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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\paragraph*{Inverse Functions}

\begin{example}
If \[
f(x)=\frac{2x+1}{x-1},\]
 find where \( f(x) \) has an inverse, and find the formula for the inverse
function \( g=f^{-1} \).
\end{example}

\subsubsection*{Solution}

Start out with the equation \( y=f(x) \), and solve that equation for \( x \).
It will automatically be the formula for the inverse, \( x=f^{-1}(y) \). But,
of course, now you are supposed to think of the inverse as just another function,
so you write it, like all generic functions, as a function of an \( x \), \( y=f^{-1}(x) \),
That last bit is accomplished after you first solve the original equation for
\( x \), \( x=f^{-1}(y) \). Then just switch the roles of \( x \) and \( y \). 

Here's how it works:\begin{eqnarray*}
y & = & f(x)\\
 & = & \frac{2x+1}{x-1}.
\end{eqnarray*}
Then, solve for \( x \):\begin{eqnarray*}
y & = & \frac{2x+1}{x-1}\\
y(x-1) & = & 2x+1\\
yx-y & = & 2x+1.\: \textrm{Now},\, \textrm{put}\, \textrm{all}\, x'\textrm{s}\, \textrm{on}\, \textrm{the}\, \textrm{left}:\\
yx-2x & = & y+1\\
x(y-2) & = & y+1,\: \textrm{or}\\
x & = & \frac{y+1}{y-2}.
\end{eqnarray*}
Then, this says that \[
f^{-1}(y)=\frac{y+1}{y-2}.\]
To take the last step, meaningless though it may be, \[
f^{-1}(x)=\frac{x+1}{x-2}\]
 is the formula for the inverse function.  

You can check to be sure that this is the inverse, since \begin{eqnarray*}
f(f^{-1}(x)) & = & f\left( \frac{x+1}{x-2}\right) \\
 & = & \frac{2\left( \frac{x+1}{x-2}\right) +1}{\left( \frac{x+1}{x-2}\right) -1}\\
 & = & \frac{2(x+1)+1(x-2)}{(x+1)-(x-2)},\: (\textrm{multiplying}\, \textrm{top}\, \textrm{and}\, \textrm{bottom}\, \textrm{by}\, (x-2)),\\
 & = & \frac{2x+2+x-2}{x+1-x+2}\\
 & = & \frac{3x}{3}\\
=x, &  & 
\end{eqnarray*}
which is what we wanted to show. You can also verify that \[
f^{-1}(f(x))=x\]
 as well, but I'll leave that to you.
\end{document}
