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Math Help - Functions?

  1. #1
    Member realintegerz's Avatar
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    Functions?



    For the j(x) one, i need to know how to put in the negative sign....

    I'm trying to find out if those functions are even or odd

    Also, how do I find the domain of a function that doesn't have a denominator?

    Like say, y + x/y = 3, or sq. root y = x^2 + 1
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  2. #2
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    Quote Originally Posted by realintegerz View Post


    For the j(x) one, i need to know how to put in the negative sign....

    I'm trying to find out if those functions are even or odd

    Also, how do I find the domain of a function that doesn't have a denominator?

    Like say, y + x/y = 3, or sq. root y = x^2 + 1
    For j(x) what you have is perfectly acceptable. If you like you could distribute the negative by multiplying everything in the numerator by -1.

    i.e. j(x) = -\frac{x^3 + 2}{x^2 + 1}= \frac{-x^3 - 2}{x^2 + 1}.

    m(-x) is fine. To find -m(-x), just take the negative of m(-x). So do exactly what I suggested for j(x).

    What you've done with n(x) is fine too. Just do what I suggested with j(x) and m(x). It's even, because if you do a little more work, you can get n(-x) = n(x).



    To find the domain of a function, it is usually easiest to get y in terms of x.

    So for the first one...

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

    Multiply everything by y

     y^2 + x = 3y

    Get all the y's on one side...

     y^2 - 3y = -x

    Complete the square...

     y^2 - 3y + \left(-\frac{3}{2}\right)^2 = \left(-\frac{3}{2}\right)^2 - x

     \left(y - \frac{3}{2}\right)^2 = \frac{9}{4} - x .

    Now solve for y...

     y - \frac{3}{2} = \pm \sqrt{\frac{9}{4} - x}

     y = \frac{3}{2} \pm \sqrt{\frac{9}{4} - x} .

    Now you should be able to analyse the right hand side to find the domain.

    Clearly what is under the square root can't be negative, so...

    \frac{9}{4} - x \geq 0

    x \leq \frac{9}{4} .

    So the domain is \left(- \infty , \frac{9}{4} \right ].


    For the second one, just square both sides and you have y = something. Analyse the right hand side.
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  3. #3
    Member realintegerz's Avatar
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    i got y = x^4 + 1 for the 2nd one

    i still dont get how i find the domain for that though.....

    do i take the 4th root of it or ?
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  4. #4
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    Quote Originally Posted by realintegerz View Post
    i got y = x^4 + 1 for the 2nd one

    i still dont get how i find the domain for that though.....

    do i take the 4th root of it or ?
    Actually it's y = (x^2 + 1)^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1

    This is a polynomial function. Polynomials are continuous over \mathbf{R}. So the domain is \mathbf{R}.
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