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%MODULE 1: Applications of IVT
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\begin_layout Title
Applications of the Intermediate Value Theorem
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The 
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{
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\backslash
hilite
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 Intermediate Value Theorem
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}
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 allows us to conclude that a continuous function takes certain values on
 an interval.
 This opens the door to a number of applications.
 We will explore some of these applications in this module.
 
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{
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Sample problems
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LatexCommand label
name "page:modf-problem"

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\backslash
begin{question}
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\end_inset

 
\end_layout

\begin_layout Itemize
Approximate a solution (if any) to 
\begin_inset Formula $\ec\cos(x)=x$
\end_inset

 to the thousandths place.
 
\end_layout

\begin_layout Itemize
Approximate the value of 
\begin_inset Formula $\ec\sqrt[5]{2}$
\end_inset

 to the ten-thousandths place.
 
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\backslash
end{question}
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%1
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{
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SAGE worksheets
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\begin_layout Standard
For this module you will need only a blank SAGE worksheet.
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\begin_layout BeginFrame
The Intermediate Value Theorem
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\begin_layout Standard
First let's recall the 
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{
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\backslash
hilite
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 Intermediate Value Theorem
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}
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 (IVT).
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\begin_layout Theorem
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[
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(Intermediate Value Theorem)
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]
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If
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\begin_deeper
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the function 
\begin_inset Formula $\ec f$
\end_inset

 is continuous on the interval 
\begin_inset Formula $\ec[a,b]$
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, and 
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\begin_layout Itemize
\begin_inset Formula $\ec f(a)\neq f(b)$
\end_inset

, 
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then 
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for any 
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 value 
\begin_inset Formula $\ec C$
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 between 
\begin_inset Formula $\ec f(a)$
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 and 
\begin_inset Formula $\ec f(b)$
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, 
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\begin_layout Itemize
there exists a value 
\begin_inset Formula $\ec c$
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 such that 
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\ec a<c<b$
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, and 
\end_layout

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\begin_inset Formula $\ec f(c)=C$
\end_inset

.
 
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\end_deeper
\end_deeper
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{
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First example
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}
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How can we apply 
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\backslash
hilite
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\end_inset

 IVT
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}
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\end_inset

 to the first sample problem? 
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\backslash
begin{question}
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\end_inset

 Approximate a solution (if any) to 
\begin_inset Formula $\ec\cos(x)=x$
\end_inset

 to the thousandths place.
 
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\backslash
end{question}
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\noindent
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%2
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\noindent
The 
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{
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\backslash
hilite
\end_layout

\end_inset

 IVT
\begin_inset ERT
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}
\end_layout

\end_inset

 helps us determine that some 
\begin_inset Formula $x$
\end_inset

 value has a 
\begin_inset Formula $y$
\end_inset

 value.
 This point of view is slightly different from the one posed by the problem.
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\begin_layout Pause

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\begin_layout Standard
Thus we have to rephrase the problem to fit into the context of the 
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{
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\backslash
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\end_inset

 IVT
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}
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.
 
\emph on
How?
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\begin_layout Standard
Try this: 
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\backslash
begin{question}
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\end_inset

Approximate a solution to 
\begin_inset Formula $\ec\cos(x)-x=0$
\end_inset

, if any such solution exists.
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\backslash
end{question}
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\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
Now the question is asking us not to find where two functions are equal,
 but to find a value of a function 
\begin_inset Formula $\ec f(x)=\cos(x)-x$
\end_inset

.
 This fits more easily into the 
\begin_inset ERT
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{
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\backslash
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\end_inset

 IVT
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}
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 point of view.
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First example
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To apply 
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 IVT
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}
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, we have to make sure that all the hypotheses are met.
 The first hypothesis is that the function 
\begin_inset Formula $\ec f(x)=\cos(x)-x$
\end_inset

 is continuous.
 You have learned in class that 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec y=\cos(x)$
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 is a continuous function everywhere; 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec y=x$
\end_inset

 is a continuous function everywhere; and 
\end_layout

\begin_layout Itemize
sums and differences of continuous functions are continuous on the intervals
 where both functions are continuous.
 
