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Thread: Verifying a function has an inverse

  1. #1
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    Verifying a function has an inverse

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

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

    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 $\displaystyle (f^{-1})'(4)$? Is this the same as $\displaystyle \frac{1}{f'(4)} = \frac{1}{-2-3(4)^2} = -\frac{1}{50} $?
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    Re: Verifying a function has an inverse

    To find the inverse function $\displaystyle f^{-1}$. If $\displaystyle y=1-2x-x^3$ then replace the $\displaystyle x$ and $\displaystyle y$ so $\displaystyle x=1-2y-y^3$, now find an expression $\displaystyle y=...$ in function of $\displaystyle x$. So you will get the inverse function.
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  3. #3
    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 $\displaystyle f(x) = 1 - 2x - x^3$ has an inverse, and find $\displaystyle (f^{-1})'(4)$.

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

    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 $\displaystyle (f^{-1})'(4)$? Is this the same as $\displaystyle \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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    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 $\displaystyle f(x) = 1 - 2x - x^3$ has an inverse, and find $\displaystyle (f^{-1})'(4)$.
    Because $\displaystyle f(-1)=4$ we need

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

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

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

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

    So
    $\displaystyle 1-2x-x^3=4$
    $\displaystyle x^3 + 2x = 3$
    $\displaystyle x = -1$.

    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
    $\displaystyle 1-2x-x^3=4$
    $\displaystyle x^3 + 2x = 3$
    $\displaystyle x = -1$.

    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 $\displaystyle x^3+2x+y-1=0$ i.e. $\displaystyle p=2,q=y-1$ . Then, $\displaystyle D=-108\left(\frac{q^2}{4}+\frac{p^3}{27}\right)<0$ which implies the equation has only one real root. So, using the well known formula for the roots:

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

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

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

    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.


    $\displaystyle f(x)=1-2x-x^3$

    $\displaystyle f'(x)=-2-3x^2$

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

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

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

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

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

    and,

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

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