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exam 2 winter 18 solutions.pdf
exam_2_winter_18_solutions.pdf
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exam 2 winter 18 solutions.pdf-Math 115 — Second M...
exam_2_winter_18_solutions.pdf-Math 115 — Second Midterm —
exam 2 winter 18 solutions.pdf-Math...
exam_2_winter_18_solutions.pdf-Math 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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