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%TCIDATA{<META NAME="Title" CONTENT="The Mean Value Theorem Speed Trap">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Wed Mar 26 00:02:22 1997}
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\section{The Mean Value Theorem Speed Trap \label{Speed trap}}

\hspace{-1in}``But officer,'' replied the motorist, ``my speedometer was
under 70 mph every time the radar-warning light came on. How can you accuse
me of speeding?''

``Your car and license number were recorded on video less than an hour ago
in Montrose,'' explained the officer, ``and that's over 80 miles away. You
certainly deserve a speeding ticket in this 75 mph zone.''

Assume that the first sentence of each speaker is true and that the speed
limit is at most 75 miles per hour along this stretch of highway. Use the
Mean Value Theorem to settle this argument.

\subsection{Janet's Solution}

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The motorist must have exceeded $80$ miles per hour at some instant during
the past hour. Because the position and velocity of the car are continuous
functions of time, the Mean Value Theorem states that at some instant of
time the instantaneous velocity must equal the average velocity, which is
greater than $\frac{80-0}{1}=\allowbreak 80$ because the elapsed time is
less than one hour.

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\subsection{Activities}

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Define $f(x)=\dfrac{x+1}{x-1}$. Plot a graph of $y=f(x)$ using the viewing
rectangle $0\leq x\leq 2$ by $-10\leq y\leq 10$. Add the secant line joining 
$\left( 0,f(0)\right) $ and $\left( 2,f(2)\right) $ to your plot. Show that
there is no value of $c$ between $0$ and $2$ such that 
\[
f^{\prime }(c)=\frac{f(2)-f(0)}{2-0} 
\]
Explain why this does not contradict the Mean Value Theorem.

\label{Ups and downs}

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