Page 134 - Calculus Workbook For Dummies

118 Part III: Differentiation

5. Find the height of the inflection point.

3
2
f x = - 2 x + 6 x - 10 x + 5
^ h
f 1 = - 1
^ h
Thus f is concave up from 3- to the inflection point at (1, –1), then concave down from there
to 3. As always, you should check your result on your graphing calculator. Hint: To get a good
feel for the look of this function, you’ll need a fairly odd graphing window — try something
like xmin = –2, xmax = 4, ymin = –20, ymax = 20.
2
4
n Find the intervals of concavity and the inflection points of g x = x - 12 x . g is concave up
^ h
from 3- to the inflection point at - , 2 - 20j; then concave down to an inflection point at
`
` , 2 - 20j; then concave up again to 3.
1. Find the second derivative.
4
g x = x - 12 x 2
^ h
3
g x = 4 x - 24 x
l ^ h
2
x =
gm ^ h 12 x - 24
2. Set to 0 and solve.
2
=
12 x - 24 0
2
x = 2
x = ! 2
3. Is the second derivative undefined anywhere? No.
4. Test the three regions and make a sign graph. See the following figure.
2
x =
gm ^ h 12 x - 24
2 =
gm ^ - h 24
0 = -
gm ^ h 24
2 =
gm ^ h 24
concave concave concave
up down up
+ – +

Because the concavity switched signs at the two zeros of gm, there are inflection points at
these two x-values.
5. Find the heights of the inflection points.
4
g x = x - 12 x 2
^ h
g - 2 = - 20
j
`
g ` 2 = - 20
j
g is concave up from 3- to the inflection point at - , 2 - 20j, concave down from there to
`
another inflection point at ` , 2 - 20j, then concave up again from there to 3.