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

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Answer 1 · 2015-05-25T14:49:54

Find and test the critical numbers for $f$ .

$f(x) =2x- 3x^(2/3) +2$

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

(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.

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