Let $f$ be continuous on $[0,1]$ and differentiable in $(0,1)$ s.t. $f(0) = f(1)$. Prove that $\exists c \ where \ 0 < c < 1 \ s.t. \ f'(1 - c) = f'(c)$.
Here is my attempt
Since $f(0) = f(1)$, Rolle's Theorem applies.
This means $\exists c \in (0,1) \ s.t. \ f'(c) = 0$
Since $c \in (0,1)$, we can see that $1 - c \in (0,1)$.
To show this is also $0$, examine the original function.
If $0 = 1 - c$, then we see $c = 1$. This will satisfy the inequality giving us $f(0) = f(1)$.
If $1 = 1 - c$, then $c = 0$. This will also satisfy the inequality giving us $f(1) = f(0)$. So whether we use $c$ or $1 - c$, Rolle's Theorem will be satisfied at both points.
I was told this is incorrect, and I can see why. Where I try to show that $1 - c$ works as well, I am only given information about the function (f(0) = f(1)). Not the derivative. So I can see why this doesn't work. I was told as a hint to examine the function $g(x) = f(x) + f(1 - x)$. However, I am not so sure how to apply this to the problem and proceed. Can anyone give me some guidance?
$g(x)=f(1-x)+f(x)$, $g(0)=g(1)$ implies there exists $c$ such that $g'(c)=0$.