I am trying to understand the solution.
Why isn't $f^{\prime}(x)$ only $= 1-sin x $. What am I missing?
$f(x)=x+\cos x$ on $[-\pi, 2 \pi]$ $f^{\prime}(x)=1-\sin x \geq 0$ and $\sin x \leq 1 ;$ this is true for any real $x$.
The maximum value > is $f(2 \pi)=2 \pi+\cos (2 \pi)=2 \pi+1$ and the minimum value is $f(\pi)=\pi+\cos (\pi)=\pi-1$
Thanks.
It's true that $f'(x)=1-\sin x$. Then, as David comments, the solution says that $f'(x)\geqslant0$ for all $x\in\mathbb{R}$, because $\sin x\leqslant1$, so $$\sin x\leqslant1\Longrightarrow-\sin x\geqslant-1\Longrightarrow f'(x)=1-\sin x\geqslant0.$$ We can conclude that $f$ is always increasing and that its minimun value is attained at the lower end of the interval. ¿$\pi$ or $-\pi$? (something is wrong there). And its maximun value can be found at the higher end, $x=2\pi$.