\end_layout

\begin_layout Standard
\noindent
Thus 
\begin_inset Formula $\ec\cos(x)-x$
\end_inset

 is continuous everywhere.
 
\end_layout

\begin_layout Standard
\align center

\emph on
You must check for continuity anytime you wish use 
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{
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 IVT
\begin_inset ERT
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}
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.
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{
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First example
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\begin_layout Standard
Now that we've verified the continuity of the function, we can try to see
 if there is any value of 
\begin_inset Formula $\ec x$
\end_inset

 such that 
\begin_inset Formula $\ec\cos(x)-x=0$
\end_inset

.
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\noindent
How does 
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 IVT
\begin_inset ERT
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}
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\end_inset

 help? If we can find 
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\begin_layout Itemize
one value of 
\begin_inset Formula $\ec x$
\end_inset

 such that 
\begin_inset Formula $\ec\cos(x)-x$
\end_inset

 is negative, and 
\end_layout

\begin_layout Itemize
one value of 
\begin_inset Formula $\ec x$
\end_inset

 such that 
\begin_inset Formula $\ec\cos(x)-x$
\end_inset

 is positive, then 
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\begin_layout Standard
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{
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\end_inset

 IVT
\begin_inset ERT
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}
\end_layout

\end_inset

 assures us that some 
\begin_inset Formula $x$
\end_inset

 value 
\emph on
between
\emph default
 those two values has a 
\begin_inset Formula $y$
\end_inset

 value of zero.
 How?
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First example
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\begin_layout Standard
An easy way to do this is to use a graph.
 Using SAGE, plot a graph of 
\begin_inset Formula $\ec f(x)=\cos(x)-x$
\end_inset

.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
We obtained the following graph by plotting on the interval 
\begin_inset Formula $[-\pi,\pi]$
\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/cos_x_minus_x_on_pis.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
This reveals the following: 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec f(0)$
\end_inset

 is positive; 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec f(1)$
\end_inset

 is negative.
 
\end_layout

\begin_layout Standard
(You can approximate their values as 
\begin_inset Formula $1$
\end_inset

 and 
\begin_inset Formula $-0.4597$
\end_inset

.)
\end_layout

\begin_layout Pause

\end_layout

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By the 
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 IVT
\begin_inset ERT
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}
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\end_inset

, we know that for 
\emph on
some
\emph default
 value 
\begin_inset Formula $\ec c$
\end_inset

 between 
\begin_inset Formula $\ec a=0$
\end_inset

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

, 
\begin_inset Formula $\ec f(c)=0$
\end_inset

.
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\begin_layout BeginFrame
First example
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\begin_layout Standard
\noindent
Why did we need 
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\end_inset

 IVT
\begin_inset ERT
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}
\end_layout

\end_inset

 to draw this conclusion? 
\emph on
Continuity.

\emph default
 If we did not know that the function is continuous, we would have to worry
 about holes.
 Graphing calculators and computers rarely, if ever, display holes in the
 graphs of their functions.
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First example
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\end_layout

\begin_layout Standard
Now we know that there is 
\emph on
some
\emph default
 value 
\begin_inset Formula $\ec c$
\end_inset

 for which 
\begin_inset Formula $\ec\cos c-c=0$
\end_inset

, and thus 
\begin_inset Formula $\ec\cos c=c$
\end_inset

.
 However, the problem asked for something more: a value of 
\begin_inset Formula $\ec c$
\end_inset

, 
\emph on
accurate to the thousandths place.

\emph default
 How can we use 
\begin_inset ERT
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{
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 IVT
\begin_inset ERT
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}
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\end_inset

 to get this kind of accuracy? Redraw the graph with smaller values of 
\begin_inset Formula $\ec a$
\end_inset

 and 
\begin_inset Formula $\ec b$
\end_inset

!
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\noindent
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First example
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\end_layout

\begin_layout OverlayArea
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{
\backslash
textwidth}{6in}
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\end_inset


\end_layout

\begin_deeper
\begin_layout Only
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\begin_layout Plain Layout

<1>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
Look at the previous graph again.
 
