Page 135 - Calculus Workbook For Dummies

119
Chapter 7: Analyzing Those Shapely Curves with the Derivative

x
o Find the intervals of concavity and the inflection points of p x = 2 . Concave down from
^ h
J N x + 9
3
K
,
3 to an inflection point at - 3 , 3 - 12 O O , then concave up till the inflection point at 0 0i,
_
K
L P J 3 N
K
O , and, finally, concave
then concave down again till the third inflection point at 3 3 , 12 O
K
up to 3. L P
1. Get the second derivative. 2 l
2 l
2
2
2
x l
2
2
^ h _ x + i x _ h x + 9i l _ 9 - x i _ x + 9 - _i 2 9 - x a i _ x + 9i k
9 - ^
x =
p x = 2 pm ^ h 4
l ^ h
2
2
_ x + 9i _ x + 9i
2
9
x + - 2 x 2 2 2 2 2
9 2
x x +
= 2 - 2 _ 9 - _i 9 - x i 2 _ x + i x
2
_ x + 9i = 2 4
_ x + 9i
9 - x 2
= 2 _ x + i9 2 _ 2 9 - 4 _ x iC
2
2
x 9 -
9 -
x x + i
2
_ x + 9i = 4
2
_ x + 9i
3
- 2 x - 18 x - 36 x + 4 x 3
= 3
2
_ x + 9i
x x -
2 _ 2 27i
= 3
2
_ x + 9i
2. Set equal to 0 and solve.
x x -
2 _ 2 27i
= 0
3
2
_ x + 9i
x x -
i
2 _ 2 27 = 0
2
2 x = 0 x - 27 0
=
or
x = 0 x ! 3 3
3. Check for undefined points of the second derivative. None.
4. Test four regions with the second derivative. You can skip the sign graph. Tip: You can do all
of this in your head because all that matters is whether the answers are positive or negative.
x x -
2 _ 2 27i
x =
pm ^ h 3
2
_ x + 9i
2 2
1 -
2 - 10 ^ a h - 10 - 27k 2 - h - h 27k
1 ^ a
^
h
^
p - 10 = 3 p - h 3
1 =
m ^
h
m ^
2 2
1 +
h
^ a - 10 + 9k ^ a - h 9k
2^ N ^h Ph 2^ N ^h N h
= = 3
P 3 P
N P
= =
P P
= N = P
2
1 -
2 1 ^ a h ^ h 27k
p 1 = 2
m ^ h
h
^
2 3 2 10 ^ a h 10 - 27k
1 +
^ a h 9k p 10 = 3
m ^
h
2
h
2^ P ^h N h ^ a 10 + 9k
=
P 3 2^ P ^h Ph
N = 3
= P
P P
= N = P
= P
The concavity goes N, P, N, P so there’s an inflection point at each of the three zeros of pm.