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\title[Functions, Tables and Graphs]
{Functions, Tables and Graphs}

\institute[College of the Holy Cross]
{
   MATH 133 -- Calculus with Fundamentals, section 1, Prof. Little\\
}

%\date[September 7, 2015]
%{September 7, 2015}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
{\bf Functions -- a conceptual view}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics[clip=true,trim = 0in 0in 0in 0in,width=2.5in]{basicfunc.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item The ``name'' of the function is $f$
\item Inputs $x$ come from some \emph{domain} (for us -- usually a set of real numbers)
\item Outputs in some other specified set (for us -- also usually a set of real numbers)
\item {\bf Key point}:  For each $x$ in the domain, there is \emph{exactly one output},
denoted by $f(x)$.
\end{itemize}
\end{columns}
\end{frame}

\begin{frame}
{\bf Functions can serve as \emph{mathematical models}}

\begin{itemize}
\item<1-> {\bf Example}:  The number $N$ of tickets to a concert that will be sold is a function of the price $P$ 
in dollars of the ticket -- $N(P)$ is the number sold if the price is $P$ dollars.
\item<2-> {\bf Example}:  The length in meters $L$ of a metal rod is a function of the temperature $T$ of the rod (in degrees Celsius) -- $L(T)$ is the length in meters if the temperature is $T$ degrees.
\item<3-> {\bf Example}:  The period $T$ in seconds of a pendulum (with a weight of a fixed mass) is a function of the length $L$ in centimeters of the pedulum -- $L(T)$ is the period in seconds if the length is $L$ centimeters.
\end{itemize}
\end{frame}

\begin{frame}
{\bf Functions described by formulas}

You have no doubt seen
this sort of function many times:  Let $f$ be the function defined by 
\[
f(x) = \frac{x^2 - 1}{x^2 + 3x}.
\]
What this means: We find $f(x)$ by ``plugging a value for $x$ into the formula''
\begin{eqnarray*}
x = -1 \Rightarrow f(-1) &=& \frac{(-1)^2 - 1}{(-1)^2 + 3\cdot (-1)} = \frac{0}{-2} = 0\\
x = 5 \Rightarrow f(5) &=& \frac{(5)^2 - 1}{(5)^2 + 3\cdot 5} = \frac{24}{40} = \frac{3}{5} = 0.6\\
x = a^3 \Rightarrow f(a^3) &=& \frac{(a^3)^2 - 1}{(a^3)^2 + 3\cdot a^3} = \frac{a^6 - 1}{a^6 + 3a^3}.
\end{eqnarray*}
\end{frame}

\begin{frame}
{\bf Example, continued}

Still considering the function 
\[
f(x) = \frac{x^2 - 1}{x^2 + 3x}.
\]

\begin{itemize}
\item<1-> You may have noticed that there are some values where ``plugging in'' leads 
to a problem, since $x^2 + 3x = x(x + 3) = 0$ when $x = 0$ or $x = -3$.
\item<2-> For instance $f(0) = \frac{(0)^2 - 1}{(0)^2 + 3\cdot 0} = \frac{-1}{0}$ is \emph{undefined}(!)
\item<3-> Similarly $f(-3)$ causes a problem:
 $f(-3) = \frac{(-3)^2 - 1}{(-3)^2 + 3\cdot (-3)} = \frac{9 - 1}{9 - 9} = \frac{8}{0}$.
\item<4-> For this reason, we must \emph{leave} $x = 0,-3$ \emph{out of the domain} of this function, 
but any other real $x \ne 0, -3$ is OK.  
\end{itemize}
\end{frame}

\begin{frame}
{\bf ``Rule of thumb'' on domains}

Unless a particular domain is specified for a function defined by a formula, the domain
is the set of \emph{all real $x$ for which the formula makes sense}. 

