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MATH 1300DMIDTERM22003
Solutions Version A
Multiple Choice Section
Question 1
 Let
f
(
x
) =
x
2

1
x
+2
, then
f
0
(
x
) =
Solution:
Use quotient rule.
f
0
(
x
) =
2
x
·
(
x
+ 2)

1
·
(
x
2

1)
(
x
+ 2)
2
=
x
2
+ 4
x
+ 1
(
x
+ 2)
2
Answer: A.
Question 2
 Find
dy
dx
for the equation
xy
3

3
y
= 2
x
.
Solution:
Implicit diﬀerentiation.
y
3
+
x
·
±
3
y
2
dy
dx
¶

3
dy
dx
= 2
(
3
xy
2

3
)
dy
dx
= 2

y
3
dy
dx
=
2

y
3
3
xy
2

3
Answer: B.
Question 3
 What are the critical points of
f
(
x
) = 3

1
x
+
1
x
2
?
Solution:
We have:
f
0
(
x
) =
1
x
2

2
x
3
=
1
x
3
(
x

2)
.
Thus,
f
0
(0) is undeﬁned and
f
0
(2) = 0. However,
x
= 0 is not in the domain of the
original
function
f
(
x
). Thus, by deﬁnition, it is not a critical point. Answer: B.
1
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View Full DocumentQuestion 4
 Let
f
(
x
) =
x
4

2
x
2
+3 and let I be the interval [

2
,
3]. Which of the following
is true for the function
f
on I?
Solution:
f
is continuous and the domain
I
is a closed interval. Thus, we only need to
plug in the critical points of
f
and endpoints of
I
into
f
and pick the largest and smallest
values. Since
f
0
(
x
) = 4
x
3

4
x
= 4
x
(
x

1)(
x
+ 1), the critical points are
x
= 0
,
±
1. The
endpoints of
I
are 3 and 2. We have:
x
2
1
0
1
3
f
(
x
)
11
2
3
2
66
Answer: C.
Question 5
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 Fall '09
 PIETERHOFSTRA
 Quotient Rule

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