆ2)ˆ2]
•
[-2*1*(2+1ˆ2)-(3-1ˆ2)*2*1]/[(2+1ˆ2)ˆ2]
58.
(1 point) If
f
(
x
) =
cos
x
-
2tan
x
, then find:
f
0
(
x
) =
f
0
(
1
) =
Correct Answers:
•
-[sin(x)]-2*[sec(x)]ˆ2
•
-[sin(1)]-2/([cos(1)]ˆ2)
59.
(1 point) Let
f
(
x
) =
5sin
(
x
)
2
+
cos
(
x
)
. Find the following:
1.
f
0
(
x
)
=
2.
f
0
(
3
)
=
Correct Answers:
•
[5*cos(x)*[2+cos(x)]+5*sin(x)*sin(x)]/([2+cos(x)]ˆ2)
•
-4.80331
60.
(1 point) Let
f
(
x
) =
4tan
(
x
)
x
. Find the following:
1.
f
0
(
x
)
=
2.
f
0
(
5
)
=
Correct Answers:
•
(4*[sec(x)]ˆ2*x-4*tan(x))/(xˆ2)
•
10.4832
11
61.
(1 point) Let
f
(
x
) =
9
x
(
sin
(
x
)+
cos
(
x
))
. Find the fol-
lowing:
1.
f
0
(
x
)
=
2.
f
0
(
-
π
4
)
=
Correct Answers:
•
9*[sin(x)+cos(x)]+9*x*[cos(x)-sin(x)]
•
-9.99649
62.
(1 point) If
f
(
x
) =
3
x
sin
(
x
)
cos
(
x
)
, find
f
0
(
x
)
.
f
0
(
x
) =
Find
f
0
(
3
)
.
f
0
(
3
) =
Correct Answers:
•
3*sin(x)*cos(x)+3*x*([cos(x)]ˆ2-[sin(x)]ˆ2)
•
8.22241
63.
(1 point) Let
f
(
x
) = (
4
x
2
-
6
)
5
(
8
x
2
-
2
)
9
f
0
(
x
) =
Correct Answers:
•
(4*xˆ2+-6)ˆ4 * (8*xˆ2+-2)ˆ8 * (896*xˆ3 + -944*x)
64.
(1 point) Let
f
(
x
) =
ln
(
x
7
)
f
0
(
x
) =
f
0
(
e
2
) =
Correct Answers:
•
7/x
•
7*eˆ-2
65.
(1 point) At noon, ship A is 50 nautical miles due west of
ship B. Ship A is sailing west at 18 knots and ship B is sailing
north at 23 knots. How fast (in knots) is the distance between
the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nauti-
cal mile per hour.)
Note: Draw yourself a diagram which shows where the ships are
at noon and where they are ”some time” later on. You will need
to use geometry to work out a formula which tells you how far
apart the ships are at time t, and you will need to use ”distance
= velocity * time” to work out how far the ships have travelled
after time t.
Correct Answers:
•
28.8054898194543
66.
(1 point) A plane flying with a constant speed of 4
km/min passes over a ground radar station at an altitude of 7
km and climbs at an angle of 25 degrees.
At what rate is the distance from the plane to the radar station
increasing 5 minutes later?
The distance is increasing at
km/min.
Hint:
Hint:
The law of cosines for a triangle is
c
2
=
a
2
+
b
2
-
2
ab
cos
(
θ
)
where
θ
is the angle between the sides of length a and b.
Correct Answers:
•
3.8555
67.
(1 point) Suppose
xy
=
2 and
dy
dt
=
-
4. Find
dx
dt
when
x
=
-
2.
dx
dt
=
Correct Answers:
•
8
68.
(1 point) Let
f
(
x
) =
4log
9
(
x
)
f
0
(
x
) =
f
0
(
3
) =
Correct Answers:
•
4/(2.19722*x)
•
0.606826
69.
(1 point) A street light is at the top of a 12 ft tall pole.
A woman 6 ft tall walks away from the pole with a speed of 5
ft/sec along a straight path. How fast is the tip of her shadow
moving when she is 35 ft from the base of the pole?
Note: You should draw a picture of a right triangle with the
vertical side representing the pole, and the other end of the hy-
potenuse representing the tip of the woman’s shadow. Where
does the woman fit into this picture?
Label her position as a
variable, and label the tip of her shadow as another variable.
You might like to use similar triangles to find a relationship be-
tween these two variables.
Correct Answers:
•
10
70.
(1 point) A spherical snowball is melting so that its di-
ameter is decreasing at rate of 0.2 cm/min.
At what is the rate is the volume of the snowball changing
when the diameter is 11 cm?
The volume is changing at a rate of
cm
3
/
min.
Correct Answers:
•
-38.0133
12
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71.
(1 point) Let
f
(
x
) =
tan
-
1
(
sin
(
2
x
))
. Find
f
0
(
x
)
.
f
0
(
x
) =
Correct Answers:
•
1/(1+[sin(2*x)]ˆ2)*2*cos(2*x)
72.
(1 point) Let
f
(
x
) =
3sec
-
1
(
4
x
)
. Find
f
0
(
x
)
.
f
0
(
x
) =
Find
f
0
(
1
)
.
f
0
(
1
) =
Correct Answers:
•
3*4*1/[|4*x|*sqrt((4*x)ˆ2-1)]
•
0.774597
73.
(1 point) Let
f
(
x
) =
x
3
tan
-
1
(
7
x
)
f
0
(
x
) =
Solution: Solution:
Using the product and chain rules, we
see
f
0
(
x
) =
3
x
2
tan
-
1
(
7
x
)+
x
3
7
1
+
49
x
2
Correct Answers:
•
3*xˆ2*atan(7*x)+xˆ3*7/(1+49*xˆ2)
74.
(1 point) Suppose
y
=
sinh
(
x
2
+
x
)
. Find
D
x
y
.
Answer:
D
x
y
=
.
Hint:
Recall the derivative of sinh
(
x
)
and use the chain rule.
Solution:
Solution:
D
x
sinh
(
x
2
+
x
) = (
2
x
+
1
)(
sinh
(
x
2
+
x
))
by the chain
rule.
Correct Answers:
•
(2*x+1)*cosh(x**2+x)
75.
(1 point) If
f
(
x
) =
6arctan
(
7sin
(
2
x
))
, find
f
0
(
x
)
.
Correct Answers:
•
6*7*2*cos(2*x)/(1+7*7*(sin(2*x))**2)
76.
(1 point) a) Find
tan
(
sin
-
1
(
1
6
)+
cos
-
1
(
5
9
)
)
=
.
(Make sure your answer is an algebraic expression with
square roots but without trigonometric or inverse trignometric
functions.)
b) Express in terms of
x
:
sin
(
2tan
-
1
(
x
)
)
=
.
Correct Answers:
•
(1/sqrt(2*1*5+5**2)+sqrt(2*5*4+4**2)/5)
/(1-1/sqrt(2*1*5+5**2)*sqrt(2*5*4+4**2)/5)
•
2*x/(1+x**2)
77.
(1 point) a) Let
f
(
x
) =
1
6
tan
-
1
x
6
.
Find
f
0
(
x
)
=
.
b) Let
g
(
x
) =
x
2
√
36
-
x
2
+
18sin
-
1
x
6
.
