Use Rolle to determine the number of possible solution of $f(x) = \sin x + x + 1$

Question

Good morning. I've found an exercise that require to find the number of possible solutions of the equation $f(x) = \sin x +x + 1$ in $[0, 2\pi]$ using Rolle's theorem. How should I proceed? I know that the theorem ask for the function to be continuous in $[a, b]$, to be differentiable in $(a, b)$ and that $f(a) = f(b)$. $f(x)$ is continuous and differentiable but $f(0) = 1$ and $f(2\pi) = 1 + 2\pi$. How can I go on from here? Should I assume that, since i cannot use Rolle, I'm sure that there cannot be more than one solution?

Answer 1

I'm assuming that you're asking about the zeroes of $f$. We have $\sin(x)+1\geq 0$ for all $x$. If $x>0$, then $f(x)>0$ so there are no positive zeroes. It remains to check that $f(0)=1\ne 0$.