Prove that $f(x) = 3-\cos(x)$ for all $x\in [0,\pi]$

Question

Suppose that $f(x)$ is differentiable everywhere such that

$f(0) = 2$, $f(\pi)=4$ and $f'(x)\ge\sin(x)\quad \forall x.$

then prove that $f(x) = 3-\cos(x)$ for all $x \in [0, \pi]$

Answer 1

Hint: let $\,g(x)=f(x)-3+\cos(x)\,$, then the problem can be restated as "given $\,g(0)=g(\pi)=0\,$ and $\,g'(x)\ge 0\,$ then prove that $\,g(x) = 0\,$ on $\,[0,\pi]\,$".