Find
g
0
(
x
)
=
.
Correct Answers:
•
1/(6ˆ2+xˆ2)
•
sqrt(6ˆ2-xˆ2)
78.
(1 point) Let
f
(
x
) =
(
x
2
+
5
)
ln
(
x
)
. Find
f
0
(
x
)
.
f
0
(
x
) =
Correct Answers:
•
[1/x*ln(xˆ2+5)+1/(xˆ2+5)*2*x*ln(x)]*exp(ln(x)*ln(xˆ2+5))
79.
(1 point) Let
f
(
x
) =
x
2
x
. Find
f
0
(
x
)
.
f
0
(
x
) =
Correct Answers:
•
[2*ln(x)+2*x*1/x]*exp(2*x*ln(x))
80.
(1 point) Let
y
=
4
x
2
+
5
x
+
4.
Find the differential
dy
when
x
=
3 and
dx
=
0
.
3
Find the differential
dy
when
x
=
3 and
dx
=
0
.
6
Correct Answers:
•
8.7
•
17.4
81.
(1 point) Let
y
=
2
x
2
.
Find the change in
y
,
Δ
y
when
x
=
3 and
Δ
x
=
0
.
1
Find the differential
dy
when
x
=
3 and
dx
=
0
.
1
Correct Answers:
•
1.22
•
1.2
82.
(1 point) Use linear approximation, i.e.
the tangent
line, to approximate
3
√
1
.
03 as follows. Let
f
(
x
) =
3
√
x
and find
the equation of the tangent line to
f
(
x
)
at
x
=
1 in the form
y
=
mx
+
b
.
Note:
The values of
m
and
b
are rational numbers which can be
computed by hand. You need to enter expressions which give
m
and
b exactly
. You may not have a decimal point in the answers
to either of these parts.
m
=
b
=
Using these values, find the approximation.
3
√
1
.
03
≈
Note:
You can enter decimals for the last part, but it will has to
be entered to very high precision (correct for 6 places past the
decimal point).
Correct Answers:
13
•
1/(3*1ˆ2)
•
1-1/(3*1ˆ2)
•
1.01
83.
(1 point) The linearization at
a
=
0 to sin
(
7
x
)
is
A
+
Bx
.
Compute
A
and
B
.
A
=
B
=
Correct Answers:
•
0
•
7
84.
(1 point) Answer the following questions for the function
f
(
x
) =
x
p
x
2
+
16
defined on the interval
[
-
7
,
5
]
.
a.)
f
(
x
)
is concave down on the region
.
b.)
f
(
x
)
is concave up on the region
.
c.)
The minimum for this function occurs at
.
d.)
The maximum for this function occurs at
.
Note:
Your answer to parts
a
and
b
must be given in
interval notation
.
Correct Answers:
•
(-7,0)
•
(0,5)
•
-7
•
5
85.
(1 point) The function
f
(
x
) =
-
2
x
3
+
36
x
2
-
210
x
+
5
has one local minimum and one local maximum.
This function has a local minimum at
x
equals
with value
and a local maximum at
x
equals
with value
Correct Answers:
•
5
•
-395
•
7
•
-387
86.
(1 point) Consider the function
f
(
x
) =
12
x
5
+
30
x
4
-
160
x
3
+
3. For this function there are four important intervals:
(
-
∞
,
A
]
,
[
A
,
B
]
,
[
B
,
C
]
, and
[
C
,
∞
)
where
A
,
B
, and
C
are the criti-
cal numbers.
Find
A
and
B
and
C
At each critical number
A
,
B
, and
C
does
f
(
x
)
have a local min,
a local max, or neither? Type in your answer as LMIN, LMAX,
or NEITHER.
At
A
At
B
At
C
Correct Answers:
•
-4
•
0
•
2
•
LMAX
•
NEITHER
•
LMIN
87.
(1 point) Find the linear approximation of
f
(
x
) =
ln
x
at
x
=
1 and use it to estimate ln
(
1
.
37
)
.
L
(
x
) =
ln1
.
37
≈
Correct Answers:
•
x-1
•
0.37
88.
(1 point) Find the absolute maximum and absolute mini-
mum of
g
(
t
) =
t
√
3
-
t
on the interval
(
0
.
1
,
2
.
3
)
.
Enter
DNE
if the absolute maximum or minimum does not exist.
The absolute max occurs at
t
=
The absolute min occurs at
t
=
Correct Answers:
•
6/3
•
DNE
89.
(1 point)
14
Graphs A and B are approximate graphs of
f
and
f
0
for
f
(
x
) =
x
2
-
6
x
+
19.
So
f
is increasing (and
f
0
is positive) on the interval
(
a
,
∞
)
for
a
=
.
Correct Answers:
•
3
90.
(1 point) Consider the function
f
(
x
) =
12
x
5
+
30
x
4
-
160
x
3
+
5.
f
(
x
)
has inflection values at (reading from left to right)
x
=
D
,
E
, and
F
where
D
is
and
E
is
and
F
is
For each of the following intervals, tell whether
f
(
x
)
is concave
up (type in CU) or concave down (type in CD).
(
-
∞
,
D
]
:
[
D
,
E
]
:
[
E
,
F
]
:
[
F
,
∞
)
:
Correct Answers:
•
-2.88600093632938
•
0
•
1.38600093632938
•
CD
•
CU
•
CD
•
CU
91.
(1 point) Consider the function
f
(
x
) =
4
(
x
-
3
)
2
/
3
. For
this function there are two important intervals:
(
-
∞
,
A
)
and
(
A
,
∞
)
where
A
is a critical number.
Find
A
For each of the following intervals, tell whether
f
(
x
)
is increas-
ing (type in INC) or decreasing (type in DEC).
(
-
∞
,
A
)
:
(
A
,
∞
)
:
For each of the following intervals, tell whether
f
(
x
)
is concave
up (type in CU) or concave down (type in CD).
(
-
∞
,
A
)
:
(
A
,
∞
)
:
Correct Answers:
•
3
•
DEC
•
INC
•
CD
•
CD
92.
(1 point) Suppose that
f
(
x
) =
ln
(
4
+
x
2
)
(A) Use interval notation to indicate where
f
(
x
)
is concave
up.
Note:
When using interval notation in WeBWorK, you use
I
for
∞
,
-I
for
-
∞
, and
U
for the union symbol. If there are no
values that satisfy the required condition, then enter ”” without
the quotation marks.
Concave up:
(B) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(C) Find all inflection points of
f
.
If there are no inflection
points, enter -1000. If there are more than one, enter them sep-
arated by commas.
Inflection point(s) at
x
=
Correct Answers:
•
(-2,2)
•
(-infinity,-2) U (2,infinity)
•
-2, 2
93.
(1 point) Let
f
(
x
) =
x
2
-
3
x
-
40.
Find the open in-
tervals on which
f
is concave up (down). Then determine the
x
-coordinates of all inflection points of
f
.
1.
f
is concave up on the intervals
2.
f
is concave down on the intervals
3.
The inflection points occur at
x
=
Notes:
In the first two, your answer should either be a single
interval, such as (0,1), a comma separated list of intervals, such
as (-inf, 2), (3,4), or the word “none”.
In the last one, your answer should be a comma separated list
of
x
values or the word “none”.
