vu (tv2894) – Homework 7 (Quest) – miner – (55096)
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001
10.0 points
Find the value of
f
′
(4) when
f
(
x
) =
2
3
x
3
/
2
+ 2
x
1
/
2
.
Correct answer: 2
.
5.
Explanation:
Since
d
dx
x
r
=
rx
r
−
1
,
we see that
f
′
(
x
) =
x
1
/
2
+ 1
x
−
1
/
2
.
At
x
= 4, therefore,
f
′
(4) =
5
2
.
002
10.0 points
Find the
x
coordinate of all points on the
graph of
f
(
x
) =
x
3

x
2

x
+ 5
at which the tangent line is horizontal.
1.
x
coord =

1
3
2.
x
coord = 1
3.
x
coords =
1
3
,

1
4.
x
coord =

1
5.
x
coords =

1
3
,
1
correct
6.
x
coord =
1
3
Explanation:
The
tangent
line
will
be
horizontal
at
P
(
x
0
,f
(
x
0
)) when
f
′
(
x
0
) = 0
.
Now
f
′
(
x
) = 3
x
2

2
x

1
= (3
x
+ 1)(
x

1)
.
Consequently,
x
0
=

1
3
,
1
.
003
10.0 points
Find an equation for the tangent line to the
graph of
f
at the point
P
= (

1
,f
(

1)) when
f
(
x
) =
4
x
+ 1
3
x
+ 1
.
1.
4
y

x
= 7
correct
2.
y

4
x
= 7
3.
y

4
x
= 9
4.
y
+ 4
x
= 9
5.
4
y
+
x
= 7
6.
4
y
+
x
= 9
Explanation:
Since
f
(

1) =
3
2
, we see that
P
=
(

1
,
3
2
)
.
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 Fall '10
 Gualdini
 Calculus, Derivative, Slope

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