Local extrema of a function

Question

I solved a question to find the local extrema of the function $$\frac{x^2 - 4}{x^2 - 1}$$ I got $$\frac{dy}{dx}= \frac{6x}{(x^2 - 1)^2}$$ And the second derivative to be $$\frac{6 - 18x^2}{(x^2 - 1)^3}$$ The critical point putting $\frac{dy}{dx}=0$ is $x=0$ Inserting $x=0$ in the second derivative, $$\frac{6-0}{0-1} = - 6$$ This should give a local maximum as the second derivative is negative but the answer in my text says its a local minimum, is there something I'm doing wrong?

Answer 1

the second derivative computation is wrong. it should be $\frac{-6-18x^2}{(x^2-1)^3}$ hope,That clarifies.