Yes/No : Is $f$ has fixed point? Given that   $f : \mathbb{R} \rightarrow \mathbb{R}$ is  a differentiable  function.
Is  the following statement  is  necessarily true or not ?
If $f '(x) \le k < 1$ for all $x \in \mathbb{R}$, then $f$ has  unique  fixed point
My attempt : I think this  statement is  false  counter example   $f(x) = x + \frac{1}{(1+e^x)}$
Is its  true ?
 A: Put $g(x) = f(x)-x$. Then $g'(x) = f'(x)-1\le k-1 < 0$. In particular, $g$ strictly decreases and is therefore injective. So, there can only be one fixed point of $f$.
If $g(y)<0$ for all $y\in\mathbb R$, then for all $y\le 0$ we have $g(0) = g(y) - g'(\xi)y < (1-k)y$, a contradiction. Hence, there is $y\in\mathbb R$ such that $g(y)\ge 0$. Set $x := y-\frac{g(y)}{k-1}\ge y$. Then $g(x) = g(y)+g'(\xi)(x-y)\le g(y)+(k-1)(x-y) = 0$. By the intermediate value theorem, there exists $c\in [y,x]$ such that $g(c) = 0$ (and thus $f(c) = c$).
A: If you set $r=\frac{|f(0)|}{1-k}$, then $f$ maps the interval $[-r,r]$ into itself, 
$$
|f(x)|=|f(0)|+|f(x)-f(0)|\le (1-k)r+k|x|\le r.
$$
The usual fixed-point arguments then apply, for instance by applying the intermediate value theorem to $g(x)=f(x)-x$ on this interval. 
Global uniqueness can be proven by contradiction, as in the proof of the Banach fixed-point theorem, as
$$
|x_*-x_{**}|=|f(x_*)-f(x_{**})|\le k|x_*-x_{**}|
$$
is a contradiction. Of course one can also apply Banach directly from the start and for the whole of $\Bbb R$.
