Exercise 5.2.11

Question

Assume that $g$ is differentiable on $[a, b]$ and satisfies $g^{\prime}(a)<$ $0<g^{\prime}(b)$.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that there exists a point $x \in(a, b)$ where $g(a)>g(x)$, and a point $y \in(a, b)$ where $g(y)<g(b)$.
  \item Now complete the proof of Darboux's Theorem started earlier.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Since $g'(a) = \lim_{x 	o a} \frac{g(x) - g(a)}{x - a} < 0$ we know
    $$
    g'(a)-\epsilon < \frac{g(x) - g(a)}{x - a} < g'(a)+\epsilon \quad	ext{when}\quad |x-a|<\delta
    $$
    Therefor
    $$
    g(x) < g(a) + (g'(a)+\epsilon)(x-a)
    $$
    Pick $\epsilon$ small enough that $g'(a)+\epsilon < 0$, and since $(x-a) > 0$ we have $g(x)<g(a)$ as desired.
    A similar argument works for $g(y) < g(b)$.
  \item By EVT $g$ obtains a minimum value over $[a,b]$. We just showed $a$ and $b$ are not minima, therefore the minimum point $c$ must be in the interior $(a,b)$ which has $g'(c) = 0$ by Theorem 5.2.6.
   \end{enumerate}

Exercise 5.2.11

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Exercise 5.2.11

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