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Math Help - Grpahing functions

  1. #1
    Member realintegerz's Avatar
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    Graphing functions

    I'm having a hard time understanding these

    Let g(x) = 3(x+2)^2 + 4
    a. Graph y=gx without a calculator (how do I graph it???, we haven't learned this yet)
    b. Why is it not invertible?
    c. Let f(x) = 3(x+2)^2 + 4, x>=-2, why is f invertible?
    d. What is the domain of f^-1
    e. Determine rule for f^-1 (x)
    Last edited by realintegerz; October 13th 2008 at 06:58 PM.
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  2. #2
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    Reply

    g(x)= 3(x+2)^2+4

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

    a) This is an upward parabola with vertex (-2,4)

    See the attached graph.

    b) since, g^{-1}(x) is a parabola, (opening to the right) is not a function (since it does not pass the vertical line test). So, g(x) is not invertible.
    domain of g(x) is R.

    c) we have restricted the domain of f(x) such that its inverse f^{-1}(x) become a function.

    e) To find inverse of f(x), interchange x and y and solve for y.

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

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

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

    (y+2)^2=\frac{x-4}{3}

    y+2=\pm \sqrt{\frac{x-4}{3}}

    y=\pm \sqrt{\frac{x-4}{3}}-2

    f^{-1}(x)=\pm \sqrt{\frac{x-4}{3}}-2

    d) for domain, the value inside the square root \ge 0

    domain of f^{-1}(x), \;\;\frac {x-4}{3}\ge 0

    \Rightarrow x -4\ge 0

    \Rightarrow x \ge 4
    Attached Thumbnails Attached Thumbnails Grpahing functions-graph3.jpg  
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