Suppose $f : [0, 1] \to [0, 1]$ is differentiable such that $|f'(x)|\ne1$ for all $x\in[0, 1]$. Prove $\exists !a,b\in[0,1]:f(a)=a,f(b)=1-b$

Question

Suppose $f : [0, 1] \to [0, 1]$ is differentiable such that $|f'(x)|\ne1$ for all $x\in[0, 1]$.

Prove there exist unique $a,b\in[0,1]$ such that $f(a)=a$ and $f(b)=1-b$.

Not sure how I'm supposed to use the $|f'(x)|\ne1$ condition.

The $a,b$ condition geometrically means that the function intersects $x=y$ and $x=1-y$ lines.
This is trivial if $f$ is a constant, but not sure how to proceed otherwise.

Mean Value Theorem gives that there exists $c$ such that $f'(c)=f(1)-f(0)$. Useful?

Answer 1

Any continuous function from $[0,1]$ into itself has a fixed point. Let $g(x)=f(x)-x$. If $g(a)=g(a')=0$ with $a\neq a'$ then $g'(x)=0$ for some $x$ which contradicts the hypothesis. Hence there is a unique $a$ with $g(a)=0$ or $f(a)=a$. Existence and uniqueness of $b$ follow by considering $1-f$.