# How can I prove/disprove that the function $f(x)$ satifies $f'(c)=1$ for certain conditions

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

$f'(c)=\frac{f(1)-f(0)}{1-0}=\frac{2-1}{1}=1$,

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

• 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. – Paul Z Oct 28 '14 at 20:23
• I think Weierstrass function can be accordingly modified to give a counterexample:en.wikipedia.org/wiki/Weierstrass_function – Timbuc Oct 28 '14 at 20:28
• @Timbuc: That is far more complicated than necessary. – Brian M. Scott Oct 28 '14 at 20:29
• Yes, now I see that, @BrianM.Scott...way more complicated. – Timbuc Oct 28 '14 at 20:46

HINT: If $f$ need not be differentiable, you can make $f$ piecewise linear with two pieces.
• are you hinting at the fact that $\;f\;$ can be constructed? Because I think the OP meant hte function is given . – Timbuc Oct 28 '14 at 20:29
• @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. – Brian M. Scott Oct 28 '14 at 20:31