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

Apr 4, 2015

Hey there :)

The critical points of $G \left(x\right) = {\left({x}^{2} - x\right)}^{\frac{1}{3}}$ occur at $x = \frac{1}{2} , x = 0$ and $x = 1$.

In order to find the critical points, you must take the derivative of $G \left(x\right)$. Using the chain rule, you should obtain $G ' \left(x\right) = \frac{2 x - 1}{3 {\left({x}^{2} - x\right)}^{\frac{2}{3}}}$ ( Try it! )

Critical points of a function are any points in the domain of the function where the value of the derivative of the function is 0 or undefined. (See http://en.wikipedia.org/wiki/Critical_point_%28mathematics%29 for more info!)

So for your function, we are looking for the $x$ values such that $G ' \left(x\right) = 0$. This is $\frac{2 x - 1}{3 {\left({x}^{2} - x\right)}^{\frac{2}{3}}} = 0$. Now the only way this is possible is if $2 x - 1 = 0$. This gives $x = \frac{1}{2}$.

But we must also see where $G ' \left(x\right)$ is undefined!
Now we can't have a zero denominator, so $G ' \left(x\right)$ is undefined if ${x}^{2} - x = 0$. Factoring, we get $x \left(x - 1\right) = 0$. This gives $x = 0$ and $x = 1$.