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Math Help - How to find the range of the following...

  1. #1
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    How to find the range of the following...

    Could you all explain how to solve these following problems in the simplest terms?

    Find the range of the following:





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  2. #2
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    f(x) = \frac{2x - 1}{x - 1}

     = \frac{2x - 2 + 1}{x - 1}

     = \frac{2x - 2}{x - 1} + \frac{1}{x - 1}

     = 2 + \frac{1}{x - 1}.

    This is a hyperbola with a horizontal and a vertical translation. Can you go from here?


    f(x) = \frac{|x-3|}{2}.

    Remember that absolute values gobble up all negatives and turn them positive. Can you go from here?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    f(x) = \frac{2x - 1}{x - 1}

     = \frac{2x - 2 + 1}{x - 1}

     = \frac{2x - 2}{x - 1} + \frac{1}{x - 1}

     = 2 + \frac{1}{x -th 1}.

    This is a hyperbola with a horizontal and a vertical translation. Can you go from here?

    Erm, where did you get the -2 and +1 from. And no, I don't think so. I already have the answer to the problem: (-infinity, 1)U(1, +infinity), so how do you go from  = 2 + \frac{1}{x -th 1} to (-infinity, 1)U(1, +infinity)?

    f(x) = \frac{|x-3|}{2}.

    Remember that absolute values gobble up all negatives and turn them positive. Can you go from here?

    I think so. So the range would be (-infinity, 3)U(3, +infinity)?
    .....
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  4. #4
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    Surely 2x - 2 + 1 = 2x - 1... I chose it because I needed to find a multiple of x - 1 in order to simplify. In this case, it's 2(x - 1) = 2x - 2.
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    For the first one, the range will be (-\infty,2)\cup(2,\infty) and not (-\infty,1)\cup(1,\infty) as written by the OP above (I can't quote it easily because it's inside another quote). Prove It's explanation is good, so try to break it down and understand Prove It's posts. For the second one, think of the range of |x-3| and then think how dividing that by 2 affects it ("hint": in this case it doesn't). For the last one, think about the range of |x-6| and then the range of |x-6| + 4, etc.

    Spoiler:

    The range of the second one is all non-negative reals.

    The range of the third one is all reals greater than or equal to 1 because the range of |x-6| is all nonnegative reals, then add 4 and you get all reals greater than or equal to 4, then take the square root and get all reals greater than or equal to 2, then divide by 2 to get all reals greater than or equal to 1.
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