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	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 4\medskip\\}
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		\mbox{}\\
		Given following function: -
		\[ f(x) = -\cos(x)-\frac{x}{2}\]
		On the interval $[0,2\pi]$:
		\begin{enumerate}
			\item What interval(s) is the function increasing?
			\item What interval(s) is the function decreasing?
			\item Where is the function neither increasing nor decreasing?
		\end{enumerate}
		\begin{soln*} \mbox{}
			Note \[ f'(x) = \sin(x)-\frac{1}{2} \]
			\begin{enumerate}
				\item When the function is increasing, the derivative will be positive. Thus we need to solve when \[ \sin(x) - \frac{1}{2} > 0\] ie when \[\sin(x) > \frac{1}{2} \]
\begin{figure}[h]
	\centering
	\includegraphics[bb=0 0 697 417,scale=.4]{quiz4.png}
	% quiz4.png: 930x556 pixel, 96dpi, 24.60x14.71 cm, bb=0 0 697 417
\end{figure}

From the picture (and from knowing when $\sin(x)=\frac{1}{2}$) one can see that this is true on the interval $(\frac{\pi}{6},\frac{5\pi}{6})$.
				\item We need to show when the derivative is negative. As you can see, it will be on the intervals $[0,\frac{\pi}{6})$ and $(\frac{5\pi}{6},2\pi]$
				\item This is when \[ \sin(x) = \frac{1}{2} \] This happens at $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$.


			\end{enumerate}

		\end{soln*}

	}
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