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Math Help - inverse function

  1. #1
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    inverse function

    f(x) = x^(1/3) + 1

    so is f^-1(x) = (x-1)^(1/3) (?)

    thank you.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by 2clients View Post
    f(x) = x^(1/3) + 1

    is f^-1(x) = (x-1)^(1/3)

    thank you.
    Nope, but not far ^^

    The inverse function of x^a is x^{\tfrac 1a}. here, a=\tfrac 13
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Nope, but not far ^^

    The inverse function of x^a is x^{\tfrac 1a}. here, a=\tfrac 13
    Thanks. I'm still unclear though:

    f(x) = x^(1/3) + 1

    so y = x^(1/3) + 1 now if I solve for x why isn't that the inverse?

    x^(1/3) = y - 1

    x = (y-1)^(1/3)

    f^-1(x) = (x-1)^(1/3)
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  4. #4
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    Quote Originally Posted by 2clients View Post
    f(x) = x^(1/3) + 1

    so is f^-1(x) = (x-1)^(1/3) (?)

    thank you.
    f(x) = \sqrt[3]{x} + 1

    y = \sqrt[3]{x} + 1

    Switch x and y, then solve for y:
    x = \sqrt[3]{y} + 1

    \sqrt[3]{y} = x - 1

    y = (x-1)^3
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  5. #5
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    ... and how do we know that such function is invertible? What about the injection and surjection, hence bijection?
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