# How do you find the critical numbers of 9x^3 - 25x^2?

Jan 4, 2016

$x = 0 , \frac{50}{27}$

#### Explanation:

Critical numbers are the $x$-values at which the derivative of a function equal $0$ or are doesn't exist.

Find the derivative of $9 {x}^{3} - 25 {x}^{2}$.

$\frac{d}{\mathrm{dx}} \left(9 {x}^{3} - 25 {x}^{2}\right) = 27 {x}^{2} - 50 x$

Set equal to $0$.

$27 {x}^{2} - 50 x = 0$
$x \left(27 x - 50\right) = 0$

$x = 0 , \frac{50}{27}$

The derivative is never undefined so there are only two critical values at $x = 0 , \frac{50}{27}$.