# Mean value theorem variation proof

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1. Jan 23, 2016

### gruba

1. The problem statement, all variables and given/known data
Let $f$ is differentiable function on $[0,1]$ and $f^{'}(0)=1,f^{'}(1)=0$. Prove that $\exists c\in(0,1) : f^{'}(c)=f(c)$.

2. Relevant equations
-Mean Value Theorem

3. The attempt at a solution

The given statement is not true. Counter-example is $f(x)=\frac{2}{\pi}\sin\frac{\pi}{2}x+10$.
Does this mean that the statement can't be proved?

2. Jan 23, 2016

### PeroK

I suspect you should also have $f(0) = 0$ and $f(1) = 1$.

Otherwise, your counterexample is fine

3. Jan 23, 2016

### HallsofIvy

Staff Emeritus
And, yes, if a statement is not true, it cannot be proved!

4. Jan 24, 2016

### Ray Vickson

There is a difference between "can't be proved" and "is false". If a statement has a counterexample, it is false; that is stronger than the claim that it 'cannot be proved' (but, of course, it cannot be proved as well).