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I had this problem on an assignment a while ago, but I don't quite understand the formulation of the problem nor the purpose:

Let $f : [a,b]→\mathbb{R}$ be a function that admits a derivative (not necessarily finite!) at any point of $[a,b]$. Prove that there exists $x_0 \in [a,b]$ such that: $|\frac{f(b)-f(a)}{b-a}|\leq |f'(x_0)|$

The original solution was a bit convoluted, and was by constructing some nested intervals, but can't we just say that if the derivative is infinite at some point then we're done, otherwise $f$ is differentiable and we're done by Lagrange's mean theorem? What's wrong with this?

N. F. Taussig
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ZNatox
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  • first state the mean value theorem – Andrew Jan 03 '24 at 22:09
  • @Andrew For a function continuous in $[a,b]$ differentiabe in $(a,b)$, there exists $x \in (a,b)$ such that: $\frac{f(b)-f(a)}{b-a}=f'(x)$ – ZNatox Jan 03 '24 at 22:12
  • those hypotheses are not satisfied here – Andrew Jan 03 '24 at 22:19
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    Well if the derivative is finite for each $x\in [a,b]$, they're satisfied no? – ZNatox Jan 03 '24 at 22:20
  • hmm, yes i think you are right then, in that case – Andrew Jan 03 '24 at 22:25
  • @ZNatox, moderators have deleted my answer because in their opinion I have not answered to your question. If you are interested in a proof that there exist $x_1,x_2\in[a,b]$ such that $f’(x_1)\leqslant\dfrac{f(b)-f(a)}{b-a}\leqslant f’(x_2)$, you should write it in your question and so perhaps moderators will undelete my answer and you will be able to see my proof. – Angelo Jan 08 '24 at 12:53

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