Can the derivative prove my function has only one root? 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?
 A: If you know the derivative is positive everywhere, then you know the function has at most one root -- for example, you can reason that if it had two different roots, then the mean value theorem says that $f'(x)=0$ for some $x$ between those roots, which you know isn't the case.
Unfortunately, in this case you need a little more finesse than that, because $f'(1)=0$, as André observed. However, the argument still shows that there can't be two roots on the same side of $1$ -- and you can easily calculate that $f(1)>0$, so another application of the MVT shows that there can't be a root greater than $1$.
But you need to argue separately that the function has at least one root. The intermediate value theorem will do that for you if you show that the function has both a positive and a negative value.
A: $g(x)=\log(1+x^2)$ 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 $\approx -1.15369622$.
