Answer
The curve, observing from left to right,
- starts at an absolute minimum at $(0,0)$
- rises on $(0,\pi)$, concave up,
- at $(\pi,\pi)$ changes concavity to "down" (inflection point),
- continues rising to an absolute maximum at $(2\pi,2,\pi).$
Work Step by Step
$y=f(x)=x-\sin x,\qquad x\in[0,2\pi]$
$f'(x)=1-\cos x$
which is never negative ($-1\leq\cos x\leq 1)$, so $f$ rises on $(0,2\pi)$
$f''(x)=-\sin x$
On $(0,2\pi)$
$f''(x)=0$ for $ x=\pi\qquad$... point of inflection.
$\left[\begin{array}{ccccc}
y'': & ++ & | & --\\
& & \pi & \\
y: & \cup & i.p. & \cap
\end{array}\right]\left\{\begin{array}{l}
f(0)=0\\
f(\pi)=\pi\\
f(2\pi)=2\pi
\end{array}\right.$
