Page 119 - Calculus Workbook For Dummies

103
Chapter 7: Analyzing Those Shapely Curves with the Derivative

Q. Find the intervals of concavity and the f m ^ h 60 x - 30 x
3
x =
5
3
^ h
inflection points of f x = 3 x - 5 x + 10.
Note that the following solution is analo- f m ^ - h 30
1 = -
gous to the solution for finding local
.5 =
extrema with the first derivative. f m ^ - h . 7 5
.5 = -
f m ^ h . 7 5
A. f is concave down from 3- to the inflec-
1 =
J N f m ^ h 30
K
tion point at - 2 2 , + 11 .24 O ; concave
O
K
L P concave concave concave concave
up from there to the inflection point at down up down up
(0, 10); concave down from there to the – + – +
J N
third inflection point at K K 2 , + . 8 76 O ; 0
O
L 2 P 2 2
and concave up from there to 3.
Because the concavity switches at all
1. Find the second derivative of f.
these “critical numbers” and because
5
3
f x = 3 x - 5 x + 10 the second derivative exists at those
^ h
4
f l ^ h 15 x - 15 x 2 numbers (from Steps 2 and 3), there are
x =
inflection points at those three x-values.
3
x =
f m ^ h 60 x - 30 x
(If the concavity switches at a point
2. Set the second derivative equal to zero where the second derivative is unde-
and solve. fined, you have to check one more thing
before concluding that you’ve got an
3
60 x - 30 x = 0 inflection point: whether you can draw a
2
x 2
1 =
30 _ x - i 0 tangent line there. This is the case when
the first derivative is defined or if there’s
2
2 x - = 0 a vertical tangent.) In a nutshell, if the
1
2
2 x = 1 concavity switches at a point where the
30 x = 0 curve is smooth, you’ve got an inflection
or x = 1
2
x = 0 2 point.
2 5. Determine the location of the three
x = !
2 inflection points.
3. Check whether there are any x-values 5 3
^ h
where the second derivative is J f x = 3 x - 5 x + 10
N
undefined. K 2 O
K
2 2 f - 2 O . 11 .24
There are none, so - , 0, and are L P
2 2
^ h
the three second derivative “critical f 0 = 10
numbers.” (Technically these aren’t f K J 2 N O
called critical numbers, but they could K 2 O . . 8 76
be because they work just like first L P
derivative critical numbers.) So f is concave down from 3- to the
J N
K
4. Plot these “critical numbers” on a inflection point at - 2 2 , + 11 .24 O , con-
K
O
number line and test the regions. L P
cave up from there to the inflection point
You can use –1, –.5, .5, and 1 as test at (0, 10), concave down from there to
numbers. The following figure shows J 2 N
O
the second derivative sign graph. the third inflection point at K K 2 , + . 8 76 O ,
L P
and, finally, concave up from there to 3.