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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Discussion Day on Inverse Functions}
\centerline{September 14, 2017}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
If a function from a domain $D$ to a range $R$ has the \emph{one-to-one} property
(that is, each $y$ in the range $R$ is $y = f(x)$ for \emph{exactly one} $x$ in the
domain $D$), then we can define a second function $f^{-1}$ with domain $R$ and range
$D$ such that 
$$x = f^{-1}(y)\ {\rm exactly\ when}\ y = f(x).$$
This new function is called the \emph{inverse function} of $f$.  As we said in 
the video for today, this $f^{-1}$ is \emph{NOT THE SAME} as $\frac{1}{f}$.  We want
to work with a few examples of this idea today.
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\noindent
{\it Questions}
\begin{enumerate} 
\item[1)]  Look at Figure 19 on page 39 of our text.
\begin{enumerate}
\item[(a)]  Which of the graphs there define functions with the one-to-one property?
Why?
\item[(b)]  For that ones that do not, could you restrict the domain to get a function
satisfying the one-to-one property?  (NOTE:  This is actually always possible!  How
would you do it here?)
\item[(c)]  Recall from Video 1.5 that there is a geometric way to get the graph of the inverse function from the graph of a one-to-one function.  Draw the graphs of the inverse functions
for each of the functions from part (a), and then do the same for each restricted domain
function from part (b).
\end{enumerate}
\item[(2)] For this problem, take it as known that the given functions have the one-to-one
property.  Find the formula for the inverse function and state what the domain of the inverse is.
\begin{enumerate}
\item[(a)]  $f(x) = -3x + 5$
\item[(b)]  $f(x) = \displaystyle{\frac{1}{x+1}}$.
\end{enumerate} 
\end{enumerate}
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