Correct Answers:
•
(-infinity,infinity)
•
NONE
•
NONE
15
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94.
(1 point) Let
f
(
x
) =
x
3
-
5
x
2
+
7
x
+
5. Find the open
intervals on which
f
is concave up (down). Then determine the
x
-coordinates of all inflection points of
f
.
1.
f
is concave up on the intervals
2.
f
is concave down on the intervals
3.
The inflection points occur at
x
=
Notes:
In the first two, your answer should either be a single
interval, such as (0,1), a comma separated list of intervals, such
as (-inf, 2), (3,4), or the word “none”.
In the last one, your answer should be a comma separated list
of
x
values or the word “none”.
Correct Answers:
•
(1.66667,infinity)
•
(-infinity,1.66667)
•
1.66667
95.
(1 point) Let
f
(
x
) =
1
5
x
2
+
6
.
Find the open inter-
vals on which
f
is concave up (down). Then determine the
x
-
coordinates of all inflection points of
f
.
1.
f
is concave up on the intervals
2.
f
is concave down on the intervals
3.
The inflection points occur at
x
=
Notes:
In the first two, your answer should either be a single
interval, such as (0,1), a comma separated list of intervals, such
as (-inf, 2), (3,4), or the word “none”.
In the last one, your answer should be a comma separated list
of
x
values or the word “none”.
Correct Answers:
•
(-infinity,-0.632456), (0.632456,infinity)
•
(-0.632456,0.632456)
•
-0.632456, 0.632456
96.
(1 point)
For the function
f
given above, determine whether the fol-
lowing conditions are true.
Input
T
if the condition is true,
otherwise input
F
.
(a)
f
0
(
-
1
) =
0;
(b)
f
0
(
1
)
does not exist;
(c)
f
0
(
x
)
<
0 if
|
x
|
<
1;
(d)
f
0
(
x
)
>
0 if
|
x
|
>
1;
(e)
f
00
(
x
)
<
0 if
x
6
=
1.
.
Correct Answers:
•
T
•
T
•
T
•
T
•
T
97.
(1 point) Answer the following questions for the function
f
(
x
) =
x
p
x
2
+
16
defined on the interval
[
-
7
,
4
]
.
a.)
f
(
x
)
is concave down on the region
.
b.)
f
(
x
)
is concave up on the region
.
c.)
The minimum for this function occurs at
.
d.)
The maximum for this function occurs at
.
Note:
Your answer to parts
a
and
b
must be given in
interval notation
.
Correct Answers:
•
(-7,0)
•
(0,4)
•
-7
•
4
98.
(1 point)
16
The graphs above are approximate graphs of
f
and
f
0
for
f
(
x
) =
x
2
(
x
-
15
)
.
So
f
is decreasing (and
f
0
is negative) on the interval
(
0
,
a
)
for
a
=
.
Correct Answers:
•
10
99.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
if the limit does not exist.
lim
x
→-
10
x
2
-
100
x
+
10
=
Correct Answers:
•
-20
100.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
t
→
0
e
-
2
t
-
1
-
1
t
=
Correct Answers:
•
2
101.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
θ
→
π
/
2
-
9
+
9sin
θ
3csc
θ
=
Correct Answers:
•
0
102.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
x
→
∞
4
e
x
9
x
=
Correct Answers:
•
INF
103.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
x
→
0
+
3ln
x
1
x
=
Correct Answers:
•
INF
104.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
x
→
∞
-
2ln
(
ln
x
)
6
x
=
Correct Answers:
•
0
105.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
t
→
0
5
t
-
4
t
7
t
=
Correct Answers:
•
0.0318776501877442
106.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
x
→
1
8ln
x
3sin
(
π
x
)
=
Correct Answers:
•
-0.848826363156776
107.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
x
→
0
4
e
x
-
4
-
4
x
6
x
2
=
Correct Answers:
•
0.333333333333333
17
108.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist.
lim
x
→
0
2arcsin
x
7
x
=
Correct Answers:
•
0.285714285714286
109.
(1 point) Suppose that
f
(
x
) =
6
x
2
-
16
.
(A) List all critical numbers of
f
. If there are no critical num-
bers, enter ’NONE’.
Critical numbers =
(B) Use interval notation to indicate where
f
(
x
)
is increas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Decreasing:
(D)List the
x
-coordinates of all local maxima of
f
. If there
are no local maxima, enter ’NONE’.
x
values of local maxima =
(E) List the
x
-coordinates of all local minima of
f
. If there
are no local minima, enter ’NONE’.
x
values of local minima =
(F) Use interval notation to indicate where
f
(
x
)
is concave
up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) List the
x
values all inflection points of
f
. If there are no
inflection points, enter ’NONE’.
Inflection points =
(I) List all horizontal asymptotes of
f
. If there are no hori-
zontal asymptotes, enter ’NONE’.
Horizontal asymptotes
y
=
(J) List all vertical asymptotes of
f
. If there are no vertical
asymptotes, enter ’NONE’.
Vertical asymptotes
x
=
(K) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
0
•
(-infinity,-4) U (-4,0)
•
(0,4) U (4,infinity)
•
0
•
none
•
(-infinity,-4) U (4,infinity)
•
(-4,4)
•
none
•
0
•
-4, 4
•
1
110.
(1 point) Suppose that
f
(
x
) =
5
x
2
x
2
+
36
.
(A) List all critical numbers of
f
. If there are no critical num-
bers, enter ’NONE’.
Critical numbers=
(B) Use interval notation to indicate where
f
(
x
)
is increas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Decreasing:
(D) List the
x
-coordinates of all local maxima of
f
. If there
are no local maxima, enter ’NONE’.
x
values of local maxima =
(E) List the
x
-coordinates of all local minima of
f
. If there
are no local minima, enter ’NONE’.
x
values of local minima =
(F) Use interval notation to indicate where
f
(
x
)
is concave
up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) List the
x
values of all inflection points of
f
. If there are
no inflection points, enter ’NONE’.
x
values of inflection points =
(I) List all horizontal asymptotes of
f
. If there are no hori-
zontal asymptotes, enter ’NONE’.
Horizontal asymptotes
y
=
(J) List all vertical asymptotes of
f
. If there are no vertical
asymptotes, enter ’NONE’.
vertical asymptotes
x
=
(K) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
0
•
(0,infinity)
•
(-infinity,0)
•
NONE
•
0
18
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•
(-3.46410161513775,3.46410161513775)
•
(-infinity,-3.46410161513775) U (3.46410161513775,infinity)
•
-3.46410161513775, 3.46410161513775
•
5
•
NONE
•
1
111.
(1 point) Suppose that
f
(
x
) =
3
x
2
-
x
3
-
2
.
(A) Find all critical numbers of
f
.
If there are no critical
numbers, enter ’NONE’.
Critical numbers =
(B) Use interval notation to indicate where
f
(
x
)
is increas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Decreasing:
(D) List the
x
-coordinates of all local maxima of
f
. If there
are no local maxima, enter ’NONE’.
x
values of local maxima =
(E) List the
x
-coordinates of all local minima of
f
. If there
are no local minima, enter ’NONE’.
x
values of local minima =
(F) Use interval notation to indicate where
f
(
x
)
is concave
up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H)List the
x
values of all inflection points of
f
. If there are
no inflection points, enter ’NONE’.
x
values of inflection points =
(I) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
0, 2
•
(0,2)
•
(-infinity,0) U (2,infinity)
•
2
•
0
•
(-infinity,1)
•
(1,infinity)
•
1
•
1
112.
