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Modulus? :/ Watch

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    Hi guys,

    There is nothing in any of my books about modulus :/
    This isn't the question that I need to answer-I've made it up to use as a 'template' to help answer the actual question-any help would be appreciated!

    |2x-4|=|3x+2|

    Thanks
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    |f(x)| = \pm\sqrt{[f(x)]^2}

    So

    |f(x)| = |g(x)|

    \pm\sqrt{[f(x)]^2} = \pm\sqrt{[g(x)]^2} \   \   \ *


    However, it's unlikely all the solutions to * will satisfy

    |f(x)| = |g(x)|

    So you'll need to substitute any found values of x back into the above equation to ensure it's satisfied.
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    (Original post by Noble.)
    |f(x)| = \pm\sqrt{[f(x)]^2}

    So

    |f(x)| = |g(x)|

    \pm\sqrt{[f(x)]^2} = \pm\sqrt{[g(x)]^2}
    I'm more confused than before
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    (Original post by dream123)
    Hi guys,

    There is nothing in any of my books about modulus :/
    This isn't the question that I need to answer-I've made it up to use as a 'template' to help answer the actual question-any help would be appreciated!

    |2x-4|=|3x+2|

    Thanks
    You need to think about the regions where:
    (i) 2x+4<0 and 3x+2<0
    (ii) 2x+4 \ge 0 and 3x+2 < 0
    (iii) 2x+4 \ge 0 and 3x+2 \ge 0

    Then use the fact that |a| = \begin{cases} a & \text{if } a \ge 0 \\ -a & \text{if } a < 0 \end{cases} to solve it. Bear in mind, though, that (i) and (iii) will actually give you the same answer.
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    (Original post by dream123)
    Hi guys,

    There is nothing in any of my books about modulus :/
    This isn't the question that I need to answer-I've made it up to use as a 'template' to help answer the actual question-any help would be appreciated!

    |2x-4|=|3x+2|

    Thanks
    Because mod f(x) is always positive, you need to add a function on both sides that makes it positive. Squaring both sides is normally used


    becomes


    NOTE: This only works if one side is purely a modulus, mixing scalars and modulus isn't as easy.
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    (Original post by dream123)
    I'm more confused than before
    Basically, when you take the modulus of an entire function, you know it cannot exist below the y axis. It's reflected on the x-axis. Graphically, when you reflect this function, you'll always go from a positive to negative gradient, or from a negative to a positive gradient. Which is essence creates two functions. The original function, and another function with a gradient of the opposite sign.

    Because the idea behind the modulus sign is to just take the magnitude of a function, irrespective of it's sign, you know that

    |6| = |-6|

    and

    |x| = |-x|

    and so

    |f(x)| = |-f(x)|

    So when you consider f(x) = 2x-4, the modulus of that f(x) produces the two functions.

    |f(x)| = |2x-4|

    |2x-4| = 2x-4 when considering x values that are \geq 2

    |2x-4| = -(2x-4) = -2x+4 when considering x values that are <2

    Because the function hits the x-axis at x=2, so it's the point at which the function is reflected. http://www.wolframalpha.com/input/?i=|2x-4|
 
 
 
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