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Solution (a) Since $f$ is continuous on the closed interval $[0,2\pi]$ and $f^\prime$ and $f^{\prime \prime}$ exist on the open interval $\left( 0,\,2\pi \right)$, we can use the Test for Concavity. The first and second derivatives of $f$ are $$f^\prime (x) =\dfrac{d}{\textit{dx}}(x-2\;\cos x) =1+2\;\sin x \qquad \hbox{and}\qquad f^{\prime \prime} (x) =\dfrac{d}{\textit{dx}}(1+2\;\sin\;x) =2\;\cos x$$

To determine concavity, we solve the inequalities $f^{\prime \prime} (x) <0$ and $f^{\prime \prime} (x) >0.$ Since $f^{\prime \prime} (x) =2\;\cos\;x,$ we have

The function $f$ is concave up on the intervals $\left( 0,\dfrac{\pi }{2}\right)$ and $\left( \dfrac{3\pi }{2},2\pi \right) ,$ and $f$ is concave down on the interval $\left( \dfrac{\pi }{2},\dfrac{3\pi }{2}\right)$.

(b) The concavity of $f$ changes at $\dfrac{\pi }{2}$ and $\dfrac{3\pi }{2}$, so the points $\left( \dfrac{\pi }{2},\dfrac{\pi }{2}\right)$ and $\left( \dfrac{3\pi }{2},\dfrac{3\pi }{2}\right)$ are inflection points of $f$.

(c) We find the critical numbers by solving the equation $f^\prime(x) =0$. $$\begin{eqnarray*} 1+2\;\sin\;x &=& 0\qquad 0\leq x\leq 2\pi \\ \sin\;x &=&-\dfrac{1}{2} \\ x=\frac{7\pi }{6}\qquad &&\hbox{ or }\qquad {x=\frac{11\pi }{6}} \end{eqnarray*}$$

Now using the Second Derivative Test, we get $$f^{\prime \prime} \left(\dfrac{7\pi }{6}\right) =2\;\cos \left( \dfrac{7\pi }{6}\right) =- \sqrt{3}<0$$

and $$f^{\prime \prime} \left( \dfrac{11\pi }{6}\right) =2\;\cos \left(\dfrac{11\pi}{6}\right) = \sqrt{3}>0$$

So, $$f\left( \dfrac{7\pi }{6}\right) =\dfrac{7\pi }{6}-2\;\cos \dfrac{7\pi }{6}=\dfrac{7\pi }{6}+ \sqrt{3}\approx 5.4$$

is a local maximum value, and $$f\left( \dfrac{11\pi }{6}\right) =\dfrac{11\pi }{6}-2\cos \dfrac{11\pi }{6}=\dfrac{11\pi }{6}- \sqrt{3}\approx 4.03$$

is a local minimum value.