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Math Help - inverse of a function

  1. #1
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    inverse of a function

    consider the function f(x)=(x-1)^2+2

    1. Determine the inverse of f(x)

    2. Determine the domain and range of f(x) and its inverse f^-1(x)
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  2. #2
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    First part, answered here.

    As for the range, since f(x)=(x-1)^2+2\implies\sqrt{f(x)-2}=|x-1|, hence, we require that f(x)\ge2, thus the range is y\ge2. (The domain is all reals, of course.)
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  3. #3
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    Quote Originally Posted by william View Post
    consider the function f(x)=(x-1)^2+2

    1. Determine the inverse of f(x)

    2. Determine the domain and range of f(x) and its inverse f^-1(x)
    y=(x-1)^2+2

    for finding inverse, interchange x and y and solve for y,

    x=(y-1)^2+2

    x-2 = (y-1)^2

    \pm \sqrt{x-2}=y-1

    y=1\pm \sqrt{x-2}

    f^{-1}(x)=1\pm \sqrt{x-2}

    For f(x)

    Domain of f(x) = \{x\in \mathbb{R}\}

    Range of f(x) = \{y\in \mathbb{R}\;| \;y\ge2\}

    Now, For f^{-1}(x)

    Domain of f^{-1}(x) =\{x\in \mathbb{R}\;|\;x\ge 2\}

    Range of f^{-1}(x)  = \{y\in \mathbb{R}\}
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  4. #4
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    Strictly speaking f(x)= (x-1)^2+ 2 does NOT have an inverse because it is not a "one to one" function.


    Yes, you can recognize that the graph of y= f(x) has a vertex at (1, 2) and then break it into TWO functions:
    g(x)= (x-1)^2+ 2 with domain 1\le x which then has range 2\le y and write g^{-1}(x)= 1+ \sqrt{x-2} or
    h(x)= (x-1)^2+ 2 with domain x\le 1 which also has range 2\le y and write h^{-1}(x)= 1- \sqt{x- 2}.

    But neither of those functions is f(x) so, strictly speaking, neither of those is the inverse of f.
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