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Math Help - Verifying a function has an inverse

  1. #1
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    Verifying a function has an inverse

    Verify f(x) = 1 - 2x - x^3 has an inverse, and find (f^{-1})'(4).

    So I take the derivative and get f'(x) = -2 -3x^2.

    It's negative for all x, so it's always decreasing, so it's 1-to-1 and has an inverse.

    Now, how do I find (f^{-1})'(4)? Is this the same as \frac{1}{f'(4)} = \frac{1}{-2-3(4)^2} = -\frac{1}{50} ?
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    MHF Contributor Siron's Avatar
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    Re: Verifying a function has an inverse

    To find the inverse function f^{-1}. If y=1-2x-x^3 then replace the x and y so x=1-2y-y^3, now find an expression y=... in function of x. So you will get the inverse function.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Verifying a function has an inverse

    Quote Originally Posted by deezy View Post
    Verify f(x) = 1 - 2x - x^3 has an inverse, and find (f^{-1})'(4).

    So I take the derivative and get f'(x) = -2 -3x^2.

    It's negative for all x, so it's always decreasing, so it's 1-to-1 and has an inverse.

    Now, how do I find (f^{-1})'(4)? Is this the same as \frac{1}{f'(4)} = \frac{1}{-2-3(4)^2} = -\frac{1}{50} ?

    For the inverse part:

    Prove that f is one-to-one an onto.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Verifying a function has an inverse

    Quote Originally Posted by Also sprach Zarathustra View Post
    For the inverse part:

    Prove that f is one-to-one an onto.
    And also look here: Inverse functions and differentiation - Wikipedia, the free encyclopedia
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    Re: Verifying a function has an inverse

    Quote Originally Posted by deezy View Post
    Verify f(x) = 1 - 2x - x^3 has an inverse, and find (f^{-1})'(4).
    Because f(-1)=4 we need

    (f^{-1})'(4)=\frac{1}{f'(-1)}.
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  6. #6
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    Re: Verifying a function has an inverse

    How do you determine that f(-1)=4?

    Switching the x and y wasn't helping because there's 2 ys in the equation.
    x=1-2y-y^3
    -2y-y^3 = x - 1???
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    Re: Verifying a function has an inverse

    Quote Originally Posted by deezy View Post
    How do you determine that f(-1)=4?
    Switching the x and y wasn't helping because there's 2 ys in the equation.
    x=1-2y-y^3
    -2y-y^3 = x - 1???
    I do not think you can solve that inverse.
    You just find f(x)=4 for x.
    See my other reply.
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  8. #8
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    Re: Verifying a function has an inverse

    So
    1-2x-x^3=4
    x^3 +  2x = 3
    x = -1.

    I can see that instinctively... but if it's a harder equation I will just have to use trial and error?
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    Re: Verifying a function has an inverse

    Quote Originally Posted by deezy View Post
    So
    1-2x-x^3=4
    x^3 +  2x = 3
    x = -1.

    I can see that instinctively... but if it's a harder equation I will just have to use trial and error?
    Well that is just the way it is. Often a graph will give you an idea of the pre-image
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Verifying a function has an inverse

    @ Deezy:

    Sorry, I just wanted to give a 'general' solution to calculate the inverse function, but indeed in this case that would be really hard because of the third degree, if you have a first degree or second degree function it's less complicated.
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Re: Verifying a function has an inverse

    In this case it's not very hard. We have the equation x^3+2x+y-1=0 i.e. p=2,q=y-1 . Then, D=-108\left(\frac{q^2}{4}+\frac{p^3}{27}\right)<0 which implies the equation has only one real root. So, using the well known formula for the roots:

    x=\sqrt[3]{-\frac{y-1}{2}+\sqrt{\frac{(y-1)^2}{4}+\frac{8}{27}}}+\sqrt[3]{-\frac{y-1}{2}-\sqrt{\frac{(y-1)^2}{4}+\frac{8}{27}}}

    We can inmediately verify that the product of the two real cubic roots of the above addends is -2/3=-p/3 . So, we can express

    f^{-1}(x)=\sqrt[3]{-\frac{x-1}{2}+\sqrt{\frac{(x-1)^2}{4}+\frac{8}{27}}}+\sqrt[3]{-\frac{x-1}{2}-\sqrt{\frac{(x-1)^2}{4}+\frac{8}{27}}}

    where we choose for each addend the real cubic root.
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  12. #12
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Verifying a function has an inverse

    Quote Originally Posted by Also sprach Zarathustra View Post
    For the inverse part:

    Prove that f is one-to-one an onto.


    f(x)=1-2x-x^3

    f'(x)=-2-3x^2

    f'(x)<0 for all x\in\mathbb{R}

    In other words f(x) is monotone decreasing for all x\in\mathbb{R}.

    We deduce from the above that f(x) is bijection.

    To show that f(x) is surjective, we note that:

    \lim_{x\to\infty}f(x)=-\infty

    and,

    \lim_{x\to-\infty}f(x)=\infty

    These two results with MVT(with a little work...) we wil get that f(x) is surjective function.
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