Page 115 - Calculus Workbook For Dummies

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Chapter 7: Analyzing Those Shapely Curves with the Derivative

Q. Determine the absolute min and absolute The derivative is undefined at x = 0
^ h
max of f x = x - x in the interval because the denominator of the deriva-
[–1, ⁄2]. tive can’t equal zero. (If you graph this
1
function (always a good idea) you’ll also
A. The absolute max is 2 and the absolute see the cusp at x = 0 and thus know
min is 0. immediately that there’s no derivative
there.)
1. Get all the critical numbers:
The critical numbers are therefore 0
First, set the derivative equal to zero and ⁄4.
1
(note, first split this function in two to
get rid of the absolute value bars): 2. Compute the function values (the
heights) at all the critical numbers.
f x = x - x ^ x $ 0h 1 1
^ h
f c m = f 0 = 0
^ h
1 4 4
x =
f l ^ h - 1
2 x It’s just a coincidence, by the way, that in
1 both cases the argument equals the
0 = - 1
2 x answer.
2 x = 1 3. Compute the function values at the two
1 edges of the interval.
x =
4 1 J 2 - 1 N
1 =
^
f - h 2 f c m = K K O O . . 0 207
2 2
x
f x = - - x ^ x < 0h L P
^ h
4. The highest of all the function values
- 1
x =
f l ^ h - 1 from Steps 2 and 3 is the absolute max;
2 - x
the lowest of all the values from Steps 2
- 1
0 = - 1 and 3 is the absolute min.
2 - x
Thus, 2 is the absolute max and 0 is the
2 - x =- 1
absolute min.
No solution
Note that finding absolute extrema
Now, determine whether the derivative is involves less work than finding local
undefined anywhere: extrema because you don’t have to use
the first or second derivative tests — do
you see why?