Page 136 - Calculus Workbook For Dummies

120 Part III: Differentiation

5. Find the heights of the inflection points.
x
p x = 2
^ h
x + 9
- 3 3 3 3
p - 3 3 = 2 p 0 = 0 p 3 3 = 2
`
^ h
j
`
j
` - 3 3 + 9 ` 3 3 + 9
j
j
- 3 3 3
= =
27 + 9 12
- 3
=
12
J N
3
K
Taking a drive on highway p, you’ll be turning right from negative 3 to - 3 , 3 - 12 O
O , where
K
L P
you’ll point straight ahead for an infinitesimal moment, then you’ll be turning left till (0, 0),
J N
3
K
then right again till 3 3 , 12 O
O , and on your final leg to 3, you’ll round a very long bend to
K
the left. L P
p Find the intervals of concavity and the inflection points of q x = 5 x - 3 x. Concave down from
^ h
- 3 till an inflection point at about (–.085, –.171), then concave up till a vertical inflection point
at (0, 0), then concave down till a third inflection point at about (.085, .171), then concave up
out to 3.
You know the routine.
q x = 5 x - 3 x
^ h
1 - / 4 5 1 - / 2 3
q x = x - x
l ^ h
5 3
4 / 9 5 2 / 5 3
x = -
qm ^ h x - + x -
25 9
- 4 / 9 5 + 2 / 5 3 = 0
25 x 9 x
Whoops, I guess this algebra’s kind of messy. Let’s get the zeros on our calculators: just graph
- 4 2
y = / 9 5 + / 5 3 and find the x-intercepts. There are two: x . - . 0 085 and x . . 0 085.
25 x 9 x
So you’ve got two “critical numbers,” right? Wrong! Don’t forget to check for undefined points
4 - / 9 5 2 - / 5 3
x = -
of the second derivative. Because qm ^ h x + x , qm is undefined at x = 0. Since q(x)
25 9
is defined at x = 0, 0 is another critical number. So you have three critical numbers and four
regions. You can test them with –1, –.01, .01, and 1:
4 / 9 5 2 / 5 3
qm ^ h x - + x -
x = -
25 9
14 14
1 = -
m ^ h
h
m ^
h
q - h q - .01 . 158 qm ^ .01 . - 158 q 1 =
m ^
225 225
Thus the concavity goes: down, up, down, up. Because the second derivative is zero at –0.085
and 0.085 and because the concavity switches there, you can conclude that there are inflection
points at those two x-values. But because both the first and second derivatives are undefined
at x = 0, you have to check whether there’s a vertical tangent there. You can see that there is
by just looking at the graph, but if want to be rigorous about it, you figure the limit of the first
derivative as x approaches zero. Since that equals infinity, you’ve got a vertical tangent at x = 0,
and thus there’s an inflection point there.
Now plug in –0.085, 0, and 0.085 into q to get the y-values and you’re done.