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Prove existence of open interval with non-zero derivative

  1. Aug 22, 2015 #1
    I'm struggling to get started with the proof that an open interval D containing x0 exists such that f'(x) ≠ 0 for all x∈D, given f'(x0)≠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?
  2. jcsd
  3. Aug 23, 2015 #2


    Staff: Mentor

    I can't believe that this is the correct statement of the problem. Consider the open interval (0, 2), with x0 = 1, and f(x) = x. For all x in this interval, f'(x) ≠ 0, and f(x0) ≠ 0. Are you sure you wrote the problem correctly?

    BTW, if this is a homework problem it should have been posted in the Homework & Coursework section, not here in the technical math sections.
  4. Aug 23, 2015 #3


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    Science Advisor

    While the derivative of an arbitrary differentiable function is not necessarily continuous, it can be proved that it has the "intermediate value property". That is, if f(a)= c and f(b)= d, then f takes on every value between c and d at some point between a and b.

    Mark44, did you misread the problem? What you give is an example for which the "theorem" works, not a counter-example.
  5. Aug 23, 2015 #4
    Assume the opposite. Would f be differentiable at x0?
  6. Aug 23, 2015 #5


    Staff: Mentor

    It wasn't clear to me what the OP was trying to prove (quoted below), and was trying to get some clarification.
  7. Aug 23, 2015 #6


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    As I understand it, we are given an arbitrary function f. If f'(x0) is non-zero we are asked to prove that there is some open interval containing x0 throughout which f'(x) is non-zero. One might assume that f is everywhere differentiable.

    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?
  8. Aug 24, 2015 #7
    This statement is false. Take [itex]x_0=0[/itex] Consider [itex] f(x)=x^2\sin(1/x)+x[/itex] for nonzero [itex]x[/itex] with [itex] f(0)=0[/itex]. Computation gives [itex] f'(0)=1[/itex] and [itex] f'(x)=2x\sin(1/x)-\cos(1/x)+1[/itex] for nonzero [itex]x[/itex]. Observing that [itex] f'(\frac{1}{2n\pi})=0[/itex] for integer [itex]n[/itex] shows that this [itex]f[/itex] is a counterexample since any open subset of [itex]\mathbb{R}[/itex] containing [itex]0[/itex] contains elements of the form [itex]\frac{1}{2n\pi}[/itex].
  9. Aug 24, 2015 #8
    Can you show that from the limit definition of the derivative?
  10. Aug 24, 2015 #9
    As you wish:

    [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.
  11. Aug 24, 2015 #10
    Yes indeed, so the original statement is in general false.

    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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