Exercise 5.3.5

Question

\begin{enumerate}[label=(\alph*)] 
  \item Supply the details for the proof of Cauchy's Generalized Mean Value Theorem (Theorem 5.3.5).
  \item Give a graphical interpretation of the Generalized Mean Value Theorem analogous to the one given for the Mean Value Theorem at the beginning of Section 5.3. (Consider $f$ and $g$ as parametric equations for a curve.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Let $h(x) = [f(b)-f(a)]g(x) - [g(b)-g(a)]f(x)$ and apply the MVT to $h$ to get $c \in (a,b)$ with
    $$
    h'(c) = \frac{h(b)-h(a)}{b-a} = \frac{[f(b)-f(a)][g(b)-g(a)] - [g(b)-g(a)][f(b)-f(a)]}{b-a} = 0
    $$
    Thus we have
    $$
    h'(c) = [f(b)-f(a)]g'(c) - [g(b)-g(a)]f'(c) = 0
    $$
    Completing the proof.
  \item Rename $x = f$, $g = y$, $t=a$, then the theorem states
    $$
    \frac{x'(t)}{y'(t)} = \frac{dx}{dy} = \frac{x(b)-x(a)}{y(b)-y(a)}
    $$
    In other words, it's the mean value theorem for parametric curves.
   \end{enumerate}

Exercise 5.3.5

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Exercise 5.3.5

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