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Math Help - Domain and signum of the following function

  1. #1
    Newbie dttah's Avatar
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    Domain and signum of the following function

    Hello, I was having some trouble with this function here...

    \frac {log{|x^2-x+2|}}{\sqrt x}

    Domain:
    We have an abs in the logarithm so all we need to do is the following:

    x^2-x+2 \ne 0
    x >= 0
    \sqrt x \ne 0

    The last two can be merged in...  x > 0
    The first one has no solution, means that I have no problem up there, so the domain is:

    (0;+ \infty)

    Now I have to study the signum...

    \frac {log{|x^2-x+2|}}{\sqrt x} >= 0

    I was used to distinguish the two cases and study two separate functions when I have the absolute value. But the stuff inside the absolute value is always positive, so the function with or without the absolute value is pretty much the same thing? So studying the function WITHOUT the abs, is the same thing? Thanks !
    Last edited by dttah; November 26th 2011 at 03:12 AM.
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  2. #2
    A Plied Mathematician
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    Re: Domain and signum of the following function

    I agree with your domain. As for the signum, what you must study is the two following cases:

    0<|x^{2}-x+2|<1, and

    1\le|x^{2}-x+2|.

    The logarithm function is negative in the first case, and non-negative in the second. When do these cases occur? That is, for which values of x do these cases occur (if they occur at all)?
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  3. #3
    Newbie dttah's Avatar
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    Re: Domain and signum of the following function

    Well, looks to me that it's never in between 0 and 1 and it is always >= 1.
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    Re: Domain and signum of the following function

    Quote Originally Posted by dttah View Post
    Well, looks to me that it's never in between 0 and 1 and it is always >= 1.
    Yes. In fact,

    x^{2}-x+2=x^{2}-x+\frac{1}{4}-\frac{1}{4}+2

    =\left(x-\frac{1}{2}\right)^{\!\!2}+\frac{7}{4}

    \ge\frac{7}{4} \quad \forall x\in\mathbb{R}.

    Conclusion?
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  5. #5
    Newbie dttah's Avatar
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    Re: Domain and signum of the following function

    That's always positive so the absolute value is well.. useless? :P
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  6. #6
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    Re: Domain and signum of the following function

    Quote Originally Posted by dttah View Post
    That's always positive so the absolute value is well.. useless? :P
    Well, it is, but I was talking about what conclusion you can achieve towards solving the problem. What can you conclude about the signum of the function?
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  7. #7
    Newbie dttah's Avatar
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    Re: Domain and signum of the following function

    always positive!
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