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

Aug 22, 2017

There are none.

#### Explanation:

This function has no critical numbers.

$f \left(x\right) = \left({x}^{2} - 4\right) \left({x}^{2} - 2 x\right) = \frac{\left(x + 2\right) \left(x - 2\right)}{x \left(x - 2\right)} = \frac{x + 2}{x} = 1 + \frac{2}{x}$

$f ' \left(x\right) = - \frac{2}{x} ^ 2$ 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.