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10 February, 02:19

What is the domain restriction that will allow you to find an inverse of the function f (x) = 3 (x - 9) 2 + 4?

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  1. 10 February, 02:43
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    f (x) = 3 (x - 9) ^2 + 4

    To find the inverse of f (x), first we denote: y = 3 (x - 9) ^2 + 4, we then find a way to express x in term of y.

    y = 3 (x - 9) ^2 + 4

    y - 4 = 3 (x - 9) ^2

    (y - 4) / 3 = (x - 9) ^2

    sqrt[ (y - 4) / 3] = x - 9

    x = sqrt[ (y - 4) / 3] + 9

    Denote left side is g (x), and variable y on the right side is x, we have:

    g (x) = sqrt[ (x - 4) / 3] + 9

    g (x) is inverse of f (x)

    Here, (x - 4) / 3 must be not a negative number (because of the definition of square root (sqrt) of a real number)

    => x - 4 > = 0 or x > = 4

    => The restriction domain of g (x), which is the inverse of f (x) is x > = 4
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