\begin{itemize}
\item<1->{\bf Example}:  The function defined by the formula
\[
f(x) = \frac{\sqrt{x + 1}}{x - 3}
\]
can be determined like this:
\item<2->  Must have $x + 1 \ge 0$ to take the square root, so $x \ge -1$.
\item<3->  But also, $x \ne 3$ or else the denominator is zero.
\item<4->  All other $x$ are OK, so the domain is the set of 
all $x$ with $-1 < x < 3$, or $x > 3$ (or the union of the two open intervals $(-1,3)$ and $(3,+\infty)$).   
\end{itemize}
\end{frame}

\begin{frame}
{\bf Tables of values}

Especially in experimental sciences, we might need to deal with an (incomplete)
description of a function by a \emph{table of values}.  Here we have the measured period 
$T$ of a pendulum as a function of the length $L$:

\[
\begin{array}{ccccccccc}
      L (cm) &\vert& 20 &\vert& 30 &\vert& 40 &\vert& 50 \\ \hline
     T (sec) &\vert& 0.9 &\vert& 1.1 &\vert& 1.27 &\vert& 1.42 
\end{array}
\]

{\bf Typical questions}: Does $T$ look \emph{linear}, or \emph{approximately linear}
as a function of $L$?  If so, what line \emph{fits the data best}?

\end{frame}

\begin{frame}
{\bf The graph of a function}

\begin{itemize}
\item<1->For a function $f$ with domain in the set of real numbers, where $f(x)$
is a real number for all $x$ in the domain, the \emph{graph} of $f$ is the 
set of points $(x,y)$ in the plane satisfying the equation:
$y = f(x)$.
\item<2->In other words, the points that are on the graph are those of the form $(x,f(x))$ -- 
the $y$-coordinate \emph{equals} the function value at the $x$-coordinate.
\item<3->If $f$ is the function with $f(x) = x^2 + 1$, for instance $(-3,10)$ is on the graph
since $f(-3) = (-3)^2 + 1 = 10$
\item<4-> But $(1,4)$ is \emph{not on the graph} of $f(x) = x^2 + 1$, since
$f(1) = (1)^2 + 1 = 2 \ne 4$.
\end{itemize}
\end{frame}

\begin{frame}
{\bf Graphs}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
%\includegraphics[scale=.2]{FirstFunc.png}
\includegraphics[scale=.3]{FirstFuncVert.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item This is the graph of $f(x) = x^5 - 3 x^3 + 10$, for $x$ with $-3 \le x \le 3$
\item For each $x = a$ in the domain, there is \emph{exactly one output} $f(x)$, so 
\emph{vertical lines $x = a$ meet the graph} just once.
\item Often say -- ``the graph passes the \emph{vertical line test}'' 
\end{itemize}
\end{columns}

\end{frame}

\begin{frame}
{\bf Other graphs}

Other graphs such the \emph{unit circle} with center at $(0,0)$, 
the set of $(x,y)$ with $x^2 + y^2 - 1 = 0$, do not ``pass the vertical
line test'' and are not graphs of functions:

\begin{figure}
\includegraphics[scale=.25]{CircVert.png}
\end{figure}

But the upper and lower semicircles, \emph{separately}, are graphs(!)
\end{frame}

\begin{frame}
{\bf Equations}

Solve for $y$ from the equation $x^2 + y^2 - 1 = 0$:
\begin{itemize}
\item<2-> First, add 1 and subtract $x^2$ from both sides: 
\[
y^2 = 1 - x^2
\]
\item<3-> Now take square roots and recall that both choices of sign give solutions:
\[
y = \pm \sqrt{1 - x^2}
\]
\item<4-> The equation $y = + \sqrt{1 - x^2}$ gives the top half of the circle, and
$y = -\sqrt{1 - x^2}$ gives the bottom half.
\item<5-> The domain of $f_{\pm}(x) = \pm\sqrt{1 - x^2}$ is the set of $x$ with $1 - x^2 \ge 0$, 
so $-1 \le x \le 1$.  
\end{itemize}
\end{frame}

\end{document}

