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 $(\sqrt{3}, 1.5\sqrt{3})$
Best Answer
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.