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

Question

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.

Answer 1

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.