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Math Help - help with sign graphs and inverse polynomial functions

  1. #1
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    help with sign graphs and inverse polynomial functions

    I have some questions that I have the answers for but no idea how to reach those answers.

    1)
    x^3 - 6x^2 +5x + 6 < 0 (use sign graph)
    the < is supposed to be "less than or equal to".
    the answer I'm given is: (-inf, 2-sqrt7] U [2, 2+sqrt7]
    I've never learned what that way of writing the answer means though.

    2) find f-1(x) (it's supposed to be find the inverse)
    f(x) = -(x-2)^2, x < 2
    the < is supposed to be "less than or equal to".
    the answer I'm given is: f-1(x) = 2-sqrt(-x)

    3) find f-1(x) (it's supposed to be find the inverse)
    f(x) = (2x+1) / (x+3), x cannot equal 3.
    the answer I'm given is: f-1(x) = (1-3x) / (x-2)

    -inf means negative infinity, sqrt means square root of

    Please help me.
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  2. #2
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    Quote Originally Posted by DarthBlood View Post
    I have some questions that I have the answers for but no idea how to reach those answers.

    1)
    x^3 - 6x^2 +5x + 6 < 0 (use sign graph)
    the < is supposed to be "less than or equal to".
    the answer I'm given is: (-inf, 2-sqrt7] U [2, 2+sqrt7]
    I've never learned what that way of writing the answer means though.[snip]
    The cubic factorises as (x^2 - 4x - 3)(x - 2). So draw the graph of y = (x^2 - 4x - 3)(x - 2) (be sure to find the three x-intercepts). The solution to the given inequality is the set of values of x such that y < 0.

    Quote Originally Posted by DarthBlood View Post
    [snip]
    2) find f-1(x) (it's supposed to be find the inverse)
    f(x) = -(x-2)^2, x < 2
    the < is supposed to be "less than or equal to".
    the answer I'm given is: f-1(x) = 2-sqrt(-x)
    [snip]
    y = f^{-1}(x) where x = -(y - 2)^2 and y < 2. Solve for y.

    Quote Originally Posted by DarthBlood View Post
    [snip]
    3) find f-1(x) (it's supposed to be find the inverse)
    f(x) = (2x+1) / (x+3), x cannot equal 3.
    the answer I'm given is: f-1(x) = (1-3x) / (x-2)

    -inf means negative infinity, sqrt means square root of

    Please help me.
    y = f^{-1}(x) where x = \frac{1 - 3y}{y - 2}. Solve for y by making it the subject:

    x = \frac{1 - 3y}{y - 2} \Rightarrow x(y - 2) = 1 - 3y \Rightarrow xy - 2x = 1 - 3y etc.
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  3. #3
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    i understand the first 2, but i still don't get the 3rd one.

    when you change the x and y, you changed using the answer. shouldn't i be changing with the original equation?

    so that the result is:

    x = \frac{2y+1}{y+3}

    not,

    x = \frac{1 - 3y}{y - 2}<br />
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  4. #4
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    Quote Originally Posted by DarthBlood View Post
    i understand the first 2, but i still don't get the 3rd one.

    when you change the x and y, you changed using the answer. shouldn't i be changing with the original equation?

    so that the result is:

    x = \frac{2y+1}{y+3}

    not,

    x = \frac{1 - 3y}{y - 2} <br />
    You're right. I misread f^-1 as the given function. The idea is exactly the same though.
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  5. #5
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    ah i got it! thanks a lot!
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