Answer
$f$ has at most one fixed point.
Work Step by Step
Let $g(x) = f(x) - x$
Then $g'(x) = f'(x)-1$
Note that for all $x$,$~~~g'(x) \neq 0~~$ since $~~f'(x) \neq 1$
If $f'(x)$ exists for all real numbers $x$, then $f(x)$ is continuous and differentiable for all $x$. Thus $g(x)$ is continuous and differentiable for all $x$.
Let's assume that $f$ has at least two fixed points $a$ and $b$.
$g(a) = f(a)-a = a-a= 0$
$g(b) = f(b)-b = b-b= 0$
According to Rolle's Theorem, there is a number $c$ in the interval $(a,b)$ such that $g'(c) = 0$.
However, this contradicts the fact that $g'(x) \neq 0$ for all $x$
Therefore, $f$ has at most one fixed point.
