How do you find the critical numbers of #y = (x^2-4)/(x^2-2x)#?

1 Answer
Mar 23, 2017

First things first, the function can be simplified as follows:

#y = ((x + 2)(x - 2))/(x(x - 2))#

#y = (x+ 2)/x#

#y = x/x + 2/x#

#y = 1 + 2/x#

The derivative of this is

#y' = -2/x^2#

There are critical numbers when the derivative is undefined or the derivative equals #0#. When the derivative is undefined at #x = 0#, the function is also undefined, so this is not a critical value.

#0 = -2/x^2#

#0 = -2#

This is obviously a contradiction, therefore the equation #-2/x^2 = 0# has no solution. This also means the given function has no critical values.

Hopefully this helps!