How do you find the critical numbers for #f(x) = x^(1/3)*(x+3)^(2/3)# to determine the maximum and minimum?

1 Answer
Dec 11, 2017

Answer:

Please see below.

Explanation:

#c# is a critical number for #f# if and only if #c# is in the domain of #f# and either #f'(c)=0# or #f'(c)# does not exist.

For #f(x) = x^(1/3)(x+3)^(2/3)#, we have

Domain of #f# is #(-oo,oo)# and

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

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

Get a common denominator and combine to make one quotient.

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

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

#f'(x) = (x+1)/(x^(2/3)(x+3)^(1/3))# is #0# at #x=-1#

and #f'(x)# does not exist for #x=-3#, and #x=0#

These are all in the domain of #f#, so the critical numbers are:

#x=-3# (local max)
#x=-1# (local min)
#x=0# (neither min nor max)

The graph of #f# is shown below.

graph{ x^(1/3)(x+3)^(2/3) [-7.024, 7.02, -3.51, 3.514]}