Question

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

Answer 1

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

Explanation:

`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`