Why is $\frac{1}{1-x}\circ\frac{1}{1-x}\circ\frac{1}{1-x}=x$ (graphically)? Of course, I can evaluate this composition of functions algebraically which is a fairly elementary exercise. But I'm curious as to whether one can find a (simple) graphical or visual argument (likely based on symmetry, perhaps due to representing this as a composition of the functions $1/x$ and $1-x$). To this extent, I've had a look at the following cobweb plot:

Which represents the composition. But this provides me with no geometric clarity or insight, perhaps because I'm not familiar enough with the geometric properties of hyperbolas? Anyways, if anyone has such an insight or ideas then I'd be grateful.
 A: There is a geometric explanation, but it's not a precalculus explanation. And the explanation does involve hyperbolic stuff, but not hyperbolas per se. Instead, the explanation involves the understanding that the formula $\frac{1}{1-x}$ represents an isometry of the upper half plane model of the hyperbolic plane $\mathbb H^2$.
If we rewrite your formula with a complex coordinate as
$$\frac{1}{-z+1}
$$
then we can see that it is a special case of a fractional linear transformation
$$f(z) = \frac{az+b}{cz+d}, \quad a,b,c,d \in \mathbb R, \quad ad-bc=1
$$
The upper half plane model of hyperbolic geometry is
$$\mathbb H^2 = \{z = x + iy \mid y > 0\}
$$
equipped with the Riemannian metric $\frac{dx^2+dy^2}{y^2}$. Every fractional linear transformation is an isometry of $\mathbb H^2$ in that metric. And here's where some group theory comes in: the fractional linear transformations are exactly the group of all  orientation preserving isometries of $\mathbb H^2$ (leaving out orientation reversing things like reflections and glide reflections).
Back to your example $f(z) = \frac{1}{-z+1}$. If you solve the fixed point equation $f(z)=z$ you'll obtain $z^2-z+1=0$ which gives $z = \frac{1}{2} + \frac{\sqrt{3}}{2} i \in \mathbb H^2$. In other words, your map fixes the point $z = \frac{1}{2} + \frac{\sqrt{3}}{2} i$.
If you compute $\frac{df}{dz}(\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ you'll get $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ (I'm not sure about the $+$ or $-$ sign) which is a cube root of unity. In other words, your map rotates around the point $z = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ by an angle $\pm 2 \pi / 3$.
It follows that $f^3$ fixes the point $\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ and, by the chain rule, has derivative $1$ at that point. In other words, $f^3$ rotates by angle $0$ around that point.
And in the hyperbolic plane (just as in the Euclidean plane), any isometry that fixes a point and rotates by angle $0$ at that point is the identity map.
A: We cube the matrix of your Mobius transformation;
$$
\left(
\begin{array}{cc}
a&b \\
c&d
\end{array}
\right)
$$
means  mapping $$   \frac{ax+b}{cx+d}$$   where we usually write $z$  instead of $x$
$$
\left(
\begin{array}{rr}
0&1 \\
-1&1
\end{array}
\right)
\left(
\begin{array}{rr}
0&1 \\
-1&1
\end{array}
\right)
\left(
\begin{array}{rr}
0&1 \\
-1&1
\end{array}
\right) = 
\left(
\begin{array}{rr}
-1&0 \\
0&-1
\end{array}
\right)
$$
The corresponding Mobius transformation is
$$ \frac{-x}{-1}  = x    $$
