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

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Answer 1 · 2016-01-04T22:08:34

$x=0,50/27$

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 $x=0,50/27$ .

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