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%MODULE 1: Mean Value Theorem
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\begin_layout Title
The Mean Value Theorem
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\begin_layout Subtitle
 You already know that the slope of the tangent line is the limit of the
 slopes of the secant lines.
 The 
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\backslash
hilite
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 Mean Value Theorem
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}
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 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.
 
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Sample problems
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\backslash
begin{question}
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\end_layout

\begin_layout Itemize
Let 
\begin_inset Formula $\ec f(x)=\sin(\frac{\pi}{2}x)$
\end_inset

.
 Is there an 
\begin_inset Formula $x$
\end_inset

 value for which 
\begin_inset Formula $\ec f'(x)=1$
\end_inset

? 
\end_layout

\begin_layout Itemize
Suppose that 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

 have the same derivative.
 What can I conclude about 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

: are they equal? if not, how close are they? 
\end_layout

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%1
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SAGE worksheets
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}
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\begin_layout Standard
\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.
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{
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The Mean Value Theorem
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\begin_layout Standard
\noindent
The 
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 Mean Value Theorem
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}
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 (MVT) gives us a new and interesting connection between the slope of a
 function's secant lines and the slope of a tangent line.
\end_layout

\begin_layout Theorem
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[
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(The Mean Value Theorem)
\begin_inset ERT
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]
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\end_inset

 Suppose that 
\begin_inset Formula $\ec f$
\end_inset

 is a function that is continuous on 
\begin_inset Formula $\ec[a,b]$
\end_inset

 and differentiable on 
\begin_inset Formula $\ec(a,b)$
\end_inset

.
 There exists 
\begin_inset Formula $\ec c\in(a,b)$
\end_inset

 such that
\begin_inset VSpace -2ex
\end_inset


\begin_inset Formula 
\[
{\ec f'(c)=\frac{f(b)-f(a)}{b-a}}.
\]

\end_inset


\end_layout

\begin_layout Standard
What exactly is this saying?
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
Notice that
\begin_inset VSpace -2ex
\end_inset


\begin_inset Formula 
\[
\ec\frac{f(b)-f(a)}{b-a}
\]

\end_inset

is the slope of the secant line connecting the points 
\begin_inset Formula $\ec(a,f(a))$
\end_inset

 and 
\begin_inset Formula $\ec(b,f(b))$
\end_inset

.
 On the other hand, 
\begin_inset Formula $\ec f'(c)$
\end_inset

 is the slope of the tangent line at 
\begin_inset Formula $\ec x=c$
\end_inset

.
 If MVT's hypotheses about 
\begin_inset Formula $\ec f$
\end_inset

, 
\begin_inset Formula $\ec a$
\end_inset

, and 
\begin_inset Formula $\ec b$
\end_inset

 are satisfied, then we can find a point 
\begin_inset Formula $\ec c$
\end_inset

 where the slope of the tangent line is the same as the slope of the secant
 line.
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Example 1
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\begin_layout Standard
\noindent
We'll illustrate MVT using the first example.
 Recall the question:
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\backslash
begin{question}
\end_layout

\end_inset

 Let 
\begin_inset Formula $\ec f(x)=\sin(\frac{\pi}{2}x)$
\end_inset

.
 Is there an 
\begin_inset Formula $x$
\end_inset

 value for which 
\begin_inset Formula $\ec f'(x)=1$
\end_inset

? 
\begin_inset ERT
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\backslash
end{question}
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\end_inset


\end_layout

\begin_layout Pause

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\begin_layout Standard
\noindent
We could answer 
\begin_inset Quotes eld
\end_inset

YES!!!
\begin_inset Quotes erd
\end_inset

 if we could find 
\begin_inset Formula $\ec a,b$
\end_inset

 satisfying MVT's hypotheses 
\emph on
and
\emph default
 having a secant line whose slope is 1.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
Such 
\begin_inset Formula $\ec a,b$
\end_inset

 exist! If 
\begin_inset Formula $\ec a=1$
\end_inset

 and 
\begin_inset Formula $\ec b=-1$
\end_inset

, then
\begin_inset Formula 
\[
{\ec\frac{f(b)-f(a)}{b-a}=\frac{\sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{2}\right)}{1-\left(-1\right)}=\frac{1-(-1)}{2}=1}.
\]

\end_inset


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{
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Example 1 illustrated
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<1>
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\end_layout

\begin_deeper
\begin_layout Standard
\noindent
How about some pictures? The plot of 
\begin_inset Formula $\ec\sin(\frac{\pi}{2}x)$
\end_inset

 looks like this: 
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
 
\end_layout

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/sinplot.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


