The function is defined on the interval $[0,1]$ with following conditions:
1) $f(0)=1$,
2) $f(1)=2$,
3) $f(x)$ is continuous on $[0,1]$,
Prove or disprove: There exists some $c$ from $(0,1)$, such that $f'(c)=1$.

My work so far:
If we assume that $f(x)$ is also differentiable on $(a,b)$ than due to Mean Value Theorem we have


therefore everything holds. But if I exclude this assumption I don't know what to do.

  • $\begingroup$ Hint: The fact that you have to assume differentiability in order to use the MVT suggests that a non-differentiable function might supply a counterexample. $\endgroup$ – Paul Z Oct 28 '14 at 20:23
  • $\begingroup$ I think Weierstrass function can be accordingly modified to give a counterexample:en.wikipedia.org/wiki/Weierstrass_function $\endgroup$ – Timbuc Oct 28 '14 at 20:28
  • 1
    $\begingroup$ @Timbuc: That is far more complicated than necessary. $\endgroup$ – Brian M. Scott Oct 28 '14 at 20:29
  • $\begingroup$ Yes, now I see that, @BrianM.Scott...way more complicated. $\endgroup$ – Timbuc Oct 28 '14 at 20:46

HINT: If $f$ need not be differentiable, you can make $f$ piecewise linear with two pieces.

  • $\begingroup$ are you hinting at the fact that $\;f\;$ can be constructed? Because I think the OP meant hte function is given . $\endgroup$ – Timbuc Oct 28 '14 at 20:29
  • $\begingroup$ @Timbuc: I take the question to be asking whether the hypotheses imply that such a $c$ exists. They don’t, since counterexamples exist, and I’m suggesting how one can easily be constructed. If the question is about a specific $f$, obviously we would have to know what that function is. $\endgroup$ – Brian M. Scott Oct 28 '14 at 20:31
  • $\begingroup$ Oh, I see @Brian. Thank you. BTW, excellent way to construct a counter example. +1 $\endgroup$ – Timbuc Oct 28 '14 at 20:46
  • $\begingroup$ @Timbuc: You’re welcome. $\endgroup$ – Brian M. Scott Oct 28 '14 at 20:46
  • $\begingroup$ @BrianM.Scott Thanks, I got it, it is actually trivial. $\endgroup$ – eyedropper Oct 28 '14 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.