Question

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

Answer 1

`x=0,50/27`

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 `9x^3-25x^2`.

`d/dx(9x^3-25x^2)=27x^2-50x`

Set equal to `0`.

`27x^2-50x=0`
`x(27x-50)=0`

`x=0,50/27`

The derivative is never undefined so there are only two critical values at `x=0,50/27`.