How do you find the critical numbers of f(x)=x^2-6x?

Dec 21, 2015

Critical point exist at $x = 3$

Explanation:

$f \left(x\right) = {x}^{2} - 6 x$

Critical points can be found at x = c where $f ' \left(c\right) = 0$ critical points are also obtained at $x = c$ if $f \left(c\right)$ is defined and $f ' \left(c\right)$ is not defined.

So first we find derivative $f ' \left(x\right)$

$f ' \left(x\right) = 2 x - 6$

We find x by solving $f ' \left(x\right) = 0$

$2 x - 6 = 0$
$\implies 2 x = 6$
$\implies \frac{2 x}{2} = \frac{6}{2}$

$\implies x = 3$ Critical point exists at this value of x.