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\end_deeper
\begin_layout Pause

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<2>
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\begin_deeper
\begin_layout Standard
\noindent
The secant line connect 
\begin_inset Formula $\ec x=-1$
\end_inset

 and 
\begin_inset Formula $\ec x=1$
\end_inset

 has slope 
\begin_inset Formula $\ec1$
\end_inset

: 
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
 
\end_layout

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/sinsecplot.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<3->
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
MVT tells us that a point 
\begin_inset Formula $\ec x=c$
\end_inset

 exists whose tangent line has the slope of this secant line: 
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
 
\end_layout

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/sinsectanplot.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
In this case, we can use calculus to show that 
\begin_inset Formula 
\[
{\ec c=\frac{2}{\pi}\arccos\frac{2}{\pi}}.
\]

\end_inset

Finding the value of 
\begin_inset Formula $\ec$
\end_inset

 is not always easy, necessary---or even possible!---so MVT allows us to
 decide whether such a point exists without making that effort.
\end_layout

\end_deeper
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{
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\end_inset

Rolle's Theorem
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}
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\begin_layout Standard
MVT can be proved using a similar fact called 
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{
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\backslash
hilite
\end_layout

\end_inset

 Rolle's Theorem
\begin_inset ERT
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}
\end_layout

\end_inset

:
\end_layout

\begin_layout Theorem
\begin_inset ERT
status open

\begin_layout Plain Layout

[
\end_layout

\end_inset

(Rolle's Theorem)
\begin_inset ERT
status open

\begin_layout Plain Layout

]
\end_layout

\end_inset

Suppose that 
\begin_inset Formula $\ec f$
\end_inset

 is a function that is continuous on 
\begin_inset Formula $\ec[a,b]$
\end_inset

 and differentiable on 
\begin_inset Formula $\ec(a,b)$
\end_inset

.
 If 
\begin_inset Formula $\ec f(a)=f(b)$
\end_inset

, then there exists 
\begin_inset Formula $\ec c\in(a,b)$
\end_inset

 such that
\begin_inset Formula 
\[
{\ec f'(c)=0}.
\]

\end_inset


\end_layout

\begin_layout Standard
You can see that Rolle's Theorem is a special case of MVT, since the slope
 of the secant line between 
\begin_inset Formula $\ec x=a$
\end_inset

 and 
\begin_inset Formula $\ec x=b$
\end_inset

 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.
\end_layout

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\noindent
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{
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Example 2
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\begin_deeper
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<1>
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\begin_deeper
\begin_layout Standard
\noindent
Recall Question 2: 
\begin_inset ERT
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\backslash
begin{question}
\end_layout

\end_inset

 Suppose that 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

 have the same derivative.
 What can I conclude about 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

: are they equal? if not, how close are they? 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset

 The answer to this question is quite important actually:
\end_layout

\end_deeper
\begin_layout Theorem
\noindent
If 
\begin_inset Formula $\ec f'(x)=g'(x)$
\end_inset

 for all 
\begin_inset Formula $\ec x\in(a,b)$
\end_inset

, then 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

 differ only by a constant; that is,
\begin_inset Formula 
\[
{\ec f(x)=g(x)+c\qquad\exists c\in\mathbb{R}}.
\]

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<2-3>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
Why is this true? Suppose that 
\begin_inset Formula $\ec f'(x)=g'(x)$
\end_inset

 for all 
\begin_inset Formula $\ec x\in(a,b)$
\end_inset

.
 First, notice that
\begin_inset Formula 
\[
{\ec f'(x)-g'(x)=0}
\]

\end_inset

 for all 
\begin_inset Formula $\ec x\in(a,b)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<3-4>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
Let 
\begin_inset Formula $\ec F=f-g$
\end_inset

; by substitution
\begin_inset Formula 
\[
{\ec F'(x)=0}
\]

\end_inset

for all 
\begin_inset Formula $\ec x\in(a,b)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

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<4->
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
Let 
\begin_inset Formula $\ec x_{1},x_{2}$
\end_inset

 be any two numbers in 
\begin_inset Formula $\ec(a,b)$
\end_inset

; by the Mean Value Theorem there exists 
\begin_inset Formula $\ec c\in(x_{1},x_{2})$
\end_inset

 such that
\begin_inset Formula 
\[
{\ec F'(c)=\frac{F(x_{1})-F(x_{2})}{x_{1}-x_{2}}}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
However, 
\begin_inset Formula $\ec F'(x)=0$
\end_inset

 for 
\emph on
all
\emph default
 
\begin_inset Formula $\ec x\in(x_{1},x_{2})\subset(a,b)$
\end_inset

, so 
\begin_inset Formula $\ec F'(c)=0$
\end_inset

.
 The equation above thus simplifies to
\begin_inset Formula 
\[
{\ec0=\frac{F(x_{1})-F(x_{2})}{x_{1}-x_{2}}\quad\Longrightarrow\quad0=F(x_{1})-F(x_{2})}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout BeginFrame
Example 2
\end_layout

