The contraction mapping theorem Let $f : \mathbb{R} \to \mathbb{R} : f(x) = 1 + x + \mathbb{e}^{-x}$ and let $ U [1,\infty)$. Firstly i need to show that $f$ maps $U$ into itself. But im can only see how $f$ maps $U$ to $[2 +1/\mathbb{e},\infty) $ as if we set $x=1$ then $f(1) = 2 + \mathbb{e}^{-1}$.
As $x \to \infty$ we get that $f(x) \to \infty$ but i can't justify how it maps $U$ to itself. 
From there i have to show that its contractive, i.e. $|f(x) - f(y)| < |x - y| $. Is it true that this would require the mean value theorem?

When someone says that $f$ maps $U$ into $U$, they don't mean that every point in $U$ must be $f(x)$ for some $x \in U$.  They merely mean that $f(x) \in U$ for every $x \in U$.  And since $[2+1/e,\infty) \subset U$, you have solved that part of the problem.
And yes, the mean value theorem would be an excellent way to do the second part.
I think you know how to do this problem - it is just that you don't trust yourself to believe in your answer.
