<s>[INST] How to find the maximum and minimum values of $\frac{8x(x^2-1)}{(x^2+1)^2}$ algebraically? The function is $f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}$.
I have tried using calculus, only to fail.
 [/INST] HINT: prove that $$-2\le \frac{8x(x^2-1)}{(x^2+1)^2}\le 2$$
we have $$2-\frac{8x(x^2-1)}{(x^2+1)^2}=2\,{\frac { \left( {x}^{2}-2\,x-1 \right) ^{2}}{ \left( {x}^{2}+1
 \right) ^{2}}}
$$
and $$2+\frac{8x(x^2-1)}{(x^2+1)^2}=2\,{\frac { \left( {x}^{2}+2\,x-1 \right) ^{2}}{ \left( {x}^{2}+1
 \right) ^{2}}}
$$
the Minimum will be attained by $$x=-1-\sqrt{2}$$ and the Maximum by $$1-\sqrt{2}$$
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