How do you find the critical points for #f(x)=(x^2)(x-2)^(1/3)#?

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Answer 1 · 2016-01-17T12:56:28

#(0, 0)# Maximum Point
#(12/7, -(144 root(3) 98)/343)# Minimum Point

To determine the critical points
Find first derivative #f' (x)# then equate to zero
that is #f' (x)=0#

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

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

#2x root(3)(x-2)+x^2/(3 (root(3) (x-2))^2)=0#

Multiply both sides by #3 (root(3) (x-2))^2#

#6x^2-12x+x^2=0#
#7x^2-12x=0#
solving for x values

#x_1=0# and #x_2=12/7#
#y=0# when #x=0#
#y=-(144 root(3) 98)/343# when #x=12/7#

The Critical Points are:

#(0, 0)# Maximum Point
#(12/7, -(144 root(3) 98)/343)# Minimum Point

Check the graph:

graph{y=x^2 (x-2)^(1/3) [-5, 5, -2.5, 2.5]}