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

Jun 8, 2017

The critical number is $x = \frac{1}{2}$.

#### Explanation:

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

${\left({x}^{2} - x\right)}^{\frac{1}{3}}$

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.

$\frac{d}{\mathrm{dx}}$ $=$ $\frac{1}{3} {\left({x}^{2} - x\right)}^{- \frac{2}{3}} \left(2 x - 1\right)$

Now we rewrite it:

$\frac{d}{\mathrm{dx}}$ $=$ (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) $= 0$ $\left(3 \sqrt[3]{{\left({x}^{2} - x\right)}^{2}}\right)$

We are left with:

$2 x - 1 = 0$

Solve:

$x = \frac{1}{2}$

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