TITLE: Can the derivative prove my function has only one root?
QUESTION [5 upvotes]: I have a function:
$$f(x)=x-\ln(x^2+1)+2$$
I want to prove my function has exactly one root. If I differentiate:
$$f'(x)=1-\frac{2x}{x^2+1}$$ I can see this value is positive for every $x$. Does this prove that my function is strictly increasing? All the theorems I know about the argument apply for a closed interval, so I don't know if my resolution is valid.  Should I also prove the function has a positive and a negative part?

REPLY [3 votes]: $g(x)=\log(1+x^2)$ is a convex function on $(-1,1)$ and a concave function on $(-\infty,-1)$ and $(1,+\infty)$, since:
$$ g''(x) = 2\frac{1-x^2}{(1+x^2)^2}. $$
By computing the bevaviour in a neighbourhood of $x=1$, we have that $g(x)\leq x$ for any $x\geq 0$, hence the only solutions of $g(x)=x+2$ have to belong to $\mathbb{R}^-$. Over $\mathbb{R}^-$ $g(x)$ is a decreasing function and $x+2$ is an increasing function: by comparing the behaviours in a left neighbourhood of $x=0$ and in a right neighbourhood of $-\infty$ we get that $g(x)=x+2$ has a unique (negative) real solution. Few steps of Newton's method with starting point $-1$ give that such a root is $\approx -1.15369622$.