How do use the first derivative test to determine the local extrema #f(x) = 3(x-4)^(2/3) +6#?

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Answer 1 · 2015-08-28T14:41:32

#6# is a local minimum. (It occurs at #4#.) There are no other local extrema.

#f(x) = 3(x-4)^(2/3) +6#

#f'(x) = 2(x-4)^(-1/3) = 2/root(3)(x-4)#

#f'# is never #0# and is not defined at #x=4#.

#4# is in the domain of #f#, so #4# is a critical number for #f#.

First derivative test:

on the interval #(-oo,4)#, we have #f'(x) <0# and

on the interval #(4,oo)#, we have #f'(x) > 0#.

Therefore, #f(4)# is a local minimum..

#f(4) = 3((4)-4)^(2/3) +6 = 6#

Because there are no other critical numbers, there are no other local extrema.