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the mean value theorem which most of us know starts with the conditions that $f$ is continuous on the closed interval $[a,b]$ and differentiable on the opened interval $(a,b)$, then there exists a $c \in (a,b)$, where $\frac{f(b)-f(a)}{b-a} = f'(c)$.

I'm guessing we're then able to define $\forall a', b' \in [a,b]$ where $c \in (a', b')$ and the mean value theorem is correspondingly valid.

However, are we then able to start with $\forall c \in (a,b)$ and then claim that there exist $a',b' \in [a,b]$, where the mean value theorem is still valid?

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3 Answers 3

up vote 1 down vote accepted

Let $[a,b]$ be any closed interval and $f$ satisfying the mean value property for every open interval $(a',b')$ contained within $[a,b]$. Let $a_n = a - \frac{1}{n}$ and $b_n = b - \frac{1}{n}$ for large values of $n$. Then we have that for each $n$, there's a point $c_n$ in $(a_n, b_n)$ so that $f'(c_n) = \frac{f(b_n) - f(a_n)}{b_n - a_n}$. It's easy to see by continuity of $f$ on $[a,b]$ that the right hand side converges to $\frac{f(b) - f(a)}{b-a}$. I guess the question then becomes:

1) Is the sequence $\{c_n\}$ convergent and,

2) Even if it is, do we have $\lim_{n\to\infty} f'(c_n) = f'(c)$.

An imposition of continuity on $f'$ would be enough for 2). Now for $1$ we argue as follows: $\{c_n\}_n$ is a sequence in $[a,b]$ and so it has a convergent subsequence by compactness. Then this new sequence, say $\{c_{n_k}\}$ satisfies

$\lim_{n_k \to \infty} f'(c_{n_k}) = \frac{f(b) - f(a)}{b-a}$, and $\lim_{n_k} c_{n_k} = c$, for some point $c \in [a,b]$. By continuity, we have that there's a point $c \in [a,b]$ so that $f'(c) = \frac{f(b) - f(a)}{b-a}$, which isn't quite the mean value theorem.

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The answer is No. Consider $y=f(x)=x^3$ and $c=0$. $f'(c)=0$ but no secant line has a zero slope as ${{f(r)-f(s)}\over{r-s}}=r^2+s^2+rs>0$.

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I haven't have your answer but that example is wrong. Because $f^{'}(x)=3x^2$. Put $x_1=a>0,x_2=-a<0$ then $f'(x_1)-f'(x_2)=0$

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