This question already has an answer here:

Let $f:[a,b]\to[a,b]$ be a continuous function which is differentiable on $(a,b)$ and $f(a)=a$ and $f(b)=b$. Prove that there exists two distinct points $c$ and $d$ such that $f'(c)f'(d)=1$.

Since the function is continuous on closed interval and differentiable on open interval therefore by mean value theorem there exists $c$ such that $f(b)-f(a)=f'(c)(b-a)$.

From there I got $f'(c)=-1$ but I am unable to find $c$ and $d$ such that $f'(c)=-1$ and $f'(d)=-1$ so that I could say $f'(c)f'(d)=1$.

And by fixed point theorem I got $f'(x)$ is greater than $1$ since $f$ does not have unique fixed point, but how to continue further?


marked as duplicate by Martin R, Winther, Robert Z real-analysis Jan 24 at 12:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.