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Math Help - Trouble with a domain problem

  1. #1
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    Trouble with a domain problem

    Hi, I'm having trouble with a domain problem, I think I'm going about it the wrong way, or maybe I'm missing something.

    f(x) = \sqrt{2x+2}
    g(x) = (x-5)/x
    (f o g)(x) = ?

    I need to find the domain of (f o g)(x). I started by finding (f o g)(x). I got:

    (f o g)(x) = \sqrt{(4x-10)/x}

    I've found that f(x) must be greater than -1 and g(x) cannot equal 0. So the ranges so far would be:

    [-1, 0)U(0,inf)

    Then I looked at the limitations of (f o g)(x). I found that it x cannot equal 0, x must be greater than or equal to \frac{5}{2}. So as I see it, the limits I listed for f(x) and g(x) can be disregarded, since x must be greater than 2.5, which is bigger than -1 or 0.

    My final answer should be:

    [ \frac{5}{2},inf)

    Or so I thought, I tried to submit that answer and it came back as incorrect.

    Am I solving this the wrong way? Thanks in advance.
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  2. #2
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    Quote Originally Posted by cb220 View Post
    Hi, I'm having trouble with a domain problem, I think I'm going about it the wrong way, or maybe I'm missing something.

    f(x) = \sqrt{2x+2}
    g(x) = (x-5)/x
    (f o g)(x) = ?

    I need to find the domain of (f o g)(x). I started by finding (f o g)(x). I got:

    (f o g)(x) = \sqrt{(4x-10)/x}

    I've found that f(x) must be greater than -1 and g(x) cannot equal 0.


    I suppose you meant that for the function f it must be that x\geq -1 , and for g it must be x\neq 0

    So the ranges so far would be:

    [-1, 0)U(0,inf)

    Then I looked at the limitations of (f o g)(x). I found that it x cannot equal 0, x must be greater than or equal to \frac{5}{2}. So as I see it, the limits I listed for f(x) and g(x) can be disregarded, since x must be greater than 2.5, which is bigger than -1 or 0.

    My final answer should be:

    [ \frac{5}{2},inf)

    Or so I thought, I tried to submit that answer and it came back as incorrect.

    Am I solving this the wrong way? Thanks in advance.

    Well, since for f it must be x\geq -1 , we then have that for f\circ g(x):=f(g(x)) it must be

    g(x)=\frac{x-5}{x}\geq -1\Longleftrightarrow \frac{2x-5}{x}\geq 0\Longleftrightarrow x\leq 0\,\,or\,\,x\geq \frac{5}{2}.

    The above, together with the first limits, gives you the answer:  -1\leq x <0\,\,or\,\,x\geq \frac{5}{2}\Longleftrightarrow [-1,\,0)\cup (5/2,\,\infty)

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    g(x)=\frac{x-5}{x}\geq -1\Longleftrightarrow \frac{2x-5}{x}\geq 0\Longleftrightarrow x\leq 0\,\,or\,\,x\geq \frac{5}{2}.
    Thanks for the quick reply. I'm with you until x\leq 0. Not sure how to arrive at that, with the direction of the sign changed.

    That said, the rest of your post makes sense, and I can see why my answer was wrong if I could only figure out how to get the x\leq 0.

    EDIT: Gah, the answer you've provided also comes back as incorrect I've tried it as [-1,0)U(5/2,inf) and [-1,0)U[5/2,inf). Can't see why it isn't working.
    Last edited by cb220; September 24th 2010 at 06:50 PM.
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  4. #4
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    Quote Originally Posted by cb220 View Post
    Thanks for the quick reply. I'm with you until x\leq 0. Not sure how to arrive at that, with the direction of the sign changed.

    That said, the rest of your post makes sense, and I can see why my answer was wrong if I could only figure out how to get the x\leq 0.

    EDIT: Gah, the answer you've provided also comes back as incorrect I've tried it as [-1,0)U(5/2,inf) and [-1,0)U[5/2,inf). Can't see why it isn't working.

    Of course: I did an elementary mistake, but it doesn't matter. you should understand the way by which we arrive to the answer and reach it...

    As before, it must be g(x)\geq -1\Longleftrightarrow x\,\,\leq 0\,\,or\,\, x\geq \frac{5}{2} , and now we must take care that for g,

    x\neq 0 , so the answer must be x< 0\,\,or\,\, x\geq\frac{5}{2}\Longleftrightarrow (-\infty,\,0)\cup [5/2,\,\infty)

    Tonio
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  5. #5
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    Hmm, okay I think I get that. {x}\geq{-1} isn't used in the final solution because that was a limiter put on f(x), but the x in f(x) isn't used in (f o g)(x), instead g(x) is used. Is that right?

    On the other hand I'm still having a hard time seeing how to get:

    g(x)\geq -1\Longleftrightarrow x\,\,\leq 0

    I got the \frac{5}{2} part of it but not getting the other half.
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  6. #6
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    Quote Originally Posted by cb220 View Post
    Hmm, okay I think I get that. {x}\geq{-1} isn't used in the final solution because that was a limiter put on f(x), but the x in f(x) isn't used in (f o g)(x), instead g(x) is used. Is that right?

    On the other hand I'm still having a hard time seeing how to get:

    g(x)\geq -1\Longleftrightarrow x\,\,\leq 0

    I got the \frac{5}{2} part of it but not getting the other half.

    \frac{x-5}{x}\geq -1\Longleftrightarrow \frac{x-5}{x}+1\geq 0\Longleftrightarrow \frac{2x-5}{x}\geq 0\Longleftrightarrow x(2x-5)\geq 0 , and

    now you've a simple quadratic inequality (draw the hiperbole of the left hand and check when it is above the x-axis...

    Tonio
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