What are the critical points of # f(x) = x^(1/3)*(x+8)#?

1 Answer
Apr 8, 2016

Answer:

The critical numbers are #0# and #-2#.

Explanation:

A critical number for #f# is a number, c, in the domain of #f# such that #f'C9)=0# or #f'(c) does not exist.

For #f(x) = x^(1/3)(x-8)#, the domain is #RR#.

And differentiating, we get

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

# = (x-8)/(3x^(2/3)) +x^(1/3)#

# = (x-8)/(3x^(2/3)) +(3x)/(3x^(2/3))#

# = (4x+8)/(3x^(2/3)#.

So, #f'(x) = 0# at #x = -2# and #f'(x)# does not exists at #x=0#.

Both are in the domain of #f#, so both are critical numbers for #f#.