Use the Mean Value Theorem to show that if $|f'(x)| ≤ C<1$, then $f$ has at most one fixed point

Use the Mean Value Theorem to show that: if $|f'(x)| ≤ C < 1$ $\forall x$, then $f(x) = x$ has at most one solution.

So using the Mean Value Theorem I know that $$-1<-C\leq \frac{f(b)-f(a)}{b-a}\leq C<1$$

I can do some manipulation with this inequality, but I am confused what "at most one solution" means. Obviously, the derivative of $f(x)=x$ is $1$ which doesn't hold for the inequality presented in the problem?

Any clarification/ hints would be appreciated.

• For some basic information about writing math at this site see e.g. here, here, here and here. – user137731 Dec 13 '14 at 1:13
• Note that "$f(x)=x$" here is not a definition of the function $f$, but merely an equation that may or may not be true for any given value of $x$. – hmakholm left over Monica Dec 13 '14 at 1:24

Say $f(x)=x$ and $f(y)=y$. Then the difference quotient $\dfrac{f(x)-f(y)}{x-y}$ equals $1$, contradicting the inequality from MVT.