\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/cos_x_minus_x_on_pis.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
It looks as if the root lies between 
\begin_inset Formula $0.6$
\end_inset

 and 
\begin_inset Formula $0.8$
\end_inset

.
 Let's plot 
\begin_inset Formula $\ec f$
\end_inset

 on that smaller interval and see if we get anything different.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
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\begin_layout Plain Layout

<2>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset

It looks as if the root lies between 
\begin_inset Formula $0.6$
\end_inset

 and 
\begin_inset Formula $0.8$
\end_inset

.
 Let's plot 
\begin_inset Formula $f$
\end_inset

 on that smaller interval and see if we get anything different.
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset

 We have 
\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/cos_x_minus_x_on_6_8.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
We're getting closer! Now try it on the interval 
\begin_inset Formula $\ec[0.71,0.75]$
\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

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\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset

Now try it on the interval 
\begin_inset Formula $[0.71,0.75]$
\end_inset

.
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset

 We have 
\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/cos_x_minus_x_on_71_75.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
We can see now that the curve intersects the axis between 
\begin_inset Formula $0.735$
\end_inset

 and 
\begin_inset Formula $0.74$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout BeginFrame
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset

First example
\begin_inset ERT
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}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Repeat this process of narrowing the graph until you have an estimate that
 is accurate to the thousandths place.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
How will you know when you're done? When the curve intersects the graph
 between two values of 
\begin_inset Formula $x$
\end_inset

 that have the same digits up to the thousandths place.
 The author, for example, obtained this graph:
\end_layout

\begin_layout Standard
\begin_inset VSpace -2ex
\end_inset


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

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/cos_x_minus_x_on_739_740.png
	lyxscale 20
	scale 30

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset VSpace -4ex
\end_inset

\SpecialChar \ldots{}
using the commands
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{
\end_layout

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\backslash
ttcolor
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\end_inset


\family typewriter
f = cos(x)-x 
\end_layout

\begin_layout Standard

\family typewriter
fplot = plot(f,0.739,0.740)
\end_layout

\begin_layout Standard

\family typewriter
fplot.show()
\family default

\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
and concluded that the root is approximately 
\begin_inset Formula $0.739$
\end_inset

.
\end_layout

\begin_layout BeginFrame
\noindent
\begin_inset ERT
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{
\end_layout

\end_inset

The method of bisection
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}
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\end_inset


\end_layout

\begin_layout Standard
\noindent
We can formalize the method shown in the previous example as follows.
 
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\backslash
begin{method}
\end_layout

\end_inset

 
\end_layout

\begin_layout Itemize
Let 
\begin_inset Formula $\ec(a,b)$
\end_inset

 be an interval and 
\begin_inset Formula $\ec f$
\end_inset

 a function such that 
\begin_inset Formula $\ec f(a)$
\end_inset

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

 have opposite signs (one is positive, the other negative).
 
\end_layout

\begin_layout Itemize
If rounding 
\begin_inset Formula $\ec a$
\end_inset

 and 
\begin_inset Formula $\ec b$
\end_inset

 to the desired precision gives the same number, you are done: the answer
 is 
\begin_inset Formula $\ec a$
\end_inset

.
 
\end_layout

\begin_layout Itemize
Let 
\begin_inset Formula $\ec c=\frac{a+b}{2}$
\end_inset

 be the midpoint of 
\begin_inset Formula $\ec[a,b]$
\end_inset

.
 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec f(c)$
\end_inset

 has the same sign as 
\begin_inset Formula $\ec f(b)$
\end_inset

, repeat these steps with the interval 
\begin_inset Formula $\ec(a,c)$
\end_inset

.
 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec f(c)$
\end_inset

 has the same sign as 
\begin_inset Formula $\ec f(a)$
\end_inset

, repeat these steps with the interval 
\begin_inset Formula $\ec(c,b)$
\end_inset

.
 
