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_{0}exists such that f'(x) ≠ 0 for all x∈D, given f'(x

_{0})≠0. It seems like it should be easy but I've been scratching around for an hour now and have gotten nowhere, could anyone give me some advice to help me along?

- Thread starter meriadoc
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Mark44

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I can't believe that this is the correct statement of the problem. Consider the open interval (0, 2), with x_{0}exists such that f'(x) ≠ 0 for all x∈D, given f'(x_{0})≠0. It seems like it should be easy but I've been scratching around for an hour now and have gotten nowhere, could anyone give me some advice to help me along?

BTW, if this is a homework problem it should have been posted in the Homework & Coursework section, not here in the technical math sections.

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HallsofIvy

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Mark44, did you misread the problem? What you give is an example for which the "theorem" works, not a counter-example.

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pbuk

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Assume the opposite. Would f be differentiable at x_{0}?

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Mark44

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It wasn't clear to me what the OP was trying to prove (quoted below), and was trying to get some clarification.

Mark44, did you misread the problem? What you give is an example for which the "theorem" works, not a counter-example.

_{0}exists such that f'(x) ≠ 0 for all x∈D, given f'(x_{0})≠0. It seems like it should be easy but I've been scratching around for an hour now and have gotten nowhere, could anyone give me some advice to help me along?

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jbriggs444

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A function f such that f(0) = 0, f'(0) = 1 and f'(x) = 0 for an infinite number of x's arbitrarily near zero would serve as a counter-example. Can you find a function like that?

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Infrared

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pbuk

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Can you show that from the limit definition of the derivative?Computation gives [itex] f'(0)=1[/itex]

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Infrared

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As you wish:Can you show that from the limit definition of the derivative?

[tex]f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{h^2\sin(1/h)+h}{h}=\lim_{h\to 0} h\sin(1/h)+1=1[/tex]

since [itex]\lim_{h\to 0} h\sin(1/h)=0[/itex] follows from the squeeze theorem.

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pbuk

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Yes indeed, so the original statement is in general false.... since [itex]\lim_{h\to 0} h\sin(1/h)=0[/itex] follows from the squeeze theorem.

I notice the OP hasn't returned; @meridoc if you are still interested does the counterexample above suggest a restriction on functions for which the statement can be proved true?

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