SOLUTIONS CHAPTER 14.1
MATH 132 WI01
2.
Function is
00
00
decreasing
in (
∞
,

1), then
00
%
00
increasing
in (

1,0),
again
00
00
decreasing
in (0,1), and ﬁnally
00
%
00
increasing
in (1,
∞
). Relative
extrema are points where the function’s behaviour changes from increasing to de
creasing 
maximum
 which in our case is (0
,
0), or from decreasing to increasing

minimum
 which in our case are (

1
,

1) and (1
,

1).
•
6.
What we are interested in is the sign of the derivative, derivative which is
already given in our case. Hence, let’s look for sign of
f
0
(
x
) = 2
x
(
x

1)
3
.
f
0
(
x
)
equals zero in
x
= 0 and in
x
= 1, so we have to check sign between these values,
and outside these values: plug in

1, we get 2(

1)(

2)
3
= 16, hence positive,
hence the original f is
increasing in the interval
(
∞
,
0); plug in
1
2
, we get
2
*
(
1
2
)(
1
2

1)
3
= (

1
2
)
3
=

1
8
, hence negative, so the original f function will be
decreasing in the interval
(0
,
1); plug in 2, we get 2
*
2(2

1)
3
= 4, hence again
positive, so now the function f is again
increasing in the interval
(1
,
∞
). As
you can see, at 0 we have a change from increasing to decreasing, so it’s a relative
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 Spring '08
 Staff
 Math, Calculus, Critical Point, Sign, between.For

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