(1 point) Suppose that
f
(
x
) =
5
x
-
7
x
+
5
.
(A) Find all critical values of
f
, compute their average, and
enter it below.
Note: If there are no critical values, enter -1000.
Average of critical values =
(B) Use interval notation to indicate where
f
(
x
)
is increasing.
Note:
Enter ’I’ for
∞
, ’-I’ for
-
∞
, and ’U’ for the union
symbol.
If you have extra boxes, fill each in with an ’x’.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreasing.
Decreasing:
(D) Find the
x
-coordinates of all local maxima of
f
, compute
their average, and enter it below.
Note: If there are no local maxima, enter -1000.
Average of
x
values =
(E) Find the
x
-coordinates of all local minima of
f
, compute
their average, and enter it below.
Note: If there are no local minima, enter -1000.
Average of
x
values =
(F) Use interval notation to indicate where
f
(
x
)
is concave up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) Find all inflection points of
f
, compute their average, and
enter it below.
Note: If there are no inflection points, enter -1000.
Average of inflection points =
(I) Find all horizontal asymptotes of
f
, compute the average of
the
y
values, and enter it below.
Note: If there are no horizontal asymptotes, enter -1000.
Average of horizontal asymptotes =
(J) Find all vertical asymptotes of
f
, compute the average of the
x
values, and enter it below.
Note: If there are no vertical asymptotes, enter -1000.
Average of vertical asymptotes =
(K) Use all of the preceding information to sketch a graph of
f
.
When you’re finished, enter a ”1” in the box below.
19
Graph Complete:
Correct Answers:
•
-1000
•
(-infinity,-5) U (-5,infinity)
•
x
•
-1000
•
-1000
•
(-infinity,-5)
•
(-5,infinity)
•
-1000
•
5
•
-5
•
1
113.
(1 point) Suppose that
f
(
x
) = (
x
2
+
10
)(
4
-
x
2
)
.
(A) Find all critical numbers of
f
. If there are no critical values,
enter ’NONE’.
Critical numbers =
(B) Use interval notation to indicate where
f
(
x
)
is increas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Decreasing:
(D) Find the
x
-coordinates of all local maxima of
f
. If there
are no local maxima, enter ’NONE’.
x
values of local maxima =
(E) Find the
x
-coordinates of all local minima of
f
. If there
are no local minima, enter ’NONE’.
x
values of local minima =
(F) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(G) List the
x
values of all inflection points of
f
. If there are
no inflection points, enter ’NONE’.
x
values of inflection points =
(H) Find all horizontal asymptotes of
f
. If there are no hori-
zontal asymptotes, enter ’NONE’.
Horizontal asymptotes
y
=
(J) Find all vertical asymptotes of
f
. If there are no vertical
asymptotes, enter ’NONE’.
Vertical asymptotes
x
=
(K) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
0
•
(-infinity,0)
•
(0,infinity)
•
0
•
NONE
•
(-infinity,infinity)
•
NONE
•
NONE
•
NONE
•
1
114.
(1 point) Suppose that
f
(
x
) =
7
x
6
-
3
x
5
.
(A) Find all critical numbers of
f
. If there are no critical num-
bers, enter ’NONE’.
Critical numbers =
(B) Use interval notation to indicate where
f
(
x
)
is increas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Decreasing:
(D) Find the
x
-coordinates of all local maxima of
f
. If there
are no local maxima, enter ’NONE’.
x
values of local maxima =
(E) Find the
x
-coordinates of all local minima of
f
. Note: If
there are no local minima, enter ’NONE’.
x
values of local minima =
(F) Use interval notation to indicate where
f
(
x
)
is concave
up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) List the
x
values of all inflection points of
f
. If there are
no inflection points, enter ’NONE’.
x
values of inflection points =
(I) Find all horizontal asymptotes of
f
. If there are no hori-
zontal asymptotes, enter ’NONE’.
Horizontal asymptotes
y
=
(J) Find all vertical asymptotes of
f
. If there are no vertical
asymptotes, enter ’NONE’.
Vertical asymptotes
x
=
(K) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
0, 0.357142857142857
•
(0.357142857142857,infinity)
•
(-infinity,0.357142857142857)
•
NONE
•
0.357142857142857
•
(-infinity,0) U (0.285714285714286,infinity)
•
(0,0.285714285714286)
•
0, 0.285714285714286
•
NONE
20
•
NONE
•
1
115.
(1 point) Suppose that
f
(
x
) =
2
x
5
-
5
x
4
.
(A) Find all critical numbers of
f
. If there are no critical val-
ues, enter ’NONE’.
critical numbers =
(B) Use interval notation to indicate where
f
(
x
)
is increas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Decreasing:
(D) Find the
x
-coordinates of all local maxima of
f
. If there
are no local maxima, enter ’NONE’.
x
values of local maxima =
(E) Find the
x
-coordinates of all local minima of
f
. If there
are no local minima, enter ’NONE’.
x
values of local minima =
(F) Use interval notation to indicate where
f
(
x
)
is concave
up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) Find all
x
values of the inflection points of
f
. If there are
no inflection points, enter ’NONE’.
x
values of inflection points =
(I) Find all horizontal asymptotes of
f
, If there are no hori-
zontal asymptotes, enter ’NONE’.
Horizontal asymptotes
y
=
=
(J) Find all vertical asymptotes of
f
. If there are no vertical
asymptotes, enter ’NONE’.
Vertical asymptotes
x
=
(K) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
0, 2
•
(-infinity,0) U (2,infinity)
•
(0,2)
•
0
•
2
•
(1.5,infinity)
•
(-infinity,1.5)
•
1.5
•
NONE
•
NONE
•
1
117.
(1 point) Find constants
a
and
b
in the function
f
(
x
) =
axe
bx
such that
f
(
1
3
) =
1 and the function has a local maximum
at
x
=
1
3
.
a
=
b
=
Solution:
SOLUTION
Using the product rule on the function
f
(
x
) =
axe
bx
, we have
f
0
(
x
) =
ae
bx
+
abxe
bx
=
ae
bx
(
1
+
bx
)
. We want
f
(
1
3
) =
1, and
since this is to be a maximum, we require
f
0
(
1
3
) =
0.
These
conditions give
f
(
1
/
3
) =
a
3
e
b
/
3
=
1
and
f
0
(
1
/
3
) =
ae
b
/
3
(
1
+
b
/
3
) =
0
.
Since
ae
b
/
3
is non-zero, we can divide both sides of the sec-
ond equation by
ae
b
/
3
to obtain 1
+
b
3
=
0. This implies
b
=
-
3.
Plugging
b
=
-
3 into the first equation gives us
a
(
1
3
)
e
-
1
=
1,
or
a
=
3
e
. How do we know we have a maximum at
x
=
1
3
and
not a minimum? Since
f
0
(
x
) =
ae
bx
(
1
+
bx
) = ()
e
-
3
x
(
1
-
3
x
)
,
and
(
3
e
)
e
-
3
x
is always positive, it follows that
f
0
(
x
)
>
0 when
x
<
1
3
and
f
0
(
x
)
<
0 when
x
>
1
3
.
