Chapter 3 - Applications of Differentiation - 3.3 Exercises - Page 183: 48

\[\begin{align} & f\left( x \right)\text{ has a relative minimum at }\left( \frac{3\pi }{2},-1 \right) \\ & f\left( x \right)\text{ has a relative maximum at }\left( \frac{\pi }{2},1 \right) \\ \end{align}\]

\[\begin{align} & f\left( x \right)=\frac{\sin x}{1+{{\cos }^{2}}x} \\ & \\ & \left( \text{a} \right)\text{Calculating the first derivative} \\ & f'\left( x \right)=\frac{d}{dx}\left[ \frac{\sin x}{1+{{\cos }^{2}}x} \right] \\ & f'\left( x \right)=\frac{\left( 1+{{\cos }^{2}}x \right)\left( \cos x \right)-\sin x\left( 2\cos x \right)\left( -\sin x \right)}{{{\left( 1+{{\cos }^{2}}x \right)}^{2}}} \\ & f'\left( x \right)=\frac{\cos x+{{\cos }^{3}}x+2{{\sin }^{2}}x\cos x}{{{\left( 1+{{\cos }^{2}}x \right)}^{2}}} \\ & f'\left( x \right)=\frac{\cos x\left( 2{{\sin }^{2}}x+1+{{\cos }^{2}}x \right)}{{{\left( 1+{{\cos }^{2}}x \right)}^{2}}} \\ & \text{Calculating the critical points}\text{, set }f'\left( x \right)=0 \\ & \cos x=0 \\ & \text{Solving for the interval }\left( 0,2\pi \right)\text{ we obtain the critical numbers} \\ & 2{{\sin }^{2}}x+1+{{\cos }^{2}}x=0,\text{ no real solutions}\text{, then} \\ & \cos x=0 \\ & x=\frac{\pi }{2},\text{ }x=\frac{3\pi }{2} \\ & \text{Set the intervals:} \\ & \left( 0,\frac{\pi }{2} \right),\left( \frac{\pi }{2},\frac{3\pi }{2} \right),\left( \frac{3\pi }{2},2\pi \right) \\ & \\ & \text{Making a table of values } \\ & \begin{matrix} \text{Interval} & \text{Test Value} & \text{Sign of }f'\left( x \right) & \text{Conclusion} \\ \left( 0,\frac{\pi }{2} \right) & x=\frac{\pi }{4} & >0 & \text{Increasing} \\ \left( \frac{\pi }{2},\frac{3\pi }{2} \right) & x=\pi & <0 & \text{Decreasing} \\ \left( \frac{3\pi }{2},2\pi \right) & x=\frac{7\pi }{4} & >0 & \text{Increasing} \\ \end{matrix} \\ & \\ & \text{Inscreasing on }\left( 0,\frac{\pi }{2} \right)\text{ and }\left( \frac{3\pi }{2},2\pi \right) \\ & \text{Decreasing on }\left( \frac{\pi }{2},\frac{3\pi }{2} \right) \\ & \\ & *f'\left( x \right)\text{ changes from positive to negative at }x=\frac{\pi }{2},\text{ then } \\ & f\left( x \right)\text{ has a relative maximum at }x=\frac{\pi }{2} \\ & f\left( \frac{\pi }{2} \right)=1 \\ & \\ & *f'\left( x \right)\text{ changes from negative to positive at }x=\frac{3\pi }{2},\text{ then } \\ & f\left( x \right)\text{ has a relative minimum at }x=\frac{3\pi }{2}\text{ } \\ & f\left( \frac{3\pi }{2} \right)=-1 \\ & \\ & \left( \text{c} \right)\text{Graph} \\ \end{align}\]