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\begin{document}
%MODULE 1: Mean Value Theorem


%USER DEFINED COMMANDS
\global\long\def\ds{{\displaystyle }}
 \global\long\def\nsp{\hspace{1em}*{9pt}}
 \global\long\def\lns{\rule{0pt}{11pt} }
 \global\long\def\xp{\mbox{\Large\sl e} }


%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{The Mean Value Theorem}


\subtitle{ You already know that the slope of the tangent line is the limit
of the slopes of the secant lines. The {\hilite Mean Value Theorem}
gives us another, very useful connection between the slopes of secant
lines and the slopes of tangent lines. We make use of the Mean Value
Theorem in future modules to justify some applications. }

\makebeamertitle

\lyxframeend{}\lyxframe{{Sample problems}}

\label{page:modf-problem} \begin{question} 
\begin{itemize}
\item Let $\ec f(x)=\sin(\frac{\pi}{2}x)$. Is there an $x$ value for which
$\ec f'(x)=1$? 
\item Suppose that $\ec f$ and $\ec g$ have the same derivative. What
can I conclude about $\ec f$ and $\ec g$: are they equal? if not,
how close are they? 
\end{itemize}
\end{question}

%1



\lyxframeend{}\lyxframe{{SAGE worksheets}}

\noindent For this module you will not need a SAGE worksheet, but
you can use a blank one to draw some of the graphs we consider.


\lyxframeend{}\lyxframe{{The Mean Value Theorem}}

\noindent The {\hilite Mean Value Theorem} (MVT) gives us a new
and interesting connection between the slope of a function's secant
lines and the slope of a tangent line.
\begin{theorem}%{}
[(The Mean Value Theorem)] Suppose that $\ec f$ is a function that
is continuous on $\ec[a,b]$ and differentiable on $\ec(a,b)$. There
exists $\ec c\in(a,b)$ such that\vspace{-2ex}
\[
{\ec f'(c)=\frac{f(b)-f(a)}{b-a}}.
\]

\end{theorem}%{}
What exactly is this saying?


\pause{}

\noindent Notice that\vspace{-2ex}
\[
\ec\frac{f(b)-f(a)}{b-a}
\]
is the slope of the secant line connecting the points $\ec(a,f(a))$
and $\ec(b,f(b))$. On the other hand, $\ec f'(c)$ is the slope of
the tangent line at $\ec x=c$. If MVT's hypotheses about $\ec f$,
$\ec a$, and $\ec b$ are satisfied, then we can find a point $\ec c$
where the slope of the tangent line is the same as the slope of the
secant line.


\lyxframeend{}\lyxframe{{Example 1}}

\noindent We'll illustrate MVT using the first example. Recall the
question:\begin{question} Let $\ec f(x)=\sin(\frac{\pi}{2}x)$. Is
there an $x$ value for which $\ec f'(x)=1$? \end{question}


\pause{}

\noindent We could answer {}``YES!!!'' if we could find $\ec a,b$
satisfying MVT's hypotheses \emph{and} having a secant line whose
slope is 1.


\pause{}

\noindent Such $\ec a,b$ exist! If $\ec a=1$ and $\ec b=-1$, then
\[
{\ec\frac{f(b)-f(a)}{b-a}=\frac{\sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{2}\right)}{1-\left(-1\right)}=\frac{1-(-1)}{2}=1}.
\]



\lyxframeend{}\lyxframe{{Example 1 illustrated}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>

\noindent How about some pictures? The plot of $\ec\sin(\frac{\pi}{2}x)$
looks like this: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/sinplot} 
\end{figure}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<2>

\noindent The secant line connect $\ec x=-1$ and $\ec x=1$ has slope
$\ec1$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/sinsecplot} 
\end{figure}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3->

\noindent MVT tells us that a point $\ec x=c$ exists whose tangent
line has the slope of this secant line: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/sinsectanplot} 
\end{figure}


\end{onlyenv}%{}

\pause{}


\noindent In this case, we can use calculus to show that 
\[
{\ec c=\frac{2}{\pi}\arccos\frac{2}{\pi}}.
\]
Finding the value of $\ec$ is not always easy, necessary---or even
possible!---so MVT allows us to decide whether such a point exists
without making that effort.

\end{overlayarea}

\lyxframeend{}\lyxframe{{Rolle's Theorem}}

MVT can be proved using a similar fact called {\hilite Rolle's Theorem}:
\begin{theorem}%{}
[(Rolle's Theorem)]Suppose that $\ec f$ is a function that is continuous
on $\ec[a,b]$ and differentiable on $\ec(a,b)$. If $\ec f(a)=f(b)$,
then there exists $\ec c\in(a,b)$ such that
\[
{\ec f'(c)=0}.
\]

\end{theorem}%{}
You can see that Rolle's Theorem is a special case of MVT, since the
slope of the secant line between $\ec x=a$ and $\ec x=b$ is zero.
We will not prove either MVT or Rolle's Theorem, but your textbook
has a proof, and your instructor can provide one if necessary.


