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Let $f: [a,b]\to R$ be a continuous function differentiable on $(a,b)$.

Define $M:=\frac{f(b)-f(a)}{b-a}$.

Show that for any $ε>0$, there are uncountably many points $x\in[a,b]$ such that $f′(x)\in(M−ε,M+ε)$

I know that by MVT, there exists $x^*\in(a,b)$ such that $f'(x^*)=M$. But I am stuck here. How should I proceed?

PS: this is a homework problem

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  • $\begingroup$ Do you know that derivatives have the IVP? $\endgroup$
    – Sorfosh
    Apr 26 '20 at 20:47
  • $\begingroup$ Yes, but doesn't it require the function to be differentiable on [a,b] instead of (a,b) $\endgroup$
    – user779519
    Apr 26 '20 at 20:49
  • $\begingroup$ @ez709, this is an assignment question. Plz note it in your question otherwise honor code is violated. $\endgroup$
    – Xiaonan
    Apr 26 '20 at 22:18
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If $f$ is linear on $[a,b]$, then $f'(x)=M$ for every point $\in(a,b)$ and we are done. So assume there is $x_0\in(a,b)$ with $\frac{f(x_0)-f(a)}{x_0-a}\ne M$. Then from continuity of $x\mapsto \frac{f(x)-f(a)}{x-a}$ on $[x_0,b]$ and the Intermediate Value Theorme, we conclude that for every $M'$ between $M$ and $\frac{f(x_0)-f(a)}{x_0-a}$ (in particular, either for every $M'\in(M,M+\epsilon)$ or for every $M\in(M-\epsilon,M)$), there exists $x\in (x_0,b)$ with $\frac{f(x)-f(a)}{x-a}=M'$. Then by the Mean Value Theorme, there exists $\xi\in (a,x)$ with $f'(\xi)=M'$.

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  • $\begingroup$ I am not getting the "from countinuity of x" part. Could you explain more? thx $\endgroup$
    – user779519
    Apr 26 '20 at 21:06
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hint

$ f $ is differentiable at $ (a,b)$, thus

the set $E= f'((a,b)) $ is an intervall.

as you said, $ M\in E$ by MVT.

then M is not an isolated point.

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