I saw an exercise that asks to prove that $f(x):=x^{2}+e^{0.1x}-1$ have a root $r<0$.
The solution stated that $f''(x)=2+(0.1)^{2}e^{0.1x}>0$ hence there is a maximum of two roots, since $0$ is a root and since $$f'(0)=0.1>0$$ there is a root $r<0$.
I know that a convex function have at most $2$ roots, but I don't understand why $f'(0)>0$ imply that there is a second root and that it is negative.
I would appreciate it if someone could explain the reasoning
The derivative is positive at $0$ (which is a root). This means the function is increasing there. Further, the second derivative is always positive, so the function will always be increasing for positive $x$. Thus if the second root exists, then it must be negative.
Unfortunately, simply knowing the derivative is positive at $0$ and that the second derivative is always positive does not tell us that there is a second root. But we can quickly verify that there is a second root, by noting that $f(-10) > 0$ trivially and seeing that the function is continuous. Now, since the derivative is positive at zero, $f$ will be negative in some neighborhood of $0$. By continuity, $f$ will be zero between $-10$ and $0$.
And that's the best you can do.