TITLE: Mean value theorem understanding
QUESTION [10 upvotes]: So I have this question here which says:
Say that f is differentiable and $f'(x)\neq1$ on $(-\infty,\infty)$. Show that there is at most one real number $a$ such that $f(a)=a$
I'm supposed to use the mean value theorem with the function $g(x)=f(x)-x$ but i'm not really sure how I'm supposed to incorporated the mean value theorem into this...
I tried to substitute for $f(b)$ with $g(b)+b$ and simplify things but i'm not getting anywhere.
Any help on this please?

REPLY [7 votes]: It is not necessary to consider $g(x)=f(x)-x$.
Suppose that there are $a,b \in \mathbb R$ such that $b>a$,  $f(a)=a$ and $f(b)=b$. Then
$\frac{f(b)-f(a)}{b-a}=1 \ne f'(c)$ for all $c \in (a,b)$, a contadiction