merino (aem2588) – HW3 – chen – (55405)
1
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19
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001
10.0points
Apply the Product Rule twice to determine
the derivative of
f
when
f
(
x
) = (
x
−
1)(
x
−
2)(
x
−
3)
.
1.
f
′
(
x
) = 3
x
2
+ 4
x
−
5
2.
f
′
(
x
) = (
x
−
2)(
x
−
3)
3.
f
′
(
x
) = 3
x
2
+ 12
x
+ 11
4.
f
′
(
x
) = (
x
−
1)(
x
−
2)
5.
f
′
(
x
) = 3
x
2
−
12
x
+ 11
correct
6.
f
′
(
x
) = 3
x
2
−
4
x
−
5
Explanation:
Applying the Product Rule to
f
once we
see that
f
′
(
x
) = (
x
−
2)(
x
−
3)+(
x
−
1)
d
dx
(
x
−
2)(
x
−
3)
.
Applying the Product a second time, but now
to the second term, we see that
d
dx
(
x
−
2)(
x
−
3) = (
x
−
3)+(
x
−
2) = 2
x
−
5
.
Thus
f
′
(
x
) = (
x
−
2)(
x
−
3) + (
x
−
1)(2
x
−
5)
= (
x
2
−
5
x
+ 6) + (2
x
2
−
7
x
+ 5)
.
Consequently,
f
′
(
x
) = 3
x
2
−
12
x
+ 11
.
002
10.0points
Find the thirddegree polynomial
Q
such
that
Q
(1) = 0
,
Q
′
(1) = 3
,
Q
′′
(1) = 10
,
Q
′′′
(1) = 18
.
1.
Q
(
x
) = 3
x
3
+ 4
x
2
−
2
x
−
1
2.
Q
(
x
) = 4
x
3
−
3
x
2
+ 2
x
−
1
3.
Q
(
x
) = 4
x
3
+ 3
x
2
−
2
x
−
1
4.
Q
(
x
) = 3
x
3
−
4
x
2
+ 2
x
−
1
correct
5.
Q
(
x
) = 2
x
3
−
4
x
2
+ 3
x
−
1
Explanation:
A degree 3 polynomial can be written as
Q
(
x
) =
ax
3
+
bx
2
+
cx
+
d
where the values of the coefficients
a, b, c
and
d
are determined by the values of
Q
and its
derivatives at
x
= 1. Now
Q
′
(
x
) = 3
ax
2
+ 2
bx
+
c
while
Q
′′
(
x
) = 6
ax
+ 2
b,
Q
′′′
(
x
) = 6
a .
From the given values
Q
′′′
(1) = 18
,
Q
′′
(1) = 10
,
Q
′
(1) = 3
,
therefore, we see that
Q
′′′
(1) = 6
a
= 18
=
⇒
a
= 3
,
while
Q
′′
(1) = 6
a
+ 2
b
= 10
=
⇒
b
=
−
4
,
and
Q
′
(1) = 3
a
+ 2
b
+
c
= 3
=
⇒
c
= 2
.
Finally,
Q
(1) =
a
+
b
+
c
+
d
= 0
=
⇒
d
=
−
1
.
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merino (aem2588) – HW3 – chen – (55405)
2
Consequently,
Q
(
x
) = 3
x
3
−
4
x
2
+ 2
x
−
1
.
003
10.0points
The graph of a differentiable function
g
is
shown in
1
2
3
1
2
3
Arrange the values of
0
,
g
′
(0)
,
g
′
(1)
,
g
′
(2)
,
g
′
(3)
in increasing order.
1.
g
′
(1)
<
0
< g
′
(3)
< g
′
(0)
< g
′
(2)
cor
rect
2.
g
′
(2)
<
0
< g
′
(3)
< g
′
(1)
< g
′
(0)
3.
g
′
(1)
<
0
< g
′
(2)
< g
′
(3)
< g
′
(0)
4.
g
′
(1)
<
0
< g
′
(0)
< g
′
(2)
< g
′
(3)
5.
g
′
(2)
<
0
< g
′
(1)
< g
′
(3)
< g
′
(0)
6.
g
′
(2)
<
0
< g
′
(1)
< g
′
(0)
< g
′
(3)
Explanation:
The sign of the derivative,
g
′
(
a
), of
g
at
x
=
a
is determined by whether the tangent
at
P
(
a, g
(
a
)) is sloping down to the right (in
which case
g
′
(
a
)
<
0) or whether it is sloping
up to the right (in which case
g
′
(
a
)
>
0). On
the other hand, the size of
g
′
(
a
) is determined
by the value of the slope of this tangent: the
inequality
g
′
(
a
)
< g
′
(
b
)
will hold when the tangent at
Q
(
b, g
(
b
)) has
greater slope than the tangent at
P
(
a, g
(
a
)).
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 Spring '10
 KNOPF
 Derivative, Differential Calculus, Product Rule, Cos, lim

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