Page 125 - Calculus Workbook For Dummies

109
Chapter 7: Analyzing Those Shapely Curves with the Derivative

x 2
2
0
b Find the local extrema of h x = + cosx - in the interval , πi with the first derivative
^ h
_
J N 2 J 2 N
test. Local max at K K π π 2 O ; local min at K K 3 π 3 π 2 - 2 O .
,
,
8 O
O
L 4 P L 4 8 P
1. Find the first derivative.
x 2
h x = + cosx -
^ h
2 2
1
h x = - sinx
l ^ h
2
2. Find the critical numbers of h.
a. Set the derivative equal to zero and solve:
1
- sinx = 0
2
2
sinx =
2
π 3 π
x = or _ These are the solutions in the given interval .i
4 4
b. Determine the x-values where the derivative is undefined.
π 3 π
The derivative isn’t undefined anywhere, so the critical numbers of h are and .
4 4
3. Test numbers from each region on your number line.
π 1 π 2
π
hl c m = - sin hl c m = - sin π 2
l ^ h
6 2 6 2 2 2 h π = 2 - sinπ
2 1 2 2
= - = - 1 = - 0
2 2 2 2
= positive = negative = positive
4. Draw a sign graph (see the following figure).
increasing decreasing increasing
+ – +
0 π 3π 2π
4 4
5. Decide whether there’s a local min, max, or neither at each of the two critical numbers.
π
Going from left to right along the function, you go up until x = and then down, so there’s a
4
π 3 π
local max at x = . It’s vice-versa for x = , so there’s a local min there.
4 4
6. Compute the y-values of these two extrema.
π 3 π
π 4 π 2 3 π 4 3 π 2
h c m = + cos - h c m = + cos -
4 2 4 2 4 2 4 2
π 2 2 3 π 2 2 2
= + - = - -
4 2 2 2 8 2 2
3 π 2
π 2 = - 2
= 8
8
J N J N
So you’ve got a max at K K π π 2 O and a min at K K 3 π 3 π 2 - 2 O .
,
,
8 O
O
L 4 P L 4 8 P