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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##.
##f(x) =2x- 3x^(2/3) +2##
Critical numbers for ##f## are values for ##x## that are:
-
In the domain of ##f##, and
-
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.