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

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Answer 1 · 2018-03-10T22:23:48

$x=0$ & $x=50/27$

Critical points = Where the derivative equals zero

If $f(x)=9x^3-25x^2$

then $f'(x)=27x^2-50x$ by the Power Rule

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

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

$x(27x-50)=0$

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

Add $50$ to both sides.

$27x=50$

Divide by $27$ from both sides.

$x=50/27$

How do you find the critical numbers for 9x^3 - 25x^29x^3 - 25x^2 to determine the maximum and minimum?