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Math Help - inverse functions

  1. #1
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    inverse functions

    When

    f(x)=x^7+2x^3+x-1

    I have found

    f^-1(3)=1

    But I need to prove this.

    Can anyone help?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse functions

    f(1)=3 and f is strictly increasing.
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  3. #3
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    Re: inverse functions

    I need to find f^-1, or the inverse of f. Not sure how to do this?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse functions

    Quote Originally Posted by Arron View Post
    I need to find f^-1, or the inverse of f. Not sure how to do this?
    By definition of f^{-1} , f^{-1}(\{y\})=\{x\in\mathbb{R}:f(x)=y\} . As f(1)=3 we can conclude that 1\in f^{-1}(\{3\}) . If we prove that f is strictly increasing then, no x\neq 1 satisfies f(x)=3 so, f^{-1}(\{3\})=\{1\} . That was your initial question. Now, if you want a closed formula f^{-1}(x) for all x\in\mathbb{R} , better forget it.
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  5. #5
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    Re: inverse functions

    Quote Originally Posted by Arron View Post
    I need to find f^-1, or the inverse of f.
    [snip]
    Why? (Read previous reply).
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: inverse functions

    Quote Originally Posted by Arron View Post
    I need to find f^-1, or the inverse of f. Not sure how to do this?
    If finding f^{-1}(*) is a 'categoric imperative' , then a method based on advanced tecniques of complex analysis will be supplied without demonstration...

    If the function the inverse of which has to be found is...

    y= x^{7} + 2\ x^{3} + x -1 (1)

    ... then first with the variable substitution z=1+y we obtain...

    z=f(x)= x^{7} + 2\ x^{3} + x (2)

    Now the inverse x=f^{-1} (z) of a function for which is f(0)=0 can be written in series expansion as...

    x= \sum_{n=1}^{\infty} a_{n}\ z^{n} (3)

    ... where the a_{n} are given by...

    a_{n}= \frac{1}{n!}\ \lim_{x \rightarrow 0} \frac{d^{n-1}}{d x^{n-1}} \{\frac{x}{f(x)}\}^{n} (4)

    In our case of course is...

    \frac{x}{f(x)} = \frac{1}{x^{6}+ 2\ x^{2} +1} (5)

    ... and the x as function of y will be...

    x= \sum_{n=1}^{\infty} a_{n}\ (1+y)^{n} (6)

    Details of computation of the a_{n} are left to the [young I suppose...] starter of the thread...

    Kind regards

    \chi \sigma
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  7. #7
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    Re: inverse functions

    Thanks for your help everyone.

    One final thing, I am trying to determine the value of (f^-1)'(3)

    Would this = 1/f'(3) = -1
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse functions

    Quote Originally Posted by Arron View Post
    Would this = 1/f'(3) = -1
    No, it is (f^{-1})'(3)=\dfrac{1}{f'(f^{-1}(3))}=\dfrac{1}{f'(1)}=\ldots
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  9. #9
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    Re: inverse functions

    would that be 1/13.
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse functions

    Quote Originally Posted by Arron View Post
    would that be 1/13.
    Right.
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  11. #11
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    Re: inverse functions

    Quote Originally Posted by Arron View Post
    would that be 1/13.
    Why do you need to even ask!? You have been told exactly what to do. Surely at this level of study you can differentiate a polynomial, substitute x = 1 and then take the reciprocal of the result!
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