I was wrestling with some of these same questions myself earlier and never really found completely solid answers for what makes a good curve. I can, however, share my functions which I wound up using. They're all constant-power ones, but with different levels of fade/cut.
To understand what makes a curve constant-power, you have to understand that the signal is a sound-pressure level signal and that power goes as sound pressure squared. So if we have input signal $w_1$ and $w_2$ and we're attenuating the signal by multiplying $w_1 \cdot f(x)$ and $w_2 \cdot f(1-x)$ then $f$ is constant-power if $f^2(x)+f^2(1-x) = 1$. So, in this case, the easiest way would be to make $f(x) = \cos(\frac{\pi}{2}x)$.
In fact, we can generalize this and say that any function $g$ with range [0,1] for domain [0,1] can be used to produce a constant-power crossfade function $f(x)=\cos(\frac{\pi}{2}g(x))$. However, it'll make the most sense if $g(0) = 0$, $g(0.5)=0.5$, $g(1)=1$, and $g$ is monotonic. With this in mind, I more looked at functions $h$ where $h(-1) = -1$, $h(0)=0$, and $h(1)=1$ and then just did a simple linear transform between [-1,1] and [0,1]. So, the first thing I tried was $h(x)=x^{2n+1}$ for non-negative integers n. This turned out to work quite well. $n=0$ gives the constant power curve you show above and then as I go to $n=1,3,10$ I get curves a lot like slow fade, slow cut, and fast cut (although not identical since these are all constant power). There's obviously a lot of room to adjust the sharpness by using other values for n.
So my final function is $f(x)=\cos(\frac{\pi}{4}((2x-1)^{2n+1}+1))$. As you can see from the graphs below, they're quite similar to slow fade, slow cut, and fast cut, except that the plateau in the middle is at about 0.7 (actually $\frac{1}{\sqrt{2}}$) rather than at 1 and the track which doesn't fall off rises to 1 at the outside edge. You might think that going down to 0.7 in the middle would have a big sound impact on the other track, but it really doesn't. Obviously, you could convert them into the exact functions by doing $\sqrt{2} \min(f(x),1/\sqrt{2})$. But I've tried them out in a software fader in Pure Data without doing that adjustment, and they sound pretty good to my ears. I was also happy because it meant that I could switch between the four different curves with only a single parameter so that keeps the logic simple.
Below: Curves for Constant Power (n=0), Constant Power Slow Fade (n=1), Constant Power Slow Cut (n=3), and Constant Power Fast Cut (n=10).