\end_layout

\begin_layout Standard
\noindent
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{method}
\end_layout

\end_inset

 This method will find the approximate value of any root, although it can
 be very slow.
\end_layout

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\begin_layout Plain Layout

{
\end_layout

\end_inset

Second example
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}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
A similar procedure will help us approximate the fifth root of two.
 This problem is a little different, inasmuch as we don't have a function
 yet! To find the function, let's reason from what we want.
\end_layout

\begin_layout BeginFrame
\noindent
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset

Second example
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}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
\begin_inset ERT
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\begin_layout Plain Layout


\backslash
begin{question}
\end_layout

\end_inset

 Approximate the value of 
\begin_inset Formula $\ec\sqrt[5]{2}$
\end_inset

 to the ten-thousandths place.
 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
Another way of saying this is that we want a value of 
\begin_inset Formula $\ec x$
\end_inset

 such that 
\begin_inset Formula $\ec x^{5}=2$
\end_inset

.
 (Raise both sides to the fifth power.)
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
In the same manner, we want a value of 
\begin_inset Formula $x$
\end_inset

 such that 
\begin_inset Formula $\ec x^{5}-2=0$
\end_inset

.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
This suggests the function 
\begin_inset Formula $\ec f(x)=x^{5}-2$
\end_inset

.
 Again, we have to check that this function is continuous, and it is.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
Use the method of bisection to approximate the solution to the equation
 
\begin_inset Formula $\ec x^{5}-2=0$
\end_inset

, accurate to the ten-thousandths digit.
 You know that 
\begin_inset Formula $\ec\sqrt[5]{2}$
\end_inset

 is larger than 1 and smaller than 2, so start with 
\begin_inset Formula $\ec a=1$
\end_inset

 and 
\begin_inset Formula $\ec b=2$
\end_inset

.
\end_layout

\begin_layout BeginFrame
Second example
\end_layout

\begin_layout Standard
\noindent
After fourteen steps, the author estimated 
\begin_inset Formula $\ec\sqrt[5]{2}\approx1.1487$
\end_inset

, with the following graph.
 
\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/fifth_root_of_2.png
	lyxscale 20
	scale 20

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

The importance of continuity
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
We cannot overemphasize the importance of continuity to the 
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
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\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 IVT
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset

.
\end_layout

\begin_layout Standard
\noindent
Suppose, for example, that you wanted to find a root of the function
\begin_inset Formula 
\[
{\ec f(x)=\frac{1}{x+1}-\frac{\lfloor{x}\rfloor}{100}-1}.
\]

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
If you merely looked at the graph, you might see this:
\end_layout

\begin_layout Standard
\align center
\begin_inset Float figure
placement h!
wide false
sideways false
status open

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

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{question}
\end_layout

\end_inset

Is there a root at the origin?
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent

\emph on
No, there is not.

\emph default
 It may look as if a root lies there, but the function is discontinuous
 at 
\begin_inset Formula $\ec x=0$
\end_inset

 (because 
\begin_inset Formula $\ec\lfloor x\rfloor$
\end_inset

 is discontinuous at 
\begin_inset Formula $\ec x=0$
\end_inset

).
\end_layout

\begin_layout Standard
You might notice this if you zoomed in a little more.
\end_layout

\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Conclusion
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset

 
\end_layout

\begin_layout Itemize
We can use the 
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 Intermediate Value Theorem
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset

 to approximate solutions to equations and irrational numbers.
 
\end_layout

\begin_layout Itemize
We must remember to check whether a function is continuous before trying
 to apply the theorem.
 
\end_layout

\begin_layout Itemize
The 
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 IVT
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 is not always the best method to solve equations or approximate irrational
 numbers.
 Later, you will learn 
\emph on
Newton's Method
\emph default
, which is usually a much more efficient method.
 
\end_layout

\begin_layout EndFrame

\end_layout

\end_body
\end_document
