Given this function $\displaystyle y=\sqrt[3]{1-x^3}$, I have to find its inverse. How do you do that, and which is its inverse?

I am clueless, and I unfortunately need this answer tomorrow. I would hugely appreciate any help with this.

Printable View

- Jun 24th 2012, 03:04 PMkkmInverting a function
Given this function $\displaystyle y=\sqrt[3]{1-x^3}$, I have to find its inverse. How do you do that, and which is its inverse?

I am clueless, and I unfortunately need this answer tomorrow. I would hugely appreciate any help with this. - Jun 24th 2012, 03:19 PMPlatoRe: Inverting a function
- Jun 24th 2012, 03:23 PMkkmRe: Inverting a function
Could you please show me how to solve for $\displaystyle y$?

- Jun 24th 2012, 03:39 PMpickslidesRe: Inverting a function
Cube both sides, subtract 1, multiply by -1 then take the cube root.

- Jun 24th 2012, 03:46 PMPlatoRe: Inverting a function
- Jun 24th 2012, 10:42 PMDevenoRe: Inverting a function
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? - Jun 25th 2012, 05:04 AMkkmRe: Inverting a function