# Math Help - Inverting a function

1. ## Inverting a function

Given this function $y=\sqrt[3]{1-x^3}$, 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.

2. ## Re: Inverting a function

Originally Posted by kkm
Given this function $y=\sqrt[3]{1-x^3}$, 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.
Swap $x~\&~y$ then solve for $y$.

3. ## Re: Inverting a function

Could you please show me how to solve for $y$?

4. ## Re: Inverting a function

Cube both sides, subtract 1, multiply by -1 then take the cube root.

5. ## Re: Inverting a function

Originally Posted by kkm
Could you please show me how to solve for $y$?
If $x = \sqrt[3]{{1 - {y^3}}}$ then cube both sides and solve for $y$.

If you cannot do this then you need to seek tutorial help.

6. ## Re: Inverting a function

another way to look at this is:

the inverse of a composition is the reverse composition of the inverses:

$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.

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: $(f \circ g)(x) = (g \circ f)(x) = x$.

so let's calculate $[(f \circ g) \circ (g^{-1} \circ f^{-1})](x)$.

$[(f \circ g) \circ (g^{-1} \circ f^{-1})](x) = (f \circ g)((g^{-1} \circ f^{-1})(x))$

$= (f \circ g)(g^{-1}(f^{-1}(x)) = f(g(g^{-1}(f^{-1}(x))))$

since $g(g^{-1}(t)) = t$ no matter what "t" is, we have (taking $t = f^{-1}(x)$),

$f(g(g^{-1}(f^{-1}(x)))) = f(f^{-1}(x)) = x$.

the proof that $[(g^{-1} \circ f^{-1}) \circ (f \circ g)](x) = x$ is entirely similar.

now let's look at "your function":

$f(x) = \sqrt[3]{1 - x^3}$.

this is the composition of 3 functions:

$f = g \circ h \circ k$, where

$k(x) = x^3$
$h(x) = 1 - x$
$g(x) = \sqrt[3]{x}$

a little thought should convince you that:

$g^{-1}(x) = x^3$ (note that means k is g's inverse)
$h^{-1}(x) = 1-x$ (h is its own inverse, this can happen)
$k^{-1}(x) = \sqrt[3]{x}$ (and as we would expect, g is k's inverse).

in short, $f^{-1} = (g \circ h \circ k)^{-1} = k^{-1} \circ h^{-1} \circ g^{-1} = g \circ h \circ k = f$

you might find this slightly amazing.

what this means is: $(f \circ f)(x) = x$.

let's "try it" with some actual number, instead of x. how about x = 5?

$f(5) = \sqrt[3]{1 - 125} = \sqrt[3]{-124}$

ok, that's kind of a weird number. so let's find f(x) when $x = \sqrt[3]{-124}$.

$f(\sqrt[3]{-124}) = \sqrt[3]{1 - (\sqrt[3]{-124})^3} = \sqrt[3]{1 - (-124)} = \sqrt[3]{125} = 5$. huh. how about that?

7. ## Re: Inverting a function

Originally Posted by Deveno
another way to look at this is:

the inverse of a composition is the reverse composition of the inverses:

$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.

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: $(f \circ g)(x) = (g \circ f)(x) = x$.

so let's calculate $[(f \circ g) \circ (g^{-1} \circ f^{-1})](x)$.

$[(f \circ g) \circ (g^{-1} \circ f^{-1})](x) = (f \circ g)((g^{-1} \circ f^{-1})(x))$

$= (f \circ g)(g^{-1}(f^{-1}(x)) = f(g(g^{-1}(f^{-1}(x))))$

since $g(g^{-1}(t)) = t$ no matter what "t" is, we have (taking $t = f^{-1}(x)$),

$f(g(g^{-1}(f^{-1}(x)))) = f(f^{-1}(x)) = x$.

the proof that $[(g^{-1} \circ f^{-1}) \circ (f \circ g)](x) = x$ is entirely similar.

now let's look at "your function":

$f(x) = \sqrt[3]{1 - x^3}$.

this is the composition of 3 functions:

$f = g \circ h \circ k$, where

$k(x) = x^3$
$h(x) = 1 - x$
$g(x) = \sqrt[3]{x}$

a little thought should convince you that:

$g^{-1}(x) = x^3$ (note that means k is g's inverse)
$h^{-1}(x) = 1-x$ (h is its own inverse, this can happen)
$k^{-1}(x) = \sqrt[3]{x}$ (and as we would expect, g is k's inverse).

in short, $f^{-1} = (g \circ h \circ k)^{-1} = k^{-1} \circ h^{-1} \circ g^{-1} = g \circ h \circ k = f$

you might find this slightly amazing.

what this means is: $(f \circ f)(x) = x$.

let's "try it" with some actual number, instead of x. how about x = 5?

$f(5) = \sqrt[3]{1 - 125} = \sqrt[3]{-124}$

ok, that's kind of a weird number. so let's find f(x) when $x = \sqrt[3]{-124}$.

$f(\sqrt[3]{-124}) = \sqrt[3]{1 - (\sqrt[3]{-124})^3} = \sqrt[3]{1 - (-124)} = \sqrt[3]{125} = 5$. huh. how about that?
Thank you soooo much for putting so much effort into your answer. Yesterday I figured out that this function was its own inverse, but your steps made it so much clearer. Thank you