How Can You Prove a Symmetric Derivative Property for a Continuous Function?

[johnson12]

Let f be continuous on [0,1] and differentiable on (0,1) such that f(0)=f(1), prove that there exist a 0 < c < 1 such that f$$\acute{}$$(1-c) = -f$$\acute{}$$(c).

thanks for any suggestions.

 

[NoMoreExams]

Just from the givens it made me think of MVT...

 

[johnson12]

That was my first impression but it seems to be a little trickier than that.

 

[Dick]

That was my first impression but it seems to be a little trickier than that.

It's a little more complicated, but not much. Define g(x)=f(x)+f(1-x). g(0)=g(1), right?