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I am wondering if the following converse (or modification) of the mean value theorem holds. Suppose $f(\cdot)$ is continuously differentiable on $[a,b]$. Then for all $c \in (a,b)$ there exists $x$ and $y$ such that $$ f'(c)=\frac{f(y)-f(x)}{y-x} $$

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2 Answers 2

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Assume that $f''(c)\ne0$. Then one can find $x_1<c<x_2$ such that $$f'(c)={f(x_2)-f(x_1)\over x_2-x_1}\ .$$ Proof. Assume $f''(c)>0$ and consider the auxiliary function $$g(t):=f(c+t)-f(c)-t f'(c)\ ,\tag{1}$$ which is defined in a full neighborhood of $t=0$. By Taylor's theorem one has $$g(t)=g(0)+t g'(0)+{t^2\over2}g''(0)+o(t^2)={t^2\over2}\bigl(f''(c)+o(1)\bigr)\qquad(t\to0)\ .$$ It follows that there is an $h>0$ with $g(t)>0$ for $0<|t|\leq h$. Assuming $g(h)\geq g(-h)$ put $t_1:=-h$, and choose $t_2\in\ ]0,h]$ such that $g(t_2)=g(t_1)$, which is possible by the intermediate value theorem.

Finally put $x_i:=c+t_i$ $(i=1,\>2)$. Then it follows from $(1)$ that $$f(x_2)-f(x_1)=g(t_2)-g(t_1)+(t_2-t_1)f'(c)=(x_2-x_1) f'(c)\ ,$$ which is equivalent to the claim.

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  • $\begingroup$ Do we need the restriction that $g(-h) \neq g(h)$? $\endgroup$
    – user103828
    May 1, 2014 at 10:08
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    $\begingroup$ @user103828: The function $g$ is not symmetric with respect to $t=0$; therefore we cannot expect $g(-h)=g(h)$. This is not a restriction but a fact we have to reckon with somehow. See my edit. $\endgroup$ May 1, 2014 at 10:50
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It is not true. Let $f(t)=t^3$. Then $f'(0)=0$, but $\frac{f(y)-f(x)}{y-x}$ is never $0$.

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  • $\begingroup$ Thanks. It seems also from your example that there is no ``easy'' way to modify the statement to make it true. $\endgroup$
    – user103828
    May 1, 2014 at 6:35
  • $\begingroup$ You are welcome. There is always a way, but offhand I cannot think of one. $\endgroup$ May 1, 2014 at 6:37
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    $\begingroup$ What if it is modified such that the derivative is nowhere vanishing? $\endgroup$ Jan 27, 2016 at 22:54
  • $\begingroup$ I am answering quickly, busy with other stuff, so may be wrong. But it looks to me as if $t^3+17$ will also give a counterexample. $\endgroup$ Jan 27, 2016 at 23:01
  • $\begingroup$ $t^3 + t$ is a counterexample. @MathematicsStudent1122 $\endgroup$
    – JOF14
    Feb 8, 2018 at 23:39

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