Fundamental differences between $e^{\frac{1}{x}}$ and $e^{-x}$, One of my students has asked for the differences between $e^{\frac{1}{x}}$ and $e^{-x}$, and I explained to her the differences in their domains, the continuity and differentiability of each function -- but also, I explained the similarities of both functions, particularly that they both have no roots and they are one-to-one functions, which means that they have inverse functions (for the purposes of Calculus I).
Is there something more fundamental and interesting that I can explain to her?  I feel I started to go off-topic in my email to her, when describing one-to-one functions, when instead she may be looking for a more basic answer about the two similar-looking exponentials.
Thanks,
 A: This is like explaining the differences and similarities between $f,g:\mathbb R_{> 0} \to \mathbb R$ given by $f(x)=x$ and $g(x)=e^x$ by saying they are both one-to-one, have the same domain, and have no roots.
The two mentioned functions are very different, even when one restricts attention to the natural numbers. Maybe consider $e^{\ln 3 x}$ and just take integers. This amounts to asking why $\frac{1}{3^n}$ and $3^{1/n}=\sqrt[n]{3}$ are different.
Graphically, as $x \to 0$, we see that $1/e^x$ just goes to $1$, whereas $e^{(1/x)}$ gets very large, heuristically since for rational numbers $e^{\frac{1}{1/n}}=e^n$
A: A fundamental difference is the asymptotic behavior: $e^{1/x}\overset{x\to\infty}{\longrightarrow}1$ while $e^{-x}\overset{x\to\infty}{\longrightarrow}0$.
Another one: $e^{-x}$ have an elementary indefinite integral while  $e^{1/x}$ does not.
A: I think of them both as $e^x$ with different things done to the domain.
With $e^{-x}$, the domain is flipped around backwards, so the usual end behaviors for plus and minus infinity are swapped. With $e^{1/x}$, the domain is turned inside out, with the end behaviors being brought in around $x=0$ and stuffed into finite intervals, and the usual behavior around $x=0$ being flung out to infinity in both directions.
The other differences between these functions are largely consequences of this. The transformation $x\mapsto -x$ is a fairly simple one, equivalent to just looking at the graph in a mirror. Thus, continuity, end behavior, derivative and antiderivative aren't changed very dramatically.
On the other hand, the transformation $x\mapsto \frac1x$ is a much more... violent action on the real line. Infinite intervals are stuffed into finite intervals, and a singularity is induced around the origin. This has much more drastic effects on the graph, its continuity, and its derivative and antiderivative.
