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

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

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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: 
Suppose $x>0$. Then the mean value theorem applied to the interval $[0,x]$ says that
$$
\frac{e^{-x}-e^{-0}}{x-0}=-e^{-c}
$$
for some $c\in(0,x)$. Since $c>0$, we have $-c<0$, so $e^{-c}<1$ and so $-e^{-c}>-1$.

Therefore
$$
e^{-x}-1 > -x
$$
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 $x\geq 0$.