How do you find critical points for #G(x)= ^3sqrt(x²-x)#?

Question

Evelyn Agard · Accepted Answer

Hi there :) The critical points of # G(x) = (x^2-x)^(1/3) # occur at # x=1/2, x=0  # and # x=1 #. In order to find the critical points, you must take the derivative of # G(x) #. Using the chain rule, you should obtain # G'(x)=(2x-1)/(3(x^2-x)^(2/3)) # ( Try it! ) Any points in the function's domain where the derivative's value is 0 or undefined are considered the function's critical points (for more information, see https://tutor.hix.ai). So for your function, we are looking for the # x # values such that # G'(x) = 0 #. This is # (2x-1)/(3(x^2-x)^(2/3)) = 0 #. Now the only way this is possible is if # 2x-1 = 0 #. This gives # x = 1/2 #.  But we must also see where # G'(x) # is undefined!
Now we can't have a zero denominator, so # G'(x) # is undefined if # x^2-x = 0#. Factoring, we get # x(x-1) = 0 #. This gives # x = 0 # and # x=1 #.

Christopher Agresta · Answer

undefined To find the critical points of the function $ G(x) = \sqrt[3]{x^2 - x} $, you need to find where the derivative is either zero or undefined. First, find the derivative of $ G(x) $ using the chain rule. Then, set the derivative equal to zero and solve for $ x $.