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find when $\frac{1}{x^2-1}$ convex and concave

Do I must look at the second derivative where $\frac{6x^2+2}{(x^2-1)^3}>0$ and $\frac{6x^2+2}{(x^2-1)^3}<0$ or can I look before and after the minimum or maximum points?

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For concavity of $C^2$ (twice differentiable) functions, you generally should consider the sign of the second derivative. The extrema do not affect concavity and you shouldn't expect it to (e.g. $f(x) = x^2$).

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here we have $$f''(x)=\frac{2(3x^2+1)}{(x-1)^3(x+1)^3}$$ thus we have $$f''(x)>0$$ if $x>1$ or $x<-1$

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$f''(x)<0$ if $$-1<x<1$$

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