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

1 Answer

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

Explanation:

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]}