Find an extreme points for $f(x) ={x\ln x\over x^2-1}$

Question

I had a hard time with finding an extreme points for $$f(x) ={x\ln x\over x^2-1}$$

We have $$f'(x)= {(x^2-1)-(x^2+1)\ln x \over (x^2-1)^2}$$ Since solving $f'(x)=0$ is the same as solving this transcendental equation $$\ln x = {x^2-1\over x^2+1}$$ for $x \ne 1$, I observe this function $$g(x) = {x^2-1\over x^2+1} -\ln x$$ which is decreasing and thus no solution.

Is there any more direct way, without involving $g(x)$?

Answer 1

You could always just graph y=lnx and y=(x^2 -1)/(x^2 +1) and look at the intersection points, which is only (1,0)