How do you find the relative extrema for f(x) =2x- 3x^(2/3) +2f(x)=2x3x23+2?

1 Answer
May 18, 2017

Relative maximum f(0) = 2f(0)=2 and relative minimum f(1) = 1f(1)=1

Explanation:

The domain of ff is RR

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

Critical numbers for f are values of x in the domain of f at which f'(x) = 0 or f'(x) does not exist.

f'(x) = 0 at x=1 and f'(x) fails to exist at x=0.
Both are in the domain of f, so both are critical numbers for f.

on the interval (-oo,0), f'(x) is positive and on (0,1) it is negative, so f(0)=2 is a relative maximum.

on (1,oo), f'(x) is again positive, so f(1) = 1 is a relative minimum.

Here is the graph of f:
graph{2x-3x^(2/3)+2 [-5.365, 5.735, -2.02, 3.53]}