Page 133 - Calculus Workbook For Dummies

117
Chapter 7: Analyzing Those Shapely Curves with the Derivative

- π
Technically x = is not one of the critical numbers; being at an endpoint, it is refused mem-
2
bership in the critical number club. It’s a moot point though, because you have to evaluate the
endpoints anyway.

π π π
c
q c m = 2 cos 2 $ m + 4 sin
6 6 6
1 1
= 2 $ + 4 $ = 3
2 2
π π π
c
q c m = 2 cos 2 $ m + 4 sin
2 2 2
4
= - 2 + = 2
5 π 5 π 5 π
q c m = 2 cos 2 $ m + 4 sin
c
6 6 6
1 1
= 2 $ + 4 $ = 3
2 2
π π π
Left endpoint: q - m = 2 cos 2 - $ m + 4 sin - m = - 2 + ^ 1 = - 6
4 - h
c
c
c
2 2 2
J N
5 π 5 π 5 π 2 O
2 0 + K
c
K
Right endpoint: q c m = 2 cos 2 $ m + 4 sin = $ 4 - . - . 2 828
4 4 4 2 O
L P
π π
Pick your winners: Absolute min at left endpoint: - , - 6m and a tie for absolute max: c , 3m
c
2 6
5 π
and c , 3m.
6
3
2
m Find the intervals of concavity and the inflection points of f x = - 2 x + 6 x - 10 x + 5. f is con-
^ h
-
cave up from 3 to the inflection point at (1, –1), then concave down from there to 3.
1. Get the second derivative.
3
2
f x = - 2 x + 6 x - 10 x + 5
^ h
2
x = -
f l ^ h 6 x + 12 x - 10
x = -
f m ^ h 12 x + 12
2. Set equal to 0 and solve.
- 12 x + 12 0
=
x = 1
3. Check for x-values where the second derivative is undefined. None.
4. Test your two regions — to the left and to the right of x = — and make your sign graph (see
1
the following figure).
x = -
f m ^ h 12 x + 12
0 =
f m ^ h 12
2 = -
f m ^ h 12
concave up concave down
+ –
1
Because the concavity switches at x = 1 and because f m equals zero there, there’s an inflection
point at x = 1.