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Graphically, it is easy to see but to analytically derive the following inequality,

$$|e^x-e^y| \le |x-y| \ \ \ \ \forall x,y \in (-\infty,0] \ \ \ \ (1)$$

do I just apply the MVT and show that the derivative of at a certain c between x and y equals $0$ hence the ratio is less than 1 and thus we derive $(1)$

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The idea is that $e^x$ is bounded on this interval since it is monotonically increasing.

By MVT, we know there exists some $c \in (-\infty,0)$ such that $$|e^x - e^y| = |e^c||x-y|$$ for any $x,y \in (-\infty,0]$. But notice that $e^c$ is bounded above by 1. This gives the desired result.

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  • $\begingroup$ Thanks Tim. I was so close. Just missed seeing it because of failure to rearrange the MVT formula. $\endgroup$
    – Quester
    Commented Dec 26, 2013 at 22:40

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