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    please help me!!

    find the domain:

    f(x) = √(3-x)/ modulus x + x

    and

    f(x) = modulus(x-1) - modulus (x+1)

    (plus range for this one!)

    plz help guys!! plz
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    (Original post by ossoss87)
    please help me!!

    find the domain:

    f(x) = √(3-x)/ modulus x + x

    and

    f(x) = modulus(x-1) - modulus (x+1)

    (plus range for this one!)

    plz help guys!! plz
    Well, for the first one, you have  0 < x < 4 as the domain. This is because for any  x \in R such that  x > 4 you have an imaginary numerator. For any number  x \in R such that  x < 0 , you are dividing by 0, which gives an undefined answer. Use this type of thinking for the second one.
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    please can u explain more? im bad in these stuff!
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    (Original post by ossoss87)
    please can u explain more? im bad in these stuff!
    No problem. Let's take a few examples.

    If  x \geq 4 , we have:

     \frac {\sqrt {-1}}{|4| + 4}

    That is

     \frac {1}{8} \sqrt {-1}

    Since we're defining the domain over real numbers only, we can't take any values for  x \geq 4 . Hence,  x < 4 or  x \leq 3 (whichever takes your fancy ).

    Now, for values of x such that x is negative (i.e.  x < 0 ), we have (taking -1 as an example):

     \frac {\sqrt {4}}{|-1| -1 }

     = [tex] \frac {2}{0}

    This is undefined. This is the case for all negative numbers, and 0. Thus, we take  x > 0 . So, our domain is the union of these two inequalities. (i.e.  0<x<4 ). Hope that helps.
 
 
 
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