# How do you find the critical numbers for 9x^3 - 25x^2 to determine the maximum and minimum?

Mar 10, 2018

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

#### Explanation:

Critical points = Where the derivative equals zero

If $f \left(x\right) = 9 {x}^{3} - 25 {x}^{2}$

then $f ' \left(x\right) = 27 {x}^{2} - 50 x$ by the Power Rule

Equate it to zero: $27 {x}^{2} - 50 x = 0$

This is a quadratic equation with $c = 0$ so if we divide $x$ from both sides we get:

$x \left(27 x - 50\right) = 0$

$27 x - 50 = 0$ Don't forget! Since we divided by $x$, $x = 0$ is also a critical point.

Add $50$ to both sides.

$27 x = 50$

Divide by $27$ from both sides.

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