How do you find the critical numbers for f(x)=x^2-6x to determine the maximum and minimum?

Oct 7, 2016

Differentiate to find the first derivative which will be zero at a max or min.

Explanation:

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

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

At a max or min, $f ' \left(x\right) = 0 \implies 2 x - 6 = 0$
$\therefore 2 x = 6 \implies x = 3$

So there is one critical point (could be max or min; in fact it is a minimum by observation as its is a +ve U shaped quadratic) when $x = 3$