EXAMPLE 6Using Calculus to Graph a Trigonometric Function

Graph $f(x)=\sin x-\cos ^{2}x,$ $0\leq x\leq 2\pi$ .

Solution

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. $$\sin x=\dfrac{-1\pm \sqrt{1-4( 1) (-1) }}{2}=\dfrac{ -1\pm \sqrt{5}}{2}$$

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 $$\cos x=0 \qquad\hbox{or}\qquad 1+2\sin x=0$$

315

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.

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 $[0,2\pi ]$.

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.