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

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Answer 1 · 2017-05-18T03:15:17

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

The domain of #f# 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]}