prove that $x \mapsto \mathrm e^{-x}$ has a unique fixed point on R

Question

Can anybody prove $x \mapsto \mathrm e^{-x}$ has a unique fixed point on R using the fixed point iteration theorem?

Answer 1

The image of the function is $(0,+\infty)$ and in this set $x\mapsto e^{-x}$ is contractive.

Answer 2

Define the function $f : x \to \mathrm e ^{-x} -x$ and remark that its derivative is strictly negative. Then $f$ is strictly decreasing, plus we have $f(0)=1$ and $f(1)=\mathrm e^{-1} -1 <0$.

It is sufficient to conclude by using the intermediate value theorem and the monotonicity that $f$ has a unique zero in $\mathbb R$ (and more precisely in $\left[0,1 \right]$)