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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{The Mean Value Theorem}

\begin{example}
Find the local maximum and minimum points and values, and the absolute maximum
minimum points and values, of the function \[
f(x)=2\sin (x)+\cos (2x),\: x\in [0,2\pi ].\]

\end{example}

\subsubsection*{Solution}

To find the critical points, we first differentiate the function,\[
f'(x)=2\cos (x)-2\sin (2x),\]
and we solve the equation \( f'(x)=0 \), \begin{eqnarray*}
0 & = & f'(x)\\
 & = & 2\cos (x)-2\sin (2x)\\
 & = & 2\cos (x)-4\sin (x)\cos (x)\: \textrm{since}\, \, \sin (2x)=2\sin (x)\cos (x)\\
 & = & 2\cos (x)(1-2\sin (x)),
\end{eqnarray*}
so that \[
f'(x)=0\iff \cos (x)=0\, or\, \sin (x)=1/2.\]
Solving for \( x \) at this stage is simply a matter of remembering where \( \cos (x)=0 \)
(which is \( x=\pi /2,\, x=3\pi /2 \) only, in the domain \( [0,2\pi ] \)),
or where \( \sin (x)=1/2 \) (\( x=\pi /6 \) or \( x=5\pi /6 \) only, in the
interval). 

According to the theorem mentioned earlier, called Fermat's Theorem, if the
derivative exists at a maximum or minimum point (local or absolute), then the
derivative has to be 0. The only other options are places where the derivative
does not exist, or the endpoints of the interval. In this case, though, the
derivative exists at all \( x \), so the only possibilities for maximum or
minimum points are at these places where the derivative is 0, and at the endpoints,
\( 0 \) and \( 2\pi  \).

So, how do you tell which is a max or a min? First off, evaluate the function
at all of these points. Since:\begin{eqnarray*}
f(x) & = & 2\sin (x)+\cos (2x)\\
f(0) & = & 2\cdot 0+1=1\\
f(\pi /6) & = & 2\cdot \frac{1}{2}+\frac{1}{2}=\frac{3}{2}\\
f(\pi /2) & = & 2\cdot 1+(-1)=1\\
f(5\pi /6) & = & 2\cdot \frac{1}{2}+\frac{1}{2}=\frac{3}{2}\\
f(3\pi /2) & = & 2\cdot (-1)+(-1)=-3\\
f(2\pi ) & = & 2\cdot 0+1=1,
\end{eqnarray*}
the absolute maximum value is \( 3/2 \), achieved at \( x=\pi /6 \) and \( x=5\pi /6 \).
The absolute minimum value is \( -3 \), achieved at \( x=3\pi /2 \). 

What about the points in between? Which are local maxima, and which are local
minima? Well, since the function has to increase up to, and decrease away from,
a local maximum,

\vspace{0.3cm}
{\par\centering \includegraphics{local-max.ps} \par}
\vspace{0.3cm}

then \( f'(x)>0 \) for \( x<x_{0} \) and \( f'(x)<0 \) for \( x>x_{0} \)
at a local maximum point \( x_{0} \), as long as the \( x \) doesn't get too
far from \( x_{0} \). Similarly, \( f'(x)<0 \) for \( x<x_{0} \) and \( f'(x)>0 \)
for \( x>x_{0} \) at a local minimum point \( x_{0} \), as long as the \( x \)
doesn't get too far from \( x_{0} \). In this case, we can look at the sign
of the first derivative to see what happens. Since \[
f'(x)=2\cos (x)(1-2\sin (x)),\]
you can see that \( f'(x) \) will have the following signs:

\vspace{0.3cm}
{\par\centering \includegraphics{derivative-signs.ps} \par}
\vspace{0.3cm}

which clearly shows that \( x=\pi /6 \) is a local maximum (not surprizing,
since it is an absolute maximum), \( x=\pi /2 \) is a local minimum, \( x=5\pi /6 \)
is a local max, and \( x=3\pi /2 \) is a local min. So, that left-over point,
\( x=\pi /2 \), is a local minimum point. We don't usually call an endpoint
a {}``local{}'' maximum or minimum, but an endpoint might be an absolute maximum
or minimum.
\end{document}
