Using the Second Derivative Test to Identify Local Extrema

  1. Determine where \(f(x)=x-2\cos x,\) \(0\) \(\leq x\leq 2\pi\), is concave up and concave down.
  2. Find any inflection points.
  3. Use the Second Derivative Test to identify any local extreme values.

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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

  • \(f^{\prime \prime} (x) > 0\qquad \hbox{ when }0<x<\dfrac{\pi }{2}\quad \hbox{ and }\quad\dfrac{3\pi }{2}<x<2\pi\)
  • \(f^{\prime \prime} (x) < 0\qquad \hbox{ when }\dfrac{\pi }{2}<x<\dfrac{3\pi }{2}\)
    •

    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.