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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}

What do we mean by an inverse function? The inverse of a function \( f \) (\emph{don't}
write as \( y=f(x) \) for now, it adds to the confusion) is another function,
\( g \), so that \( g(f(x))=x \), that is, \( g \) undoes what \( f \) does.
Also, it works the other way for the same functions, \( f(g(x))=x \). We usually
call that \( g \) by \( f^{-1} \), even though that is not the same as \( 1/f(x) \)
as you might think. Another way of writing the relationship between \( f \)
and \( f^{-1} \) is that \[
x=f^{-1}(y)\, \iff \, y=f(x).\]


The square root is the standard example of an inverse function. \[
f^{-1}(x)=\sqrt{x}\]
 certainly undoes what \( f(x)=x^{2} \) does, since \[
\sqrt{x^{2}}=x.\]
 Right? Not exactly. If \( x \) is positive, then yes, it works that way. This
example points out one confusing fact about some inverse functions. Often they
don't really make sense. In order for an inverse function to \( f \) to exist,
not only for each \( x \) can there only be one \( y \) with \( y=f(x) \)
(the vertical line test for it to be a function!), but it also has to work the
other way around, that is, for each \( y \) there can be only one \( x \).
This is the horizontal line test. 

Now, for the square function \( f(x)=x^{2} \), it fails the horizontal line
test. So, no inverse, move on. But, if you restrict the domain of \( f \) to
be only the nonnegative reals, \( x\geq 0 \), then, for each \( y=x^{2} \),
there is only one \( x \), and that \( x \) is the inverse applied to \( y \).
This described the square root function, the inverse of the squaring function
restricted to positive numbers.


\subsection*{\protect\( x\protect \) and \protect\( y\protect \)}

There is a cheap trick the textbooks have at this point. and that is to mess
with what is \( x \) and what is \( y. \) If \( y=f(x) \) is the formula
for \( f \), then how do you write the formula for \( f^{-1} \), the inverse
of \( f \), if it exists? One way is to just treat the inverse \( g \) as
any other function, \[
y=g(x)=f^{-1}(x),\]
so that the \( x \) is the independent variable, \( g(x)=y \). But \( x \)
and \( y \) play different roles here, for the function \( f \), given \( x \)
you describe the \( y \), and it's the same for \( g \). But, \( g \) ought
to be really defined on the range of \( f \) (domain of \( g \) is the range
of \( f \) , and the other way around). So maybe it makes more sense to deal
with \( g \) as \( g(y)=x \). This has the added benefit of making sense when
you solve. 


\subsection*{Finding the inverse}

Given \( y=f(x) \), how do you find the inverse \( g \), if it does exist
(after you straighten out the domain of \( f \) so it works)? Well, the easy
way is to find it as \( x=f^{-1}(y) \). That way, \( x \) plays the same role
throughout. Thinking along those lines, if \( y=f(x) \), then for that \( x \)
and that \( y \), \( f^{-1}(y)=x \), since \( f^{-1} \) undoes what \( f \)
does, so you go forward and back, \( x\rightarrow y,\, y\rightarrow x \). The
method is to simply solve the equation \( y=f(x) \) for \( y \) in terms of
\( x. \) If the solution makes sense, the inverse exists, and there is the
formula.

\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.

\begin{xca}
Find the inverse of \( f(x)=x^{3}-1 \).\[
f^{-1}(x)=\]

\end{xca}

\subsection*{Derivatives of inverse functions; implicit differentiation}

The basic technique to find derivatives of an inverse function \( f^{-1}(x) \)
in terms of the derivative of the original function \( f(x) \) is one case
of a more general trick, called \emph{implicit differentiation}. 

