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Math Help - domain and 1 to 1..ps dan you rock

  1. #1
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    domain and 1 to 1..ps dan you rock

    id like to thank you dan for your previous help.....it really helped me understand how the process worked

    I have 2 questions if anyone could help me

    with F(x) = square root of x - 3 and G(x) = 4x^2 + 3
    a) write the function (G of F)(X) thats G of F of X
    I came up with G(square root of x - 3) = 4(square root of x - 3)^2 + 3
    G(x) = 4x + 12
    b) find the domain...of the G of F of X ....
    i have a few ideas but am kinda lost on this one




    SECOND QUESTION
    Given the one to one function G(x)= x^2+4 where X >
    write g^-1(x)
    write domain and range of G(x) G^-1(x)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mazdatuner
    F(x) = square root of x - 3 and G(x) = 4x^2 + 3
    a) write the function (G of F)(X) thats G of F of X
    I came up with G(square root of x - 3) = 4(square root of x - 3)^2 + 3
    G(x) = 4x + 12
    b) find the domain...of the G of F of X ....
    i have a few ideas but am kinda lost on this one
    You are very welcome!

    F(x)=\sqrt{x-3} and G(x)=4x^2+3

    a) You need to find: (G \circ F)(x)=G(F(x))
    So.
    G(F(x)) = G(\sqrt{x-3})=4(\sqrt{x-3})^2+3
    =4(x-3)+3=4x-9
    (You forgot to add the last 3.)

    b) You need to be a bit careful when considering the domain. At first blush it looks like the domain is  (-\infty,\infty) because there are no (apparent) restrictions on the domain. However, there IS a restriction given by the F(x): x \ge 3. So we can't actually have any x less than 3 in G(F(x)). This is subtle and implies that there actually IS a distinction between 4x-9 and 4(\sqrt{x-3})^2+3, which of course there is.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mazdatuner
    Given the one to one function G(x)= x^2+4 where X >
    write g^-1(x)
    write domain and range of G(x) G^-1(x)
    I usually like to skip the G(x) notation here and set y=x^2+4. It's a tad more convenient. To get the inverse function simply switch the position of x and y, then solve for y.

    x=y^2+4

    y^2=x-4

    y=\sqrt{x-4}
    Technically we should have a \pm in front of this, but we have to choose one or the other, else we no longer have a function. The choice of + is standard. So G^{-1}(x)=\sqrt{x-4}.

    (A good way to check this is to calculate G(G^{-1}(x)) and G^{-1}(G(x)), which both should be equal to x.)

    The domain of this is x such that the expression is real (and not otherwise undefined.) So the domain is  [4, \infty) .

    The range is all the values of the function over its domain, so I get  [0, \infty) .

    -Dan
    Last edited by topsquark; March 31st 2006 at 05:24 AM. Reason: Addendum
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  4. #4
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    I realized many people have difficulty finding the range of the function. The usual method is to sketch the function and see what values y-takes. But, the problem is it non-rigorous. Here is an algebraic method for which you can find the range. Since you found what G^{-1}=\sqrt{x-4} you need to find its range. By definition a certain value of y is in the range of the function if and only if there is such an x when mapped by the function gives y. Thus, you need to find all y such the the equation:
    y=\sqrt{x-4}
    has a real solution for x.
    Notice that \sqrt{x-4}\geq 0 thus, there would be no solutions if y<0. Now, when y\geq 0 then you can square both sides to get.
    y^2=x-4 and notice this equation has a solution x=y^2+4. Thus, all and only those values of y that are non-negative are in the range. Thus, y\geq 0 or another way of writing this is y\in [0,\infty]
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