Page 130 - Calculus Workbook For Dummies

114 Part III: Differentiation

Do you see the trouble you’re going to run into with y(x)? The first derivative is undefined at
x = ! 2 2. And the second derivative will also be undefined at those x-values, because when
you take the second derivative with the quotient rule, squaring the bottom, the denominator
2
will contain that same factor, x - 8i. The second derivative test will thus fail at 2 2! , and
_
you’ll have to use the first derivative test. In contrast to y(x), the second derivative test works
great with s(t):
1. Get the critical numbers.
21 21 2
s = l - t
4 4
21 21 2
0 = - t
4 4
21 t = 21
2
4 4
t = ! 1
sl is not undefined anywhere, so –1 and 1 are the only critical numbers.
2. Do the second derivative.
21 21
s = l - t 2
4 4
21
m
s = - t
2
3. Plug in the critical numbers.
21
1 =
m ^
s - h _ concave up : mini
2
21
s 1 = - ^ concave down : maxh
m ^ h
2
4. Get the heights of the extrema.
3
21 - 1h 7 - 1h
^
^
1 = +
^
s - h 8 - = . 4 5
4 4
3
21 1 ^ h 7 1 ^ h
8
s 1 = + - = 11 .5
^ h
4 4
You’re done. s has a local min at (–1, 4.5) and a local max at (–1, 11.5).
2
i Find the absolute extrema of f x = sinx + cosx on the interval , πA. Absolute max at
^ h
0 7
π 5 π
c , 2m; absolute min at c , - 2m.
4 4
1. Find critical numbers.
f x = sinx + cosx
^ h
x =
f l ^ h cosx - sinx
0 = cosx - sinx
sinx = cosx _ divide both sides by cosxi
tanx = 1
π 5 π
x = , _ the solutions in the given intervali
4 4
The derivative is never undefined, so these are the only critical numbers.