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Math Help - Domain question

  1. #1
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    Domain question

    f(x)=(5x+sqrt(x^2+2x))/4+x


    I figure that the domain can't include -4 and  x^2+2 is greater than or equal to zero

    working from there I get

    x^2+2x , is greater than or equal to zero so

    x^2 is greater than or equal to -2x

    and therefore x is greater than or equal to -2

    however I have been told the answer is actually x equal to or LESS than -2

    I don't understand, where did x become less than -2?
    Last edited by kingsolomonsgrave; October 4th 2012 at 04:33 PM.
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  2. #2
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    Re: Domain question

    that is not the correct method to solve a quadratic inequality.

    x^2+2x \ge 0

    x(x+2) \ge 0

    now think about the graph of the parabola y = x(x+2) ... its zeros are x = 0 and x = -2 ... where (x-values) are the y-values > 0 ?
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  3. #3
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    Re: Domain question

    Quote Originally Posted by kingsolomonsgrave View Post
    f(x)=(5x+sqrt(x^2+2x))/4+x


    I figure that the domain can't include -4 and  x^2+2 is greater than or equal to zero

    working from there I get

    x^2+2x , is greater than or equal to zero so

    x^2 is greater than or equal to -2x

    and therefore x is greater than or equal to -2

    however I have been told the answer is actually x equal to or LESS than -2

    I don't understand, where did x become less than -2?
    You need to use brackets where they're needed. Anyway, I assume what you meant was \displaystyle \begin{align*} f(x) = \frac{5x + \sqrt{x^2 + 2x}}{4 + x} \end{align*}.

    Yes, you are correct that \displaystyle \begin{align*} x \neq -4 \end{align*}. Yes, you are correct that the function is only defined where \displaystyle \begin{align*} x^2 + 2x \geq 0 \end{align*}. Solving this inequality we have

    \displaystyle \begin{align*} x^2 + 2x &\geq 0 \\ x^2 + 2x + 1 &\geq 1 \\ (x + 1)^2 &\geq 1 \\ |x + 1| &\geq 1 \\ x + 1 \leq -1 \textrm{ or } x + 1 &\geq 1 \\ x \leq -2 \textrm{ or } x &\geq 0  \end{align*}


    So the domain of this function is \displaystyle \begin{align*} x \in (-\infty, -4) \cup (-4, -2] \cup [0, \infty)  \end{align*}.
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