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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

\maketitle

\subsection{Corollaries of the Mean Value Theorem}

\begin{thm}
\textbf{\emph{{[}Generalized MVT{]}}}. If \( f(x) \) and \( g(x) \) are continuous
on a closed interval \( [a,b] \), differentiable on the open interval \( (a,b) \),
and if \( g'(x)\neq 0 \) for every \( x\in (a,b) \), then there is some point
\( c\in (a,b) \) so that \[
\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.\]

\end{thm}
\begin{proof}
This result uses a variation of the same trick as for the MVT. Construct from
\( f \) and \( g \) another function, one which satisfies Rolle's theorem,
and see what it tells us. Set \[
h(x):=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a)),\]
which is well-defined (that is, the definition makes sense, in this case meaning
that the denominator is not 0) since \( g(a)\neq g(b) \), because (by Rolle's
theorem) \( g'(x) \) would have to be 0 somewhere in between were \( g(b)=g(a) \). 

Then, because \( f \) and \( g \) are continuous on the closed interval \( [a,b] \),
and differentiable on the open interval \( (a,b) \), so is \( h(x) \). Also,
\( h(a)=h(b)=0 \), so we satisfy Rolle's theorem's hypothesis. That means that
there is some \( c\in (a,b) \) so that \( h'(c)=0 \). But, \[
h'(x)=f'(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(x),\]
 so \[
0=h'(c)=f'(c)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)\Rightarrow \frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)},\]
 where that last division makes sense because \( g'(x) \) is never 0. \end{proof}

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