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28 January, 10:36

For the function f (x) = square root (x-5), find f^-1. What is the range of f^-1? Any explanation and answer is appreciated!

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  1. 28 January, 11:03
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    F (x) = sqrt (x-5)

    f^-1 (x) will be this ...

    y=sqrt (x-5)

    x=sqrt (y-5) (switch x and y)

    x^2 = [sqrt (y-5) ]^2 (solve for y)

    x^2=y-5

    x^2+5=y

    f^-1 (x) = x^2+5

    The range of f^-1 (x) is the same as the domain of f (x) so ...

    Find the domain of f (x)

    x-5≥0

    x≥5

    interval notation: [5,∞)

    That's the domain of f (x) and therefore it's the range of f^-1 (x)

    If you want me to explain this further, tell me in the comments.

    Best wishes!
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