# What are the critical points of (x^2/(x^2-1))?

Mar 28, 2015

I will put f(x) in front of $\left({x}^{2} / \left({x}^{2} - 1\right)\right)$

To find the critical number, you must get the first derivative of
f(x) = $\left({x}^{2} / \left({x}^{2} - 1\right)\right)$

The first derivative is ${f}^{'} \left(x\right) = \frac{- 2 x}{{x}^{2} - 1} ^ 2$

Now you must set ${f}^{'} \left(x\right) = 0$, and you must also find where the ${f}^{'} \left(x\right)$ does not exist (dne).

${f}^{'} \left(x\right) = 0$ ------------------------- ${f}^{'} \left(x\right)$ dne

$\frac{- 2 x}{{x}^{2} - 1} ^ 2$ = 0 ------------------${f}^{'} \left(x\right)$ dne at x = 1 and x = -1

-2x = 0
x = 0 ------------------------------now you must plug them back to the
---------------------------------------- back to the original equation.
---------------------------------------Since x = 1 and x = -1 are undefined, they --------------------------------------------------------are not critical numbers

Since x = 0 is defined in the original equation, it is a critical number.