Some functions are not given \textit{explicitly} as a formula, but \textit{implicitly}
in that the function (usually called \( y \)) appears in an equation involving
it and the variable \( x \). The prototype is the equation \[
x^{2}+y^{2}=1.\]


\( y \) is defined by this as the function which takes a given \( x \) to
the number \( y \) that fits with \( x \) in that equation. Of course, in
this case we can solve for \( y \) explicitly, \[
y=\pm \sqrt{1-x^{2}}.\]


\( \pm  \)? That's not a function. But, in a small enough region, (from \( x=-1 \)
to \( x=1 \)), starting at \( (0,1) \), say, \[
y=\sqrt{1-x^{2}}.\]
 On the other hand, in a small enough region near \( (0,-1) \), \[
y=-\sqrt{1-x^{2}}.\]
 This is a standard situation with functions defined implicitly: an equation
in \( x \) and \( y \) will define a function \( y=f(x) \) for \( x \) in
a neighborhood so the curve passes the vertical line test for \( x \)'s in
that neighborhood with some value of \( y \) starting with a given point \( (x_{0},y) \).

It's really quite easy to differentiate these functions; in fact it's often
easier to differentiate them than it is to solve for them explicitly. Just differentiate
both sides of the equation, using the chain rule. \begin{eqnarray*}
x^{2}+y^{2} & = & 1\, \, \mathrm{Think}\, \mathrm{of}\, \mathrm{this}\, \mathrm{as}:\\
x^{2}+(y(x))^{2} & = & 1.\, \, \mathrm{Now},\, \mathrm{differentiate}\, \mathrm{both}\, \mathrm{sides}:\\
2x+2(y(x))\frac{dy}{dx} & = & 0\\
2x+2y\frac{dy}{dx} & = & 0\\
2y\frac{dy}{dx} & = & -2x\\
\frac{dy}{dx} & = & -2x/2y=-x/y.
\end{eqnarray*}


\begin{xca}
Find \( \frac{dy}{dx} \) at \( (2,1) \) using implicit differentiation, if:
\( y^{3}x+yx^{2}+y^{5}=11 \). 
\end{xca}
\[
\left. \frac{dy}{dx}\right| _{(2,1)}=\]


In the particular case of an inverse \[
y=f^{-1}(x)\]
to a function \( f(x) \), which we assume we already know how to differentiate,
we can write the equation \[
y=f^{-1}(x)\]
 as \[
f(y)=x\]
which is an equation defining \( y \) implicitly as a function of \( x \),
and it defined \( y \) as \( f^{-1}(x) \), of course. 

Now differentiate both sides implicitly, \begin{eqnarray*}
f(y) & = & x\\
f'(y)\frac{dy}{dx} & = & 1,
\end{eqnarray*}
and solve for \( dy/dx \)--which is the derivative of the inverse:\[
\left( f^{-1}(x)\right) '=\frac{dy}{dx}=\frac{1}{f'(y)},\]
where that \( y \) in the last denominator can be re-written as \( f^{-1}(x) \),\[
\left( f^{-1}(x)\right) '=\frac{1}{f'\left( f^{-1}(x)\right) }.\]
In the applications of this to particular functions, later in this chapter,
we will often be able to simplify that last expression considerably.

\begin{example}
Use implicit differentiation to find the derivative of \[
\sqrt{x}.\]

\end{example}

\subsubsection*{Solution}

Start off with the {}``frontwards{}'' function, \( f(x)=x^{2} \). Then the
square root is the inverse, \[
f^{-1}(x)=\sqrt{x}.\]
Following the method (not just plugging into the formula) we just went through
in general, \begin{eqnarray*}
f^{-1}(x)=y & = & \sqrt{x}\\
y^{2} & = & x
\end{eqnarray*}
Differentiating both sides,\begin{eqnarray*}
y^{2} & = & x\\
2y\frac{dy}{dx} & = & 1,\: \textrm{or}\\
\frac{dy}{dx} & = & \frac{1}{2y}\\
 & = & \frac{1}{2\sqrt{x}}.
\end{eqnarray*}
Of course, we knew that already, but this is a new way to approach it. 

\begin{xca}
If \( f(x) \) is a function whose derivative satisfies the equation \[
f'(x)=\left( f(x)\right) ^{2},\]
find the derivative of \( f^{-1}(x) \).\[
\left( f^{-1}(x)\right) '=\]
\end{xca}

\end{document}
