# How do you find the relative extrema for f(x) =2x- 3x^(2/3) +2 on the interval [-1,3]?

Find and test the critical numbers for ##f##.

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How do you find the relative extrema for f(x) =2x- 3x^(2/3) +2 on the interval [-1,3]?
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##f(x) =2x- 3x^(2/3) +2##

Critical numbers for ##f## are values for ##x## that are:

1. In the domain of ##f##, and

2. at which ##f'(x) = 0## or ##f'(x)## does not exist.

(This is a wonderful example of why you cannot ignore the second kind of critical numbers.)

For this function, we have:

##f'(x) = 2-2x^(-1/3) = 2(1-x^(-1/3)) = 2((root(3)x -1)/root(3)x)##

##f'(x) = 0## ##color(white)”sssssssss”## ##f'(x)## does not exist

##x=1## ##color(white)”ssssssssssssss”## ## x =0##

Using the first derivative we find that ##f## is

increasing on ##(-oo, 0)## (Yes, I know the domain has been restricted, but it is not necessary.)

decreasing on (0,1) so

##f(0) = 2## is a relative maximum.

##f## is increasing on (1,oo), so

##f(1) = 1## is a relative minimum.

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