Roots of a function with a logarithm / Graph of a function So I'm trying to draw the graph (manually) of a function:
$$f(x)=x-2\ln{(x^2+1)}$$
Finding the first/second derivatives is fairly easy and so is proving that there are no asymptotes, but for the life of me I can't find the roots of the function, AKA where it intersects with $Ox$. Without them I don't know it's behavior when $x>0$, apart from the fact that it reaches a local maximum at $x=2-\sqrt{3}$ and a minimum at $x=2+\sqrt{3}$, but those could be reached while $y=f(x)>0$ when $x>0$, which is clearly not the case if you use a plot graph to check. WolframAlpha doesn't give me any good roots, no steps either. Any ideas?

You can't find these roots analitically because $e^x = -2(x^2+1)$ is transcedental. However, you know the only extrema of $f$ are a local maximum $x_1 >0$, (then) a local minimum $x_2<0$, and that $\lim_{x\to\pm\infty} f(x) = \pm\infty$. The sign changes, the intermediate value theorem and the fact that you found all local extrema of $f$ guarantee that there are only three roots. It reminds me of cubic polynomials which have this exact shape (up to changes in $f'(x)$).

So even if you can't find the roots, you know how many there are and an interval containing them (e.g. there is a root in $(-\infty,x_1)$. What can be furthemore done with nothing but a calculator is approximating the position of these roots, by the bisection method, for example.
