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Math Help - Inverse functions

  1. #1
    Member realintegerz's Avatar
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    Inverse functions

    How do I find f^-1 (2), x for f^-1 (x) = 0

    I have a graph with a function f on it,

    f(x) = [ 2x^3 + 1/2 x + 5 ]/2



    And also, it would be great if someone could find the equation for the inverse of f(x)
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by realintegerz View Post
    How do I find f^-1 (2), x for f^-1 (x) = 0

    I have a graph with a function f on it,

    f(x) = [ 2x^3 + 1/2 x + 5 ]/2



    And also, it would be great if someone could find the equation for the inverse of f(x)
    Remember that for inverses the domain and range switch. For example

    f(x) has domain A and range B than f^{-1}(x) has domain B and range A

    Ok so using this info we can solve what f^{-1}(x) = 0 and f^{-1}(2) is since we know the function is

    f(x) = \frac{2x^3 +\frac{x}{2} +5}{2}



    1) f^{-1}(x) = 0 is f(0)=?

    so

    x = \frac{5}{2}



    2)To solve the other problem set

    2 = \frac{2x^3 +\frac{x}{2} +5}{2}

    and solve for x
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  3. #3
    Member realintegerz's Avatar
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    hmmm..

    im stuck on that f^-1 (2)...

    So far i simplified it to...

    -2 = 4x^3 + x
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by realintegerz View Post
    hmmm..

    im stuck on that f^-1 (2)...

    So far i simplified it to...

    -2 = 4x^3 + x
    i tried using the rational root theorem but couldn't get any nice roots so i used the cubic equation. You could also do it numerically.
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  5. #5
    Member realintegerz's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    i tried using the rational root theorem but couldn't get any nice roots so i used the cubic equation. You could also do it numerically.
    I know the quadratic equation..but not cubic...

    could you explain that?
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