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Math Help - [SOLVED] Domain

  1. #1
    Riley123
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    [SOLVED] Domain

    What is the domain of these problems??

    R(x) = 5x squared over 10-x


    Q(x) = -x(8-x) over 3xsquared + 7x+2


    H(x) = x-7 over 2x to the 4th +7
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Riley123
    What is the domain of these problems??

    R(x) = 5x squared over 10-x


    Q(x) = -x(8-x) over 3xsquared + 7x+2


    H(x) = x-7 over 2x to the 4th +7
    These would be clearer if (remaining with ASCII) you write them:

    R(x) = 5 x^2/(10-x)

    Q(x) = -x (8-x)/(3 x^2 + 7 x + 2)

    H(x) = (x-7)/(2 x^4 + 7)

    where I have to make several guesses at what you actual mean (note
    here ^ denotes raising to a power, so x^2 is x squared).

    Better yet would be:
    R(x)\ =\ \frac{5\ x^2}{(10-x)}

    Q(x)\ =\ \frac{-x\ (8-x)}{(3x^2\ +\ 7x\ +\ 2)}

    H(x)\ =\ \frac{(x-7)}{(2x^4\ +7)}

    The domain of a function is the set on which the function is defined.
    From the look of this problem it looks as through we are actually being
    asked for the set of real numbers on which the functions are defined.
    Though what follows would be equally applicable if we were working
    with the complex numbers.

    In all of these function definitions both the numerators and
    denominators can be evaluated for all real numbers x, but
    the functions are undefined when the denominators are zero.

    In the first case the denominator is (10-x) which is zero when
    x is equal to 10, so the domain of R is the set of all real number
    except 10.

    The rest are similar, in each case the domain is the set of all
    of the reals except the points at which the denominator are
    zero (it at all).

    RonL
    Last edited by CaptainBlack; November 15th 2005 at 10:15 AM.
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