I have an formula $\frac{x^3}{x^2-1}$. How can I find the local extrema of it without calculus or graphing it? Additionally, is there a general approach for formula like $\frac{x^a}{x^b-c}$, where $a,b,c$ are some real values.
I know that that is an odd function, so if the local minima for $x>1$ is at $(x,y)$, then the local maxima for $x<-1$ is $(-x, -y)$.
I have tried to use AM-GM to find the local minima for $x>1$, $$\frac{x^3}{x^2-1} = 0.5*\left((x +1) + (x-1) + \frac{1}{x-1} + \frac{1}{x+1} \right)\geq 2$$
However, the equality condition cannot be met here since $x-1$ is never equal to $x+1$. The actual local minima should be at $(\sqrt{3}, 1.5\sqrt{3})$
Without calculus it may be difficult to find the extrema, but knowing where they are it can be proved they are indeed extrema without the calculus. For example for $x_0=\sqrt{3}$ we have $$f(\sqrt{3}) = \frac{3\sqrt{3}}{2}$$ $$f(x) - f(\sqrt{3}) = \frac{x^3}{x^2-1} - \frac{3\sqrt{3}}{2} = \frac{x^3- \frac{3\sqrt{3}}{2}(x^2-1)}{x^2-1} = \\ = \frac{(x-\sqrt{3})^2(x+\frac{\sqrt{3}}{2})}{x^2-1}$$ so for $x>1$ we have $f(x)-f(\sqrt{3}) \ge 0$, which means that $\sqrt{3}$ is a local minimum.