Question: I need to prove (using the mean value theorem) that for all $x\in(0,\infty)$, the following inequality holds:

$\exp(-x)\geq 1 - x$

---

I don't know how the mean value theorem is applicable here, since we don't have a closed interval.

How do I prove the statement?

Answer: 
Hint: let $f(t)=e^{-t}$, and (for a fixed $x$) try using the mean value theorem on the interval $[0,x]$.
Let $f(x)=e^{-x}-1+x$. Fix a value $b \in [0,\infty)$ and consider the interval $[0,b]$. Note that $f(0)=0$.

Now according to the Mean Value Theorem, there exist a $c\in (0,b)$ such that $$f'(c)=\frac{f(b)-f(0)}{b-0}=\frac{e^{-b}-1+b}{b}>0.$$

This shows us that $f$ is increasing on any interval $(0,b)$ where $b$ is some fixed value in $[0,\infty)$. So $f(x)\geq f(0)$ for all $x\geq 0$. Thus $e^{-x}\geq 1-x$ for all $x\geq 0$.