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Math Help - Inverse functions help?

  1. #1
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    Inverse functions help?

    the questino is
    A)find domain and range
    b)find domain and range of inverse function
    c)Find an explicit formula for the inverse function as a function of x.
    f(x) (x-2)^2-9 x is bigger than or equal to 4

    a)domain of original function is [4, infinity)
    How do I find the range?
    b)domain of inverse function is range of original function, range of inverse function is domain of original function so [4, infinity)
    c) the formula for the inverse function is
    f(x)=(x-2)^2-9
    x=(f(x)-2)^2-9
    x+9 = (f(x)-2)^2
    \sqrt{x+9} = f(x)-2
    2+\sqrt{x+9} = f(x)
    i would think that the range would be [9, infinity), since 9 would be the lowest number than can still make the square root non negative, but the book's answer says the range of the original function/domain of inverse is [-5, infinity) does this have something to do with the restricted domain? if so, what do I do to get this range?
    Last edited by dorkymichelle; December 17th 2009 at 08:56 PM.
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  2. #2
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    I like the work you have done here. It might be useful to know that

    the domain of f(x) is the range of f'(x) and

    the range of f(x) is the domain of f'(x)

    To make your square roots look better use \sqrt{} instead of \sqrt()

    This is the difference between \sqrt{x+9} and \sqrt(x+9)
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  3. #3
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    Thanks for the tip about the square roots, i couldn't figure out how to make it look right.

    the problem is I don't know how to find the range of the original f(x), so I can't find the domain of f'(x)
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  4. #4
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    Graph the function to get a rough idea of the shape and turning point.

    Here's a hint, the range for your function is [f(4),\infty)

    Now find f(4)
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  5. #5
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    f(4) = -5
    why do I plug in the 4 to get the range?
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  6. #6
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    Quote Originally Posted by dorkymichelle View Post
    f(4) = -5
    why do I plug in the 4 to get the range?
    The point of using "4" is that your domain is [4, \infty).

    Another important point is that f(x)= (x-2)^2- 9 is equal to -9 when x= 2 and is larger than -9 for all other x, because (x- 2)^2 is positive for all x except 2. That is, the graph is a parabola with vertex at (2, 9). But your domain is x\ge 4- what does the parabola look like past x= 4? What does that tell you about the domain?

    Now, as for the inverse, a standard way of finding inverse functions is to "swap" x and y: if y= f(x) then x= f^{-1}(y). From y= (x-2)^2- 9 you get x= (y- 2)^2- 9. Solve for y. When taking the square root, to 'invert' the square, remember that y must be larger than or equal to 4.
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