# How to prove that there exists two distinct points $c$ and $d$ such that $f'(c)f'(d)=1$ where $f$ is continuous on [a,b] [duplicate]

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?