Answer
The curve, observing from left to right,
- starts at an absolute minimum at $(0,0)$
- rises on $(0,\pi/2)$, concave up,
- at $(\pi/2,2.721)$ changes concavity to "down" (inflection point),
- continues rising to a local maximum at $(4\pi/3, 8.255)$ turns, falls to
- $(3\pi/2,8.162)$ changes concavity to "up" (inflection point),
-falls to a local minimum at $(5\pi/3, 8.069)$ turns,
- and rises to the absolute maximum at $(2\pi,8.883)$
Work Step by Step
$y=f(x)=\sqrt{3}x-2\cos x,\qquad x\in[0,2\pi]$
$f'(x)=\sqrt{3}+\sin x$
$ f'(x)=0\quad$when $\left[\begin{array}{l}
\sin x=-\frac{\sqrt{3}}{2}\\
x=\frac{4\pi}{3},\frac{4\pi}{3}
\end{array}\right]$,
$\left[\begin{array}{cccccccc}
y': & & ++ & | & -- & | & ++\\
& & & \frac{4\pi}{3} & & \frac{5\pi}{3} & \\
y: & & \nearrow & \max & \searrow & \min & \nearrow
\end{array}\right]\left\{\begin{array}{l}
f(0)=-2\\
f(\frac{4\pi}{3})\approx 8.255\\
f(\frac{5\pi}{3})\approx 8.069\\
f(2\pi)\approx 8.883
\end{array}\right\}$
$f''(x)=-\cos x$
On $(0,2\pi)$
$f''(x)=0$ for $ x=\displaystyle \frac{\pi}{2},\frac{3\pi}{2}\qquad$... points of inflection.
$\left[\begin{array}{ccccccc}
y'': & ++ & | & -- & | & ++\\
& & \pi/2 & & 3\pi/2 & \\
y: & \cup & i.p. & \cap & i.p. & \cup
\end{array}\right],\quad \left\{\begin{array}{l}
f(\pi/2)\approx 2.721\\
f(3\pi/2)\approx 8.162\\
\end{array}\right\}$
