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

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Answer 1 · 2016-10-07T13:59:49

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

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

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

At a max or min, $f'(x)=0 => 2x-6=0$
$:. 2x=6=>x=3$

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 $x=3$

How do you find the critical numbers for f(x)=x^2-6xf(x)=x2−6xf(x)=x^2-6x to determine the maximum and minimum?