Since
f
0
is positive to the
left of
x
=
1
3
and negative to the right of
x
=
1
3
,
f
(
1
3
)
is a local
maximum.
Correct Answers:
•
3*e
•
-1*3
118.
(1 point) Find a formula for a curve of the form
y
=
e
-
(
x
-
a
)
2
/
b
for
b
>
0 with a local maximum at
x
=
6 and
points of inflection at
x
=
2 and
x
=
10.
y
=
Solution:
SOLUTION
The maximum of
y
=
e
-
(
x
-
a
)
2
/
b
occurs at
x
=
a
. (This is be-
cause the exponent
-
(
x
-
a
)
2
/
b
is zero when
x
=
a
and negative
for all other
x
-values. The same result can be obtained by taking
derivatives.) Thus we know that
a
=
6.
Points of inflection occur where
d
2
y
/
dx
2
changes sign,
that is,
where
d
2
y
/
dx
2
=
0.
Differentiating gives
dy
dx
=
-
2
(
x
-
6
)
b
e
-
(
x
-
6
)
2
/
b
, so
d
2
y
dx
2
=
-
2
b
e
-
(
x
-
6
)
2
/
b
+
4
(
x
-
6
)
2
b
2
e
-
(
x
-
6
)
2
/
b
=
2
b
e
-
(
x
-
6
)
2
/
b
(
-
1
+
2
b
(
x
-
6
)
2
)
.
Since
e
-
(
x
-
6
)
2
/
b
is never zero,
d
2
y
/
dx
2
=
0 where
-
1
+
2
b
(
x
-
6
)
2
=
0
.
. We know
d
2
y
/
dx
2
=
0
at
x
=
10, so substituting
x
=
10 gives
-
1
+
2
b
(
10
-
6
)
2
=
0
.
Solving for
b
gives
b
=
32.
Since
a
=
6, the function is
y
=
e
-
(
x
-
6
)
2
/
32
.
You can check that at
x
=
6, we have
d
2
y
dx
2
=
2
32
e
-
0
(
-
1
+
0
)
<
0
so the point
x
=
6 does indeed give a maximum.
Correct Answers:
21
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•
eˆ(-1*(x-6)ˆ2/32)
119.
(1 point) The number,
N
, of people who have heard a
rumor spread by mass media at time,
t
, is given by
N
(
t
) =
a
(
1
-
e
-
kt
)
.
There are 150000 people in the population who hear the rumor
eventually. 23 percent of them heard it on the first day. Find
a
and
k
, assuming
t
is measured in days.
a
=
k
=
Solution:
SOLUTION
Since as
t
→
∞
,
N
→
a
, we have
a
=
150000. Note that while
N
(
t
)
will never actually reach 150000, it will become arbitrarily
close to 150000. Since
N
represents the number of people, it
makes sense to round up long before
t
→
∞
. When
t
=
1, we
have
N
=
0
.
23
(
150000
) =
34500 people, so plugging into our
formula gives
N
(
1
) =
34500
=
150000
1
-
e
-
k
(
1
)
.
Solving for
k
gives 0
.
23
=
1
-
e
-
k
, so that
k
=
-
ln
(
1
-
0
.
23
)
≈
0
.
261.
Correct Answers:
•
150000
•
-1*ln(1 - 23/100)
120.
(1 point) Answer the following questions for the func-
tion
f
(
x
) =
x
p
x
2
+
6
x
+
18
+
3
p
x
2
+
6
x
+
18
defined on the interval
[
-
8
,
1
]
.
All answers refer to
x
-
coordinates.
A.
f
(
x
)
is concave down on the region
to
B.
f
(
x
)
is concave up on the region
to
C. The inflection value for this function is
D. The minimum for this function occurs at
E. The maximum for this function occurs at
Correct Answers:
•
-8
•
-3
•
-3
•
1
•
-3
•
-8
•
1
121.
(1 point) Answer the following questions for the func-
tion
f
(
x
) =
x
3
x
2
-
36
.
Enter ”INF” for
∞
and ”-INF” for
-
∞
.
A. The function
f
(
x
)
has two vertical asympototes:
x
=
B.
f
(
x
)
has one local maximum and one local minimum oc-
curing at
x
values :
x
max
=
and
x
min
=
C. For each interval, tell whether
f
(
x
)
is increasing (type in
INC) or decreasing (type in DEC).
(
-
∞
,
max
)
(
max
,
-
6
)
(
-
6
,
0
)
(
0
,
6
)
(
6
,
min
)
(
min
,
+
∞
)
D.
f
(
x
)
is
concave
up
on
the
open
intervals:
.
E. The inflection point for this function occurs at
x
=
F. Sketch the graph of
f
(
x
)
and bring it to class.
Correct Answers:
•
-6, 6
•
-10.3923
•
10.3923
•
INC
•
DEC
•
DEC
•
DEC
•
DEC
•
INC
•
(-6,0), (6,infinity)
•
0
122.
(1 point) Consider the function
f
(
x
) =
3
x
+
3
x
-
1
.
For this function there are four important intervals:
(
-
∞
,
A
]
,
[
A
,
B
)
,
(
B
,
C
)
, and
[
C
,
∞
)
where
A
, and
C
are the critical numbers
and the function is not defined at
B
.
Find
A
and
B
and
C
For each of the following intervals, tell whether
f
(
x
)
is increas-
ing (type in INC) or decreasing (type in DEC).
(
-
∞
,
A
]
:
[
A
,
B
)
:
(
B
,
C
]
:
[
C
,
∞
)
:
Correct Answers:
•
-1
•
0
•
1
•
INC
•
DEC
•
DEC
•
INC
22
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123.
(1 point) Please answer the following questions about
the function
f
(
x
) =
ln
(
11
x
2
+
6
)
.
Instructions:
If you are asked to find
x
- or
y
-values, enter either
a number, a list of numbers separated by commas, or
None
if
there aren’t any solutions.
Use
interval notation
if you are
asked to find an interval or union of intervals, and enter
{ }
if
the interval is empty.
(a) Find the critical numbers of
f
, where it is increasing and
decreasing, and its local extrema.
Critical numbers
x
=
Increasing on the interval
Decreasing on the interval
Local maxima
x
=
Local minima
x
=
(b) Find where
f
is concave up, concave down, and has in-
flection points.
Concave up on the interval
Concave down on the interval
Inflection points
x
=
(c) Find any horizontal and vertical asymptotes of
f
.
Horizontal asymptotes
y
=
Vertical asymptotes
x
=
(d) The function
f
is ?
because ?
for all
x
in the domain
of
f
, and therefore its graph is symmetric about the ?
(e) Sketch a graph of the function
f
without having a graph-
ing calculator do it for you.
Plot the
y
-intercept and the
x
-
intercepts, if they are known. Draw dashed lines for horizon-
tal and vertical asymptotes. Plot the points where
f
has local
maxima, local minima, and inflection points.
Use what you
know from parts (a) and (b) to sketch the remaining parts of
the graph of
f
. Use any symmetry from part (d) to your advan-
tage. Sketching graphs is an important skill that takes practice,
and you may be asked to do it on quizzes or exams.