\lyxframeend{}\lyxframe{\noindent {Example 2}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>

\noindent Recall Question 2: \begin{question} Suppose that $\ec f$
and $\ec g$ have the same derivative. What can I conclude about $\ec f$
and $\ec g$: are they equal? if not, how close are they? \end{question}
The answer to this question is quite important actually:

\end{onlyenv}%{}
\begin{theorem}%{}
\noindent If $\ec f'(x)=g'(x)$ for all $\ec x\in(a,b)$, then $\ec f$
and $\ec g$ differ only by a constant; that is,
\[
{\ec f(x)=g(x)+c\qquad\exists c\in\mathbb{R}}.
\]

\end{theorem}%{}

\pause{}
\begin{onlyenv}%{}
<2-3>

\noindent Why is this true? Suppose that $\ec f'(x)=g'(x)$ for all
$\ec x\in(a,b)$. First, notice that
\[
{\ec f'(x)-g'(x)=0}
\]
 for all $\ec x\in(a,b)$.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3-4>

\noindent Let $\ec F=f-g$; by substitution
\[
{\ec F'(x)=0}
\]
for all $\ec x\in(a,b)$.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4->

\noindent Let $\ec x_{1},x_{2}$ be any two numbers in $\ec(a,b)$;
by the Mean Value Theorem there exists $\ec c\in(x_{1},x_{2})$ such
that
\[
{\ec F'(c)=\frac{F(x_{1})-F(x_{2})}{x_{1}-x_{2}}}.
\]


\end{onlyenv}%{}

\pause{}


\noindent However, $\ec F'(x)=0$ for \emph{all} $\ec x\in(x_{1},x_{2})\subset(a,b)$,
so $\ec F'(c)=0$. The equation above thus simplifies to
\[
{\ec0=\frac{F(x_{1})-F(x_{2})}{x_{1}-x_{2}}\quad\Longrightarrow\quad0=F(x_{1})-F(x_{2})}.
\]


\end{overlayarea}

\lyxframeend{}\lyxframe{Example 2}
\begin{overlayarea}
{\textwidth}{6in}
\begin{theorem}%{}
\noindent If $\ec f'(x)=g'(x)$ for all $\ec x\in(a,b)$, then $\ec f$
and $\ec g$ differ only by a constant; that is,
\[
{\ec f(x)=g(x)+c\qquad\exists c\in\mathbb{R}}.
\]
 \end{theorem}%{}
\begin{onlyenv}%{}
<1>

\noindent Why is this true? \ldots{}However, $\ec F'(x)=0$ for \emph{all}
$\ec x\in(x_{1},x_{2})\subset(a,b)$, so $\ec F'(c)=0$. The equation
above thus simplifies to
\[
{\ec0=\frac{F(x_{1})-F(x_{2})}{x_{1}-x_{2}}\quad\Longrightarrow\quad0=F(x_{1})-F(x_{2})}.
\]
Since $\ec x_{1}$ and $\ec x_{2}$ are arbitrary, and the difference
between their $y$ values is zero, $\ec F$ must be constant everywhere
on $\ec(a,b)$.

\end{onlyenv}%{}

\pause{}


\noindent Since $\ec F=f-g$, we conclude that $\ec f$ and $\ec g$
differ only by a constant.


\pause{}


\noindent For example, since the derivative of $\ec\sin x$ is $\ec\cos x$,
we know now that \emph{every} function whose derivatrive is $\ec\cos x$
has the form
\[
\ec\sin x+C
\]
 where $\ec C$ is a constant.

\end{overlayarea}

\lyxframeend{}\lyxframe{{Conclusion} }

\noindent In this module, we discussed: 
\begin{itemize}
\item The {\hilite Mean Value Theorem} (MVT):\setlength{\shadeboxwidth}{4in}

\begin{theorem}%{}
Suppose that $\ec f$ is a function that is continuous on $\ec[a,b]$
and differentiable on $\ec(a,b)$. There exists $\ec c\in(a,b)$ such
that
\[
{\ec f'(c)=\frac{f(b)-f(a)}{b-a}}.
\]
 
\end{theorem}%{}
\item As a consequence of MVT

\begin{theorem}%{}
If $\ec f'(x)=g'(x)$ for all $\ec x\in(a,b)$, then $\ec f$ and
$\ec g$ differ only by a constant; that is,
\[
{\ec f(x)=g(x)+c\qquad\exists c\in\mathbb{R}}.
\]
 
\end{theorem}%{}
\end{itemize}

\lyxframeend{}
\end{document}
