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Thread: Inverting a function

  1. #1
    kkm
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    Inverting 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.
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    Re: Inverting a function

    Quote Originally Posted by kkm View Post
    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.
    Swap $\displaystyle x~\&~y$ then solve for $\displaystyle y$.
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  3. #3
    kkm
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    Re: Inverting a function

    Could you please show me how to solve for $\displaystyle y$?
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    Re: Inverting a function

    Cube both sides, subtract 1, multiply by -1 then take the cube root.
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    Re: Inverting a function

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

    If you cannot do this then you need to seek tutorial help.
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  6. #6
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    Re: 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?
    Last edited by Deveno; Jun 24th 2012 at 10:52 PM.
    Thanks from kkm
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  7. #7
    kkm
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    Re: Inverting a function

    Quote Originally Posted by Deveno View Post
    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?
    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
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