How do you find the critical numbers for #root3((x^2-x))# to determine the maximum and minimum?

1 Answer
Jun 8, 2017

Answer:

The critical number is #x=1/2#.

Explanation:

First we take the derivative using the chain rule, to make things easier for us we rewrite the problem using powers.

#(x^2-x)^(1/3)#

Now we apply the chain rule we take the derivative of the outside and multiple it by the derivative of the inside. It's important that you know the power rule.

#d/dx# #=# #1/3(x^2-x)^(-2/3)(2x-1)#

Now we rewrite it:

#d/dx# #=# #(2x-1)/(3root(3)((x^2-x)^2)#

Set it equal to zero and solve:

#(2x-1)/(3root(3)((x^2-x)^2)# #=0#

#cancelcolor(green)(3root(3)((x^2-x)^2))/1**(2x-1)/cancelcolor(green)(3root(3)((x^2-x)^2)# #=0# #(3root(3)((x^2-x)^2))#

We are left with:

#2x-1=0#

Solve:

#x=1/2#

By looking at the graph you can see that #x=1/2# is a critical number.
graph{(x^2-x)^(1/3) [-1.912, 4.248, -1.023, 2.057]}