Correct Answers:
•
0
•
(0,infinity)
•
(-infinity,0)
•
NONE
•
0
•
(-0.738548945875996,0.738548945875996)
•
(-infinity,-0.738548945875996) U (0.738548945875996,infinit
•
-0.738548945875996, 0.738548945875996
•
NONE
•
NONE
•
even
•
f(-x) = f(x)
•
y-axis
124.
(1 point) Please answer the following questions about
the function
f
(
x
) =
5
x
2
ln
(
x
)
,
x
>
0
.
Instructions:
If you are asked to find
x
- or
y
-values, enter either
a number, a list of numbers separated by commas, or
None
if
there aren’t any solutions.
Use
interval notation
if you are
asked to find an interval or union of intervals, and enter
{ }
if
the interval is empty.
(a) Find the critical numbers of
f
, where it is increasing and
decreasing, and its local extrema.
Critical numbers
x
=
Increasing on the interval
Decreasing on the interval
Local maxima
x
=
Local minima
x
=
(b) Find where
f
is concave up, concave down, and has in-
flection points.
Concave up on the interval
Concave down on the interval
Inflection points
x
=
(c) Find any horizontal and vertical asymptotes of
f
.
Horizontal asymptotes
y
=
Vertical asymptotes
x
=
(d) The function
f
is ?
because ?
for all
x
in the domain
of
f
, and therefore its graph is symmetric about the ?
(e) Sketch a graph of the function
f
without having a graph-
ing calculator do it for you.
Plot the
y
-intercept and the
x
-
intercepts, if they are known. Draw dashed lines for horizon-
tal and vertical asymptotes. Plot the points where
f
has local
maxima, local minima, and inflection points.
Use what you
know from parts (a) and (b) to sketch the remaining parts of
23
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the graph of
f
. Use any symmetry from part (d) to your advan-
tage. Sketching graphs is an important skill that takes practice,
and you may be asked to do it on quizzes or exams.
Correct Answers:
•
0.606530659712633
•
(0.606530659712633,infinity)
•
(0,0.606530659712633)
•
NONE
•
0.606530659712633
•
(0.22313016014843,infinity)
•
(0,0.22313016014843)
•
0.22313016014843
•
NONE
•
NONE
•
neither
•
not applicable
•
not applicable
125.
(1 point) Suppose that
f
(
x
) =
x
1
/
3
(
x
+
3
)
2
/
3
(A) Find all critical values of
f
. If there are no critical values,
enter
None
. If there are more than one, enter them separated
by commas.
Critical value(s) =
(B) Use interval notation to indicate where
f
(
x
)
is increasing.
Note:
When using interval notation in WeBWorK, you use
I
for
∞
,
-I
for
-
∞
, and
U
for the union symbol. If there are no
values that satisfy the required condition, then enter ”” without
the quotation marks.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreasing.
Decreasing:
(D) Find the
x
-coordinates of all local maxima of
f
. If there are
no local maxima, enter
None
. If there are more than one, enter
them separated by commas.
Local maxima at
x
=
(E) Find the
x
-coordinates of all local minima of
f
. If there are
no local minima, enter
None
. If there are more than one, enter
them separated by commas.
Local minima at
x
=
(F) Use interval notation to indicate where
f
(
x
)
is concave up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) Find all inflection points of
f
.
If there are no inflection
points, enter
None
. If there are more than one, enter them sep-
arated by commas.
Inflection point(s) at
x
=
(I) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a
1
in the box below.
Graph Complete:
Correct Answers:
•
0, -3, -1
•
(-infinity,-3) U (-1,infinity)
•
(-3,-1)
•
-3
•
-1
•
(-infinity,-3) U (-3,0)
•
(0,infinity)
•
0
•
1
126.
(1 point) Suppose that
f
(
x
) =
ln
(
3
+
x
2
)
(A) Find all critical values of
f
. If there are no critical values,
enter
None
. If there are more than one, enter them separated
by commas.
Critical value(s) =
(B) Use interval notation to indicate where
f
(
x
)
is increasing.
Note:
When using interval notation in WeBWorK, you use
I
for
∞
,
-I
for
-
∞
, and
U
for the union symbol. If there are no
values that satisfy the required condition, then enter ”” without
the quotation marks.
Increasing:
(C) Use interval notation to indicate where
f
(
x
)
is decreasing.
Decreasing:
(D) Find the
x
-coordinates of all local maxima of
f
. If there are
no local maxima, enter
None
. If there are more than one, enter
them separated by commas.
Local maxima at
x
=
(E) Find the
x
-coordinates of all local minima of
f
. If there are
no local minima, enter
None
. If there are more than one, enter
them separated by commas.
Local minima at
x
=
(F) Use interval notation to indicate where
f
(
x
)
is concave up.
24
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Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) Find all inflection points of
f
.
If there are no inflection
points, enter
None
. If there are more than one, enter them sep-
arated by commas.
Inflection point(s) at
x
=
(I) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a
1
in the box below.
Graph Complete:
Correct Answers:
•
0
•
(0,infinity)
•
(-infinity,0)
•
None
•
0
•
(-1.73205080756888,1.73205080756888)
•
(-infinity,-1.73205080756888) U (1.73205080756888,infinity)
•
-1.73205080756888, 1.73205080756888
•
1
127.
(1 point)
For the function
f
given above, determine whether the fol-
lowing conditions are true.
Input
T
if the condition is ture,
otherwise input
F
.
(a)
f
0
(
x
)
<
0 if 0
<
x
<
2;
(b)
f
0
(
x
)
>
0 if
x
>
2;
(c)
f
00
(
x
)
<
0 if 0
≤
x
<
1;
(d)
f
00
(
x
)
>
0 if 1
<
x
<
4.
(e)
f
00
(
x
)
<
0 if
x
>
4;
(f) Two inflection points of
f
(
x
)
are, the smaller one is
x
=
and the other is
x
=
Correct Answers:
•
T
•
F
•
T
•
T
•
T
•
1
•
4
128.
(1 point)
Find the limit. Use l’Hospital’s Rule if appropriate. Use INF
to represent positive infinity, NINF for negative infinity, and D
for the limit does not exist. Assume
b
6
=
0 .
lim
x
→
1
x
a
-
1
x
b
-
1
=
Correct Answers:
•
a/b
129.
(1 point) For the given cost function
C
(
x
) =
54
√
x
+
x
2
421875
find
a) The cost at the production level 1800
b)
The
average
cost
at
the
production
level
1800
c)
The
marginal
cost
at
the
production
level
1800
d) The production level that will minimize the average cost.
e) The minimal average cost.
Solution:
Solution:
Part (a)
To calculate the cost at the production level 1800, we need only
plug 1800 into the cost function
C
(
x
)
to calculate
C
(
1800
) =
54
√
1800
+
(
1800
)
2
421875
=
2298
.
7060.
Part (b)
The average cost at the production level 1800 is the average cost
of producing each unit. In part (a) we saw that the total cost
was 2298
.
7060
,
and we know that we are producing 1800 units.
Hence, the average cost per unit is
2298
.
7060
1800
=
1
.
2771
Part (c)
The marginal cost function is the derivative of the cost func-
tion.
Hence, we differentiate the cost function
C
(
x
)
and get
C
0
(
x
) =
54
2
√
x
+
2
x
421875
.
