Suppose that $f$ is continuous and differentiable on $[0, 1]$, that $0\leq f (x) \leq 1$ for each $x\in [0, 1]$, and that $f '(x) \ne 1$ for all $x \in [0, 1]$. Show that there exists exactly one number $x > \in [0, 1]$ such that $f (x) = x$.
So far I have done proving there is at least one $x \in [0,1]$ by IVT.
If $f(0)=0$ and $f(1)=1$ then we are done else, if $f(0)>0$ and $f(1)<1$. Let $g(x)=x-f(x)$ then $g(1)= 1 -f(1)>0$ and $g(0)= - f(0)<0$
Then I am trying to prove there is exactly one $x \in [0,1]$ s.t $f(x)=x$
So am I suppose to prove this by contradiction with mean value theorem or Rolle's? If so, how should I set this up?
You have shown that $g(0)g(1)<0$ so there must be a $a\in[0,1]$ such that $g(a)=0$ , by IVT, thus $f(a)=a$.
If there is another $b\neq a$ (WLOG $b>a$) such that $f(b)=b$, then applying MVT on $[a,b]$ givves $f'(c)[b-a]=f(b)-f(a)\Rightarrow f'(c)=1$, for some $c\in(0,1)$, a contradiction.