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$f$ is differentiable in $[a,b]$, and satisfies $f(a)<f(b)$, $f'(a)=f'(b)=0$. Show that there exists $c$ in $[a,b]$

\begin{equation} \frac{f(c)-f(a)}{c-a}=f'(c) \end{equation}

How is the mean value theorem \begin{equation} \frac{f(b)-f(a)}{b-a}=f'(c) \end{equation} applicable here?

stph
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2 Answers2

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Consider the auxiliary function $$g(x):={f(x)-f(a)\over x-a} \quad(a<x\leq b), \quad g(a):=0\ .$$ Then $g$ is continuous on $[a,b]$ and differentiable in $(a,b)$. One computes $$g'(x)={(x-a) f'(x)-\bigl(f(x)-f(a)\bigr)\over (x-a)^2}\qquad(0<x\leq b)\ .$$ If $g(b)=0$, then this is a consequence of Rolle's Theorem.

If $g(b)>0$, then $g'(b)<0$; therefore $g(x)>g(b)>0$ immediately to the left of $b$. This allows us to conclude that $g$ assumes its maximum on $[0,b]$ in an interior point $c$. From $g'(c)=0$ it then follows that $${f(c)-f(a)\over c-a}=f'(c)\ .$$

The case for $g(b)<0$ is similar.

CarlSagan
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As mentioned in the comments, this is Flett's version of the MVT. A proof of this can be found at this link, also from the comments.

apnorton
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  • In regards to the link, what is Flett reverencing when he says "then there exists an $x_1$ in $a<x_1<b$ such that $\phi(x_1)>\phi(b)$"? – CarlSagan Nov 27 '19 at 04:34