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

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Answer 1 · 2015-12-21T07:23:06

Critical point exist at $x=3$

$f(x) = x^2 - 6x$

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

So first we find derivative $f'(x)$

$f'(x) = 2x - 6$

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

$2x - 6 = 0$
$=> 2x = 6$
$=> (2x)/2 = 6/2$

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

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