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exam 2 winter 18 solutions.pdfMath 115 — Second M...
exam_2_winter_18_solutions.pdfMath 115 — Second Midterm —
exam 2 winter 18 solutions.pdfMath...
exam_2_winter_18_solutions.pdfMath 115 — Second Midterm —
Page 8
Math 115 / Exam 2 (March 20, 2018.)
page 8
6
. [5 points]
The function
P
(
t
) is given by the equation
P
(
t
)=
t
+4
t<
2
t
2

3
t
+8
2
≤
t
≤
3
1
9
(
t
3
+ 44)
t>
3
For which values of
t
is
P
(
t
)
diﬀerentiable
? Show all your work to justify your answer.
Solution:
•
At
t
= 2:
–
Continuity:
lim
t
→
2

t
+4=6
and
lim
t
→
2
+
t
2

3
t
+8=6
and
P
(2) = 6
.
Hence
P
(
t
) is continuous at
t
= 2.
–
Diﬀerentiability: The function
g
(
t
)=
t
+ 4 satisﬁes
g
0
(
t
) = 1 and hence
g
0
(2) = 1.
Similarly, it
h
(
t
)=
t
2

3
t
+ 8 satisﬁes
h
0
(
t
)=2
t

3 and
h
0
(2) = 1.
So
lim
t
→
2

P
0
(
t
) = 1 = lim
t
→
2+
P
0
(
t
)
.
Hence
P
(
t
) is diﬀerentiable at
t
= 2 (with
P
0
(2) = 1).
•
At
t
= 3:
–
Continuity:
lim
t
→
3

t
2

3
t
+8=8
and
lim
t
→
3
+
1
9
(
t
3
+ 44) =
71
9
6
=8
.
Hence
P
(
t
) is not continuous at
t
= 3 and therefore not diﬀerentiable at
t
= 3.
•
All the functions (polynomials) involved in the formula of
P
(
t
) are diﬀerentiable on the
domains assigned to them.
Hence
P
(
t
) is diﬀerentiable for all
t
6
= 3.
Answer:
The function
P
(
t
) is diﬀerentiable for the following values of
t
: all
t
6
= 3.
Page 9
Math 115 / Exam 2 (March 20, 2018.)
page 9
7
. [10 points]
The amount of chlorine in a chemical reaction
C
(
t
) (in gallons)
t
seconds after it
has been added into a solution is given by the function
C
(
t
)=2

3(
t

5)
4
5
(
t

1)
e

t
for
t
≥
0
.
Notice that
C
0
(
t
)=
3(
t

6)(5
t

9)
e

t
5(
t

5)
1
/
5
.
a
. [8 points] Use calculus to ﬁnd the time(s) (if any) at which the amount of chlorine in the
solution is the greatest and the smallest. If the function has no global maximum or global
minimum write “
None
” in the appropriate space. Show all your work.
Solution:
Critical points:
•
C
0
(
t
) = 0:
t
= 6 and
t
=1
.
8.
•
C
0
(
t
) undeﬁned:
t
= 5.
Finding the output values of
C
(
t
) at critical points and the behavior of the function at
the endpoints:
t
0
1.8
5
6
C
(
t
)
2 + 3(5)
4
5
≈
12.87
≈
0
.
994
2
≈
1
.
96
lim
t
→∞
2

3(
t

5)
4
5
(
t

1)
e

t
=2
.
Answer:
Global maximum(s) at
t
=0
Global minimum(s) at
t
=1
.
8.
b
. [2 points] What is the maximum amount of chlorine in the solution? If there is never a
maximum amount of chlorine in the solution, write “
None
”.
Solution:
Answer:
2 + 3(5)
4
5
≈
12
.
87 gallons.
Page 10
Math 115 / Exam 2 (March 20, 2018.)
page 10
8
. [14 points]
The graph of the
derivative
g
0
(
x
) of the function
g
(
x
) with domain

5
<x<
10
is shown below.
The function
g
0
(
x
) has cor
ners at
x
= 5 and
x
= 7, and
it is linear on the intervals
(5
,
7) and (7
,
10).
If there is not enough infor
mation given to answer the
question,
write
“
NEI
”.
If
the
answer
is
none,
write
“
None
”.

5

4

3

2

1
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
x
y
=
g
0
(
x
)
a
. [3 points] Estimate the interval(s) on which the function
g
(
x
) is concave up.
Solution:
Answer:
(3,3) and (7,10)
b
. [3 points] Estimate all the
x
coordinates of the inﬂection points of
g
(
x
).
Solution:
Answer:
x
=

3
,
3
,
7.
c
. [2 points] Estimate the values of
x
in

5
<x<
10 for which
g
00
(
x
) is not deﬁned.
Solution:
Answer:
x
=5
,
7.
d
. [2 points] Estimate the interval(s) on which
g
000
(
x
)
>
0. Recall that
g
000
(
x
) is the derivative
of
g
00
(
x
).
Solution:
Answer:
(approximately) (

5
,

2) and (0
,
1
.
8).
e
. [4 points] Let
P
(
x
) be the quadratic approximation of
g
(
x
) at
x
= 8. Find the formula
of
P
(
x
) in terms of only the variable
x
if
g
(8) =

2. Your answer should not include the
letter
g
.
Solution:
g
(8) =

2,
g
0
(8) = 1 +
4
3
=
7
3
and
g
00
(8) =
4
3
. Then
Answer:
P
(
x
)=

2+
7
3
(
x

8) +
2
3
(
x

8)
2
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