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Math Help - Finding f^-1(x)

  1. #1
    Member M670's Avatar
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    Finding f^-1(x)

    Let . Find .

    make its y=15x^3-3
    y+3=15x^3

    I think up to this point I am right,

    My final answer is this y= but its wrong
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  2. #2
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    Re: Finding f^-1(x)

    Quote Originally Posted by M670 View Post
    Let . Find .

    make its y=15x^3-3
    y+3=15x^3

    I think up to this point I am right,

    My final answer is this y= but its wrong
    When finding an inverse function, the x and y values swap. So if \displaystyle \begin{align*} f(x) \end{align*} is given by \displaystyle \begin{align*} y = 15x^3 - 3 \end{align*} then \displaystyle \begin{align*} f^{-1}(x) \end{align*} will be defined by

    \displaystyle \begin{align*} x &= 15y^3 - 3 \\ x+3 &= 15y^3 \\ \frac{x + 3}{15} &= y^3 \\ \sqrt[3]{\frac{x + 3}{15}} &= y \end{align*}

    So the inverse function is \displaystyle \begin{align*} f^{-1}(x) = \sqrt[3]{\frac{x + 3}{15}} \end{align*}
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  3. #3
    Member M670's Avatar
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    Re: Finding f^-1(x)

    I just tired that answer and the program is still telling me I am wrong??
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  4. #4
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    Re: Finding f^-1(x)

    You must be entering it wrong then. Try ((x + 3)/(15))^(1/3).
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  5. #5
    Member M670's Avatar
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    Re: Finding f^-1(x)

    thank you I entered it like 3sqrt((x+3)/15) which I guess was telling it 3 times the sqrt
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  6. #6
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    Re: Finding f^-1(x)

    Quote Originally Posted by M670 View Post
    thank you I entered it like 3sqrt((x+3)/15) which I guess was telling it 3 times the sqrt
    cube root, not square root.
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