Finally, to to determine the marginal
cost at production level 1800, we plug 1800 into
C
0
(
x
)
and get
C
0
(
1800
) =
54
2
√
1800
+
2
(
1800
)
421875
=
0
.
6449.
Part (d)
From part (b), the average cost function is
A
(
x
) =
C
(
x
)
x
where
x
is the production level. To minimize this, we find the derivative
and solve for zero. Simplifying, we have
A
(
x
) =
54
√
x
+
x
421875
.
Hence, the derivative is
A
0
(
x
) =
-
(
54
)
2
√
x
3
+
1
421875
.
Solving for
25
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zero, we get:
A
0
(
x
) =
0
-
(
54
)
2
√
x
3
+
1
421875
=
0
(
54
)
2
√
x
3
=
1
421875
2
√
x
3
= (
54
)(
421875
)
x
3
/
2
=
(
54
)(
421875
)
2
x
=
(
54
)(
421875
)
2
2
/
3
=
50625
.
0000
Part (e)
Finally, the minimal average cost is found by plugging the value
for
x
which we just found into the average cost function
A
(
x
)
.
This yields
A
(
50625
.
0000
) =
54
√
50625
.
0000
+
50625
.
0000
421875
=
0
.
3600.
Correct Answers:
•
2298.70597104441
•
1.27705887280245
•
0.644929436401226
•
50625
•
0.36
130.
(1 point) A manufacture has been selling 1200 televi-
sion sets a week at $360 each. A market survey indicates that
for each $11 rebate offered to a buyer, the number of sets sold
will increase by 110 per week.
a) Find the function representing the demand
p
(
x
)
, where
x
is the number of the television sets sold per week and
p
(
x
)
is
the corresponding price.
p
(
x
) =
b) How large rebate should the company offer to a buyer, in
order to maximize its revenue?
dollars
c) If the weekly cost function is 72000
+
120
x
, how should it set
the size of the rebate to maximize its profit?
dollars
Correct Answers:
•
(1200-x)/10 +360
•
120
•
60
131.
(1 point) A fence 5 feet tall runs parallel to a tall build-
ing at a distance of 7 feet from the building. We want to find the
the length of the shortest ladder that will reach from the ground
over the fence to the wall of the building.
Here are some hints for finding a solution:
Use the angle that the ladder makes with the ground to define
the position of the ladder and draw a picture of the ladder lean-
ing against the wall of the building and just touching the top of
the fence.
If the ladder makes an angle 1.03 radians with the ground,
touches the top of the fence and just reaches the wall, calculate
the distance along the ladder from the ground to the top of the
fence.
The distance along the ladder from the top of the fence to the
wall is
Using these hints write a function
L
(
x
)
which gives the to-
tal length of a ladder which touches the ground at an angle
x
,
touches the top of the fence and just reaches the wall.
L
(
x
)
=
.
Use this function to find the length of the shortest ladder
which will clear the fence.
The length of the shortest ladder is
feet.
Correct Answers:
•
5.83227096153733
•
13.5970158598381
•
5/sin(x)+7/cos(x)
•
16.8914833600749
132.
(1 point) Let
Q
= (
0
,
2
)
and
R
= (
7
,
6
)
be given points
in the plane. We want to find the point
P
= (
x
,
0
)
on the
x
-axis
such that the sum of distances
PQ
+
PR
is as small as possible.
(Before proceeding with this problem, draw a picture!)
To solve this problem, we need to minimize the following func-
tion of
x
:
f
(
x
) =
over the closed interval
[
a
,
b
]
where
a
=
and
b
=
.
We find that
f
(
x
)
has only one critical number in the interval at
x
=
where
f
(
x
)
has value
Since this is smaller than the values of
f
(
x
)
at the two endpoints,
we conclude that this is the minimal sum of distances.
Correct Answers:
•
sqrt(x**2+2**2)+sqrt((7-x)**2+6**2)
•
0
•
7
•
1.75
•
10.6301458127346
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133.
(1 point) Centerville is the headquarters of Greedy Ca-
blevision Inc. The cable company is about to expand service to
two nearby towns, Springfield and Shelbyville. There needs to
be cable connecting Centerville to both towns. The idea is to
save on the cost of cable by arranging the cable in a Y-shaped
configuation.
Centerville is located at
(
7
,
0
)
in the
xy
-plane,
Springfield is at
(
0
,
5
)
, and Shelbyville is at
(
0
,
-
5
)
. The cable
runs from Centerville to some point
(
x
,
0
)
on the
x
-axis where
it splits into two branches going to Springfield and Shelbyville.
Find the location
(
x
,
0
)
that will minimize the amount of cable
between the 3 towns and compute the amount of cable needed.
Justify your answer.
To solve this problem we need to minimize the following
function of
x
:
f
(
x
) =
We find that
f
(
x
)
has a critical number at
x
=
To verify that
f
(
x
)
has a minimum at this critical number we
compute the second derivative
f
00
(
x
)
and find that its value at
the critical number is
, a positive number.
Thus the minimum length of cable needed is
Correct Answers:
•
2*sqrt(x**2+5**2)+7 -x
•
2.88675134594813
•
0.259807621135332
•
15.6602540378444
134.
(1 point) A Norman window has the shape of a semi-
circle atop a rectangle so that the diameter of the semicircle is
equal to the width of the rectangle.
What is the area of the
largest possible Norman window with a perimeter of 48 feet?
Solution:
Solution:
To solve this maximization problem, we must find a formula
for the area of the window in terms of one of its dimensions. Let
h
be the height of the rectangular portion of the window, and let
x
be half of the width of the rectangular portion (which is then
the radius of the semicircular top). Then the area is given by
2
xh
+
π
x
2
2
(since we have half of a circle on top of the rectan-
gle).
Now, we need to find an expression for the height of the rect-
angular portion of the window in terms of the width, 2
x
. This
is where we use the information that the entire window should
have a perimeter of 48 feet.
The perimeter of the window is
given by 2
x
+
2
h
+
π
x
, again since we are only using half of a
circle. Setting this equation equal to 48 and solving for
h
, we
get:
2
x
+
2
h
+
π
x
=
48
2
h
=
48
-
2
x
-
π
x
h
=
48
-
2
x
-
π
x
2
Plugging this into our expression, we get a formula for the area
of the window in terms of the width of the rectangular portion.
That is:
a
(
x
) =
2
x
48
-
2
x
-
π
x
2
+
π
x
2
2
=
48
x
-
2
x
2
-
π
2
x
2
Taking the derivative of this and solving for 0, we get:
a
0
(
x
) =
45
-
4
x
-
π
x
a
0
(
x
) =
0
45
-
4
x
-
π
x
=
0
4
x
+
π
x
=
45
x
(
4
+
π
) =
45
x
=
45
4
+
π
≈
6
.
3011
Checking will show that this is in fact a local maximum, so
that the maximum area of the window can be found by plugging
this value of
x
into the formula for area. That is:
a
(
6
.
3011
)
≈
161
.
31
Correct Answers:
•
161.309
135.
(1 point) How would you divide a 19 inch line into two
parts of length
A
and
B
so that
A
+
B
=
19 and the product
AB
is maximized? (Assume that
A
≤
B
.
