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Math Help - Determine the domain of this function

  1. #1
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    Determine the domain of this function

    Determine the domain of this function

    f(x)= square root (x + 6)/ x^2-4x+3


    please explain how to do this and please put answer in interval form.

    ok so i factor out the denominator (x-3)=0 (x-1)=0
    Last edited by refresh; November 11th 2009 at 10:23 PM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by refresh View Post
    Determine the domain of this function

    f(x)= square root x + 6/ x^2-4x+3


    please explain how to do this and please put answer in interval form.
    f(x)=\sqrt{x}+\frac{6}{x^2-4x+3}

    To find the domain of a function, you start with a "default" domain of \mathbb{R} and remove all points where the function is not defined. \sqrt{x} can't take any x<0, and \frac{6}{x^2-4x+3} is undefined where x^2-4x+3=0. Every other input value is fine.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    f(x)=\sqrt{x}+\frac{6}{x^2-4x+3}

    To find the domain of a function, you start with a "default" domain of \mathbb{R} and remove all points where the function is not defined. \sqrt{x} can't take any x<0, and \frac{6}{x^2-4x+3} is undefined where x^2-4x+3=0. Every other input value is fine.
    i got x=3 and x=1 how do i put this in interval form???
    and the question was f(x)= square root (x+6)/x^2-4x+3
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by refresh View Post
    i got x=3 and x=1 how do i put this in interval form???
    Since x\geq0 and x\neq 1 and x\neq3, the interval form is:

    \text{dom}(f)=[0,1)\cup(1,3)\cup(3,\infty)
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  5. #5
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    Quote Originally Posted by redsoxfan325 View Post
    Since x\geq0 and x\neq 1 and x\neq3, the interval form is:

    \text{dom}(f)=[0,1)\cup(1,3)\cup(3,\infty)
    hmmm thats odd the back of the bok has a different answer??? maybe the book is wrong.
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  6. #6
    Super Member redsoxfan325's Avatar
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    What does the book say?

    Edit 1: Just noticed that you said \sqrt{\frac{x+6}{(x-1)(x-3)}}. In this case you need to find the places where \frac{x+6}{(x-1)(x-3)}<0.

    On (-\infty,-6), (x-1)(x-3)>0 and x+6<0, so \frac{x+6}{(x-1)(x-3)}<0.

    On [-6,1), (x-1)(x-3)>0 and x+6>0, so \frac{x+6}{(x-1)(x-3)}>0.

    On [1,3], (x-1)(x-3)\leq0 and x+6>0, so \frac{x+6}{(x-1)(x-3)}<0. (Note that this interval includes the two points where the denominator is zero.)

    On (3,\infty), (x-1)(x-3)>0 and x+6>0, so \frac{x+6}{(x-1)(x-3)}>0.

    So the domain is [-6,1)\cup(3,\infty).

    Edit 2: It's definitely not all real numbers. Either the book is wrong or you're reading the answer to the wrong problem.
    Last edited by redsoxfan325; November 11th 2009 at 10:44 PM. Reason: Misssed something
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  7. #7
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    Quote Originally Posted by redsoxfan325 View Post
    What does the book say?
    all real numbers x
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  8. #8
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    Quote Originally Posted by redsoxfan325 View Post
    What does the book say?

    Edit 1: Just noticed that you said \sqrt{\frac{x+6}{(x-1)(x-3)}}. In this case you need to find the places where \frac{x+6}{(x-1)(x-3)}<0.

    On (-\infty,-6), (x-1)(x-3)>0 and x+6<0, so \frac{x+6}{(x-1)(x-3)}<0.

    On [-6,1), (x-1)(x-3)>0 and x+6>0, so \frac{x+6}{(x-1)(x-3)}>0.

    On [1,3], (x-1)(x-3)\leq0 and x+6>0, so \frac{x+6}{(x-1)(x-3)}<0. (Note that this interval includes the two points where the denominator is zero.)

    On (3,\infty), (x-1)(x-3)>0 and x+6>0, so \frac{x+6}{(x-1)(x-3)}>0.

    So the domain is [-6,1)\cup(3,\infty).

    Edit 2: It's definitely not all real numbers. Either the book is wrong or you're reading the answer to the wrong problem.
    im sorry i read the wrong answer ok this is what the book said.

    [-6,1) U (1,3) U (3,infinity)


    by the way the only the (x+6) is in the squareroot.
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  9. #9
    Super Member redsoxfan325's Avatar
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    That makes it easier then. (-\infty, 6) is still off limits, but now only the two points 1 and 3 (as opposed to the interval [1,3]) are problematic, so the answer is what the books says:

    [-6,1)\cup(1,3)\cup(3,\infty)
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  10. #10
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    Quote Originally Posted by redsoxfan325 View Post
    That makes it easier then. (-\infty, 6) is still off limits, but now only the two points 1 and 3 (as opposed to the interval [1,3]) are problematic, so the answer is what the books says:

    [-6,1)\cup(1,3)\cup(3,\infty)
    hey thanks for all your help. im new to this site so i don't really know how everything works yet so i have a quick question. there is a donate button on the bottom, if i donate will you get the money or what?
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  11. #11
    Super Member redsoxfan325's Avatar
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    You would be donating money to the website, not me.
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