What are the critical values, of #f(x) = x^(1/3)*(8-x) in [0,8]#?

1 Answer
Dec 10, 2015

They are #0# and #2#.

Explanation:

#f(x) = x^(1/3)*(8-x) = 8x^(1/3)-x^(4/3)# #" "# for #0 <= x <= 8#

#f'(x) = 8/3 x^(-2/3)-4/3x^(1/3) = 8/(3root(3)(x^2))-(4root(3)x)/3#

# = (8-4x)/(3root(3)(x^2))#

#f'(x)# does not exist at #x=0# which is in the domain of #f#, so #0# is a critical number for #f#.

#f'(x) = 0# at #x=2# which is in the domain of #f#, so #2# is a critical number for #f#.