Originally Posted by
Deveno another way to look at this is:
the inverse of a composition is the reverse composition of the inverses:
$\displaystyle (f \circ g)^{-1} = g^{-1} \circ f^{-1}$.
to see that this is so, recall that an inverse of a function (provided one does exist) is a function g so that f(g(x)) = g(f(x)) = x.
we can write this as: $\displaystyle (f \circ g)(x) = (g \circ f)(x) = x$.
so let's calculate $\displaystyle [(f \circ g) \circ (g^{-1} \circ f^{-1})](x)$.
$\displaystyle [(f \circ g) \circ (g^{-1} \circ f^{-1})](x) = (f \circ g)((g^{-1} \circ f^{-1})(x))$
$\displaystyle = (f \circ g)(g^{-1}(f^{-1}(x)) = f(g(g^{-1}(f^{-1}(x))))$
since $\displaystyle g(g^{-1}(t)) = t$ no matter what "t" is, we have (taking $\displaystyle t = f^{-1}(x)$),
$\displaystyle f(g(g^{-1}(f^{-1}(x)))) = f(f^{-1}(x)) = x$.
the proof that $\displaystyle [(g^{-1} \circ f^{-1}) \circ (f \circ g)](x) = x$ is entirely similar.
now let's look at "your function":
$\displaystyle f(x) = \sqrt[3]{1 - x^3}$.
this is the composition of 3 functions:
$\displaystyle f = g \circ h \circ k$, where
$\displaystyle k(x) = x^3$
$\displaystyle h(x) = 1 - x$
$\displaystyle g(x) = \sqrt[3]{x}$
a little thought should convince you that:
$\displaystyle g^{-1}(x) = x^3$ (note that means k is g's inverse)
$\displaystyle h^{-1}(x) = 1-x$ (h is its own inverse, this can happen)
$\displaystyle k^{-1}(x) = \sqrt[3]{x}$ (and as we would expect, g is k's inverse).
in short, $\displaystyle f^{-1} = (g \circ h \circ k)^{-1} = k^{-1} \circ h^{-1} \circ g^{-1} = g \circ h \circ k = f$
you might find this slightly amazing.
what this means is: $\displaystyle (f \circ f)(x) = x$.
let's "try it" with some actual number, instead of x. how about x = 5?
$\displaystyle f(5) = \sqrt[3]{1 - 125} = \sqrt[3]{-124}$
ok, that's kind of a weird number. so let's find f(x) when $\displaystyle x = \sqrt[3]{-124}$.
$\displaystyle f(\sqrt[3]{-124}) = \sqrt[3]{1 - (\sqrt[3]{-124})^3} = \sqrt[3]{1 - (-124)} = \sqrt[3]{125} = 5$. huh. how about that?