Question

How do you find the critical numbers for `y = (x^2-4)/(x^2-2x)` to determine the maximum and minimum?

Answer 1

There are none.

Explanation:

This function has no critical numbers.

`f(x) = (x^2-4)(x^2-2x) = ((x+2)(x-2))/(x(x-2)) = (x+2)/x = 1+2/x`

`f'(x) = -2/x^2` is never `0` and is defined for all `x` in the domain of `f`.

Therefore the function has no critical numbers, so it has no local extrema.