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status open

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{
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\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Theorem
\noindent
If 
\begin_inset Formula $\ec f'(x)=g'(x)$
\end_inset

 for all 
\begin_inset Formula $\ec x\in(a,b)$
\end_inset

, then 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

 differ only by a constant; that is,
\begin_inset Formula 
\[
{\ec f(x)=g(x)+c\qquad\exists c\in\mathbb{R}}.
\]

\end_inset

 
\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<1>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
Why is this true? \SpecialChar \ldots{}
However, 
\begin_inset Formula $\ec F'(x)=0$
\end_inset

 for 
\emph on
all
\emph default
 
\begin_inset Formula $\ec x\in(x_{1},x_{2})\subset(a,b)$
\end_inset

, so 
\begin_inset Formula $\ec F'(c)=0$
\end_inset

.
 The equation above thus simplifies to
\begin_inset Formula 
\[
{\ec0=\frac{F(x_{1})-F(x_{2})}{x_{1}-x_{2}}\quad\Longrightarrow\quad0=F(x_{1})-F(x_{2})}.
\]

\end_inset

Since 
\begin_inset Formula $\ec x_{1}$
\end_inset

 and 
\begin_inset Formula $\ec x_{2}$
\end_inset

 are arbitrary, and the difference between their 
\begin_inset Formula $y$
\end_inset

 values is zero, 
\begin_inset Formula $\ec F$
\end_inset

 must be constant everywhere on 
\begin_inset Formula $\ec(a,b)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
Since 
\begin_inset Formula $\ec F=f-g$
\end_inset

, we conclude that 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

 differ only by a constant.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
For example, since the derivative of 
\begin_inset Formula $\ec\sin x$
\end_inset

 is 
\begin_inset Formula $\ec\cos x$
\end_inset

, we know now that 
\emph on
every
\emph default
 function whose derivatrive is 
\begin_inset Formula $\ec\cos x$
\end_inset

 has the form
\begin_inset Formula 
\[
\ec\sin x+C
\]

\end_inset

 where 
\begin_inset Formula $\ec C$
\end_inset

 is a constant.
\end_layout

\end_deeper
\begin_layout BeginFrame
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{
\end_layout

\end_inset

Conclusion
\begin_inset ERT
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}
\end_layout

\end_inset

 
\end_layout

\begin_layout Standard
\noindent
In this module, we discussed: 
\end_layout

\begin_layout Itemize
The 
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\end_layout

\end_inset


\begin_inset ERT
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\backslash
hilite
\end_layout

\end_inset

 Mean Value Theorem
\begin_inset ERT
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}
\end_layout

\end_inset

 (MVT):
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\backslash
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\end_layout

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\end_layout

\begin_deeper
\begin_layout Theorem
Suppose that 
\begin_inset Formula $\ec f$
\end_inset

 is a function that is continuous on 
\begin_inset Formula $\ec[a,b]$
\end_inset

 and differentiable on 
\begin_inset Formula $\ec(a,b)$
\end_inset

.
 There exists 
\begin_inset Formula $\ec c\in(a,b)$
\end_inset

 such that
\begin_inset Formula 
\[
{\ec f'(c)=\frac{f(b)-f(a)}{b-a}}.
\]

\end_inset

 
\end_layout

\end_deeper
\begin_layout Itemize
As a consequence of MVT
\end_layout

\begin_deeper
\begin_layout Theorem
If 
\begin_inset Formula $\ec f'(x)=g'(x)$
\end_inset

 for all 
\begin_inset Formula $\ec x\in(a,b)$
\end_inset

, then 
\begin_inset Formula $\ec f$
\end_inset

 and 
\begin_inset Formula $\ec g$
\end_inset

 differ only by a constant; that is,
\begin_inset Formula 
\[
{\ec f(x)=g(x)+c\qquad\exists c\in\mathbb{R}}.
\]

\end_inset

 
\end_layout

\end_deeper
\begin_layout EndFrame

\end_layout

\end_body
\end_document
