Page 131 - Calculus Workbook For Dummies

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Chapter 7: Analyzing Those Shapely Curves with the Derivative

If you divide both sides of an equation by something that can equal zero at one or more x-values
(like you do above when dividing both sides by cosx), you may miss one or more solutions. You
π 3 π
have to check whether any of those x-values is a solution. In this problem, cosx = 0 at and ,
2 2
and it’s easy to check (in line 4 above) that sinx does not equal cosx at either of those values. If
it did, you’d have one or two more solutions and one or two more critical numbers. Note that you
have to check these values before dividing out the “dividing thing” (cosx here).

2. Evaluate the function at the critical numbers.
π π π 5 π 5 π 5 π
f c m = sin + cos f c m = sin + cos
4 4 4 4 4 4
2 2 2 2
= + = - -
2 2 2 2
= 2 = - 2
3. Evaluate the function at the endpoints of the interval.
f 0 = sin0 + cos0 = 1
^ h
2
2
h
f 2 ^ π = sin π + cos π = 1
4. The largest of the four answers from Steps 2 and 3 is the absolute max; the smallest is the
absolute min.
π 5 π
The absolute max is at c , 2m. The absolute min is at c , - 2m.
4 4
2
3
j Find the absolute extrema of g x = 2 x - 3 x - 5 on the interval - . , .5A. Absolute min at
5
^ h
7
(–.5, –6); absolute max at (0, –5).
1. Find critical numbers.
2
3
g x = 2 x - 3 x - 5
^ h
2
g x = 6 x - 6 x
l ^ h
2
0 = 6 x - 6 x
x x -
0 = 6 ^ 1h
x = , 0 1
x = 1 is neglected because it’s outside the given interval; x = 0 is your only critical number.
2. Evaluate the function at x = 0.
3 2
5
h
^
g 0 = ^h 2 0 - ^h 3 0 - = - 5
3. Do the endpoint thing.
3 2
g - .5 = ^h 2 - .5 - ^h 3 - h 5
.5 -
^
= 2 $ ^ - .125 - $ .25 - 5
3
h
= - 6
3
3
.5 = $
2
g ^ h 2 .5 - $ .5 - 5
= 2 $ .125 - $ .25 - 5
3
= - . 5 5
4. Pick the smallest and largest answers from Steps 2 and 3.
The absolute min is at the left endpoint, (–.5, –6). The absolute max is smack dab in the
middle, (0, –5).