A
=
B
=
Correct Answers:
•
9.5
•
9.5
136.
(1 point) Find two positive numbers
A
and
B
(with
A
≤
B
) whose sum is 44 and whose product is maximized.
A
=
B
=
Correct Answers:
•
22
•
22
137.
(1 point) Find the length
L
and width
W
(with
W
≤
L
)
of the rectangle with perimeter 60 that has maximum area, and
then find the maximum area.
L
=
W
=
Maximum area =
Correct Answers:
•
15
•
15
27
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•
225
138.
(1 point) An open box is to be made out of a 8-inch
by 20-inch piece of cardboard by cutting out squares of equal
size from the four corners and bending up the sides. Find the
dimensions of the resulting box that has the largest volume.
Dimensions of the bottom of the box:
x
Height of the box:
Correct Answers:
•
16.4785 x 4.47853
•
1.76073
139.
(1 point) The owner of a used tire store wants to con-
struct a fence to enclose a rectangular outdoor storage area
adjacent to the store, using part of the side of the store (which
is 230 feet long) for part of one of the sides. (See the figure
below.)
There are 450 feet of fencing available to complete
the job.
Find the length of the sides parallel to the store and
perpendicular that will maximize the total area of the outdoor
enclosure.
Note:
you can click on the image to get a enlarged view.
Length of parallel side(s) =
Length of perpendicular sides =
Correct Answers:
•
225
•
112.5
140.
(1 point) A small resort is situated on an island that lies
exactly 5 miles from
P
, the nearest point to the island along a
perfectly straight shoreline. 10 miles down the shoreline from
P
is the closest source of fresh water. If it costs 2.1 times as much
money to lay pipe in the water as it does on land, how far down
the shoreline from
P
should the pipe from the island reach land
in order to minimize the total construction costs?
Distance from
P
=
Correct Answers:
•
2.70765180536941
141.
(1 point) Find two numbers differing by 40 whose prod-
uct is as small as possible.
Enter your two numbers as a comma separated list, e.g. 2, 3.
The two numbers are
.
Correct Answers:
•
-20, 20
142.
(1 point) Find the dimensions of the rectangle with area
361 square inches that has minimum perimeter, and then find
the minimum perimeter.
1.
Dimensions:
2.
Minimum perimeter:
Enter your result for the dimensions as a comma separated
list of two numbers. Do not include the units.
Correct Answers:
•
19, 19
•
76
143.
(1 point) An open box is to be made out of a 6-inch
by 14-inch piece of cardboard by cutting out squares of equal
size from the four corners and bending up the sides. Find the
dimensions of the resulting box that has the largest volume.
Dimensions of the bottom of the box:
x
Height of the box:
Correct Answers:
•
11.3885 x 3.38851
•
1.30575
144.
(1 point) A parcel delivery service will deliver a pack-
age only if the length plus the girth (distance around, taken per-
pendicular to the length) does not exceed 104 inches. Find the
maximum volume of a rectangular box with square ends that
satisfies the delivery company’s requirements.
Maximum Volume =
in
3
.
Correct Answers:
•
10415.4
145.
(1 point) A fence is to be built to enclose a rectangu-
lar area of 280 square feet. The fence along three sides is to be
made of material that costs 3 dollars per foot, and the material
for the fourth side costs 14 dollars per foot. Find the dimensions
of the enclosure that is most economical to construct.
Dimensions:
x
Correct Answers:
•
28.1662 x 9.941
146.
(1 point) A rectangle has its two lower corners on the
x
-axis and its two upper corners on the parabola
y
=
8
-
x
2
.
What are the dimensions of such a rectangle with the greatest
possible area?
1.
Width
=
2.
Height
=
Correct Answers:
•
3.26599
•
5.33333
28
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147.
(1 point) A rancher wants to fence in an area of 500000
square feet in a rectangular field and then divide it in half with a
fence down the middle, parallel to one side.
What is the shortest length of fence that the rancher can use?
Length of fence =
feet.
Correct Answers:
•
3464.1
148.
(1 point) A Union student decided to depart from Earth
after graduation to find work on Mars.
Careful calculations
made regarding the space shuttle to be built used the follow-
ing mathematical model for the velocity (in ft/sec) of the shuttle
from liftoff at
t
=
0 seconds until the solid rocket boosters are
jettisoned at
t
=
62
.
6 seconds:
v
(
t
) =
0
.
00125483
t
3
-
0
.
08905
t
2
+
30
.
21
t
+
3
.
25
Using this model, consider the acceleration of the shuttle. Find
the absolute maximum and minimum values of acceleration be-
tween liftoff and the jettisoning of the boosters.
1.
Absolute maximum of acceleration
=
2.
Absolute minimum of acceleration
=
Correct Answers:
•
33.8131
•
28.1035
149.
(1 point) The linear approximation at
x
=
0 to
√
4
+
3
x
is
A
+
Bx
where
A
=
and where
B
=
Correct Answers:
•
2
•
0.75
150.
(1 point) Let
f
(
x
) =
4
x
3
-
7. Find the open intervals
on which
f
is increasing (decreasing). Then determine the
x
-
coordinates of all relative maxima (minima).
1.
f
is increasing on the intervals
2.
f
is decreasing on the intervals
3.
The relative maxima of
f
occur at
x
=
4.
The relative minima of
f
occur at
x
=
Notes:
In the first two, your answer should either be a single
interval, such as (0,1), a comma separated list of intervals, such
as (-inf, 2), (3,4), or the word “none”.
In the last two, your answer should be a comma separated list
of
x
values or the word “none”.
Correct Answers:
•
(-infinity,infinity)
•
NONE
•
NONE
•
NONE
151.
(1 point) Suppose that
f
(
x
) =
5
x
x
2
-
25
.
(A) List all critical numbers of
f
. If there are no critical num-
bers, enter ’NONE’.
Critical numbers =
(B) Use interval notation to indicate where
f
(
x
)
is decreas-
ing.
Note:
Use ’INF’ for
∞
, ’-INF’ for
-
∞
, and use ’U’ for the
union symbol.
Decreasing:
(C)List the
x
-values of all local maxima of
f
. If there are no
local maxima, enter ’NONE’.
x
values of local maxima =
(D) List the
x
-values of all local minima of
f
. If there are no
local minima, enter ’NONE’.
x
values of local minima =
(E) List the
x
values of all inflection points of
f
. If there are
no inflection points, enter ’NONE’.
Inflection points =
(F) Use interval notation to indicate where
f
(
x
)
is concave
up.
Concave up:
(G) Use interval notation to indicate where
f
(
x
)
is concave
down.
Concave down:
(H) List all horizontal asymptotes of
f
. If there are no hori-
zontal asymptotes, enter ’NONE’.
Horizontal asymptotes
y
=
(I) List all vertical asymptotes of
f
.
If there are no vertical asymptotes, enter ’NONE’.
vertical asymptotes
x
=
(J) Use all of the preceding information to sketch a graph of
f
. When you’re finished, enter a ”1” in the box below.
Graph Complete:
Correct Answers:
•
NONE
•
(-infinity,-5) U (-5,5) U (5,infinity)
•
NONE
•
NONE
•
0
•
(-5,0) U (5,infinity)
•
(-infinity,-5) U (0,5)
•
0
•
-5, 5
•
1
Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America
29
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