Graph \(f(x)=\sin x-\cos ^{2}x,\) \(0\leq x\leq 2\pi\) .
Step 1 The domain of \(f\) is \(\{ x|0\leq x\leq 2\pi \} \). Since \(f( 0) =\sin 0-\cos ^{2}0=-1\), the \(y\)-intercept is \(-1\). The \(x\)-intercepts satisfy the equation \[ \sin x-\cos ^{2}x=\sin x-( 1-\!\sin ^{2}x) =\sin ^{2}x+\sin x-1=0 \]
This trigonometric equation is quadratic in form, so we use the quadratic formula. \[ \sin x=\dfrac{-1\pm \sqrt{1-4( 1) (-1) }}{2}=\dfrac{ -1\pm \sqrt{5}}{2} \]
Trigonometric equations are discussed in Section P. 7, pp. 61–63.
Since \(\dfrac{-1-\sqrt{5}}{2}<-1\) and \(-1\leq \sin x\leq 1\), the \(x\) -intercepts occur at \(x=\sin ^{-1}\left( \dfrac{-1+\sqrt{5}}{2}\right) \approx 0.67\) and at \(x=\pi -\sin ^{-1}\left( \dfrac{-1+\sqrt{5}}{2}\right) \approx 2.48.\) Plot the intercepts.
Step 2 The function \(f\) has no asymptotes.
Step 3 \[ \begin{array}{@{}rcl} f' (x)&=&\dfrac{d}{\textit{dx}}( \sin x-\cos ^{2} x) =\cos x+2\, \cos\, x\, \sin x=\cos x(1+2\sin x)\\ f'' (x)& =&\dfrac{d}{\textit{dx}}[ {\cos x(1+2\sin x)}] =2\, \cos ^{2}x-\!\sin x(1+2\sin x)\\ &=&2\cos ^{2}x-\!\sin x-2\sin ^{2}x=-4\sin ^{2}x-\!\sin x+2 \end{array} \]
The critical numbers occur where \(f^\prime (x) =0\). That is, where \[ \cos x=0 \qquad\hbox{or}\qquad 1+2\sin x=0 \]
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In the interval \([0,2\pi ] \): \(\cos x=0\) if \(x=\dfrac{\pi }{2}\) or if \(x=\dfrac{3\pi }{2}\); and \(1+2\sin x=0\) when \(\sin x=-\dfrac{1}{2}\), that is, when \(x=\dfrac{7\pi }{6}\) or \(x=\dfrac{11\pi }{6}\). So, the critical numbers are \(\dfrac{\pi }{2}\), \(\dfrac{7\pi }{6}\), \(\dfrac{3\pi }{2}\), and \(\dfrac{11\pi }{6}\). At each of these numbers, the tangent lines are horizontal.
Step 4 To apply the Increasing/Decreasing Function Test, we use the numbers \(0,\) \(\dfrac{\pi }{2}\), \(\dfrac{7\pi }{6}\), \(\dfrac{3\pi }{2}\), \( \dfrac{11\pi }{6},\) and \(2\pi\) to form five intervals.
| Interval | Sign of fs\(^\prime\) | Conclusion |
|---|---|---|
| \(\left( 0,\dfrac{\pi }{2}\right)\) | positive | \(f\) is increasing on \(\left( 0,\dfrac{\pi }{2}\right)\) |
| \(\left( \dfrac{\pi }{2},\dfrac{7\pi }{6}\right)\) | negative | \(f\) is decreasing on \(\left( \dfrac{\pi }{2},\dfrac{7\pi}{6}\right)\) |
| \(\left(\dfrac{7\pi}{6},\dfrac{3\pi}{2}\right)\) | positive | \(f\) is increasing on \(\left( \dfrac{7\pi }{6},\dfrac{3\pi}{2}\right)\) |
| \(\left( \dfrac{3\pi }{2},\dfrac{11\pi }{6}\right)\) | negative | \(f\) is decreasing on \(\left( \dfrac{3\pi }{2},\dfrac{11\pi }{6}\right)\) |
| \(\left( \dfrac{11\pi }{6},2\pi \right)\) | positive | \(f\) is increasing on \(\left( \dfrac{11\pi }{6},2\pi \right)\) |
Step 5 By the First Derivative Test, \(f\left( \dfrac{\pi }{2} \right) =1\) and \(f\left( \dfrac{3\pi }{2}\right) =-1\) are local maximum values, and \(f\left(\dfrac{7\pi}{6}\right) =-\dfrac{5}{4}\) and \(f\left(\dfrac{11\pi}{6}\right) =-\dfrac{5}{4}\) are local minimum values. Plot these points.
Step 6 We apply the Test for Concavity. To solve \(f^{\prime \prime} (x) >0\) and \(f^{\prime \prime} (x) <0\), we first solve the equation \(f^{\prime \prime} (x) =-4\sin ^{2}x-\!\sin x+2=0,\) or equivalently, \[ \begin{eqnarray*} 4\sin ^{2}x+\sin x-2 &=&0 \qquad 0\leq x\leq 2\pi \\[3pt] \sin x &=&\dfrac{-1\pm \sqrt{1+32}}{8} \\[3pt] \sin x &\approx &0.593\qquad \hbox{or}\qquad \sin x\approx -0.843 \\ x &\approx &0.63 \qquad x\approx 2.51 \qquad x\approx 4.14 \qquad x\approx 5.28 \end{eqnarray*} \]
We use these numbers to form five subintervals of \([0,2\pi ]\).
| Interval | Sign of f\(^{\prime \prime}\) | Conclusion |
|---|---|---|
| \((0,0.63)\) | positive | \(f\) is concave up on the interval \(( 0,0.63)\) |
| \(( 0.63,2.51)\) | negative | \(f\) is concave down on the interval \(( 0.63,2.51)\) |
| \(( 2.51,4.14)\) | positive | \(f\) is concave up on the interval \(( 2.51,4.14)\) |
| \(( 4.14,5.28)\) | negative | \(f\) is concave down on the interval \(( 4.14,5.28)\) |
| \(( 5.28,2\pi )\) | positive | \(f\) is concave up on the interval \(( 5.28,2\pi )\) |
The function \(f\) in Example 6 is periodic with period \(2\pi\). To graph this function over its unrestricted domain, the set of real numbers, repeat the graph in Figure 51 over intervals of length \(2\pi\).
The inflection points are \(( 0.63,-0.06) \), \((2.51,-0.06) \), \(( 4.14,-1.13) \), and \(( 5.28,-1.13) .\) Plot the inflection points.
Step 7 The graph of \(f\) is given in Figure 51.