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Math Help - Interval Form?

  1. #1
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    Interval Form?

    how do i write this in interval form??

    (x+2)/(x-5)≤0
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  2. #2
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    Quote Originally Posted by VNVeteran View Post
    how do i write this in interval form??

    (x+2)/(x-5)≤0
    You should first note that x \neq 5.



    \frac{x + 2}{x - 5} \leq 0

    1 + \frac{7}{x - 5} \leq 0

    \frac{7}{x - 5} \leq 1


    Now we will have two alternatives, depending on whether the denominator is positive or negative...

    Case 1: x - 5 is positive, which means x > 5.


    7 \leq x - 5

    14 \leq x

    x \geq 14.



    Case 2: x - 5 is negative, which means x < 5.

    7 \geq x - 5

    14 \geq x

    x \leq 14.


    It's pretty obvious that the ONLY x which can not satisfy this equation are x = 5.


    So the domain is x < 5 \cup x > 5.

    Alternatively, it can be written x \neq 5 or (-\infty , 5) \cup (5, \infty).
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by VNVeteran View Post
    how do i write this in interval form??

    (x+2)/(x-5)≤0
    please be specific. are we talking about writing the domain in interval notation or the solution to the inequality in interval notation?

    I suspect the latter. In which case set the numerator and denominator to zero. You get x = -2 or x = 5 (now note that x cannot be 5, because the denominator cannot in fact be zero, we just do this to figure out the signs). Now you can draw a number line and plot x = -2 and x = 5 on it. Then plug in numbers in each of the three intervals that you have. you will find that the middle interval is the only one that works. the resulting solution is [-2, 5)
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  4. #4
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    Quote Originally Posted by VNVeteran View Post
    how do i write this in interval form??

    (x+2)/(x-5)≤0
    A fraction is negative if and only if numerator and denominator are of different signs. That is:

    1) x+2\ge 0 and x- 5< 0
    which lead to x\ge -2 and x< 5: [-2, 5)
    or

    2) x+2\le 0 and x- 5> 0
    which lead to x\le -2 and x> 5 which are impossible together.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    You should first note that x \neq 5.



    \frac{x + 2}{x - 5} \leq 0

    1 + \frac{7}{x - 5} \leq 0

    \frac{7}{x - 5} \leq 1


    Now we will have two alternatives, depending on whether the denominator is positive or negative...

    Case 1: x - 5 is positive, which means x > 5.


    7 \leq x - 5

    14 \leq x

    x \geq 14.



    Case 2: x - 5 is negative, which means x < 5.

    7 \geq x - 5

    14 \geq x

    x \leq 14.


    It's pretty obvious that the ONLY x which can not satisfy this equation are x = 5.


    So the domain is x < 5 \cup x > 5.

    Alternatively, it can be written x \neq 5 or (-\infty , 5) \cup (5, \infty).
    ok do i put any brackets in the interval form????? thats mainly what im confused about. when do i use brackets and when do i not?
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    A fraction is negative if and only if numerator and denominator are of different signs. That is:

    1) x+2\ge 0 and x- 5< 0
    which lead to x\ge -2 and x< 5: [-2, 5)
    or

    2) x+2\le 0 and x- 5> 0
    which lead to x\le -2 and x> 5 which are impossible together.
    no disrespect but i think your wrong. my book says the answer is (-infin, 5] U [5, infin)
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  7. #7
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    no, he's correct, and i do think you should reconsider twice on determining when an answer is wrong or not, see the procedure first.

    i wouldn't be surprised of another typo on a book, so in this case, book's answer is wrong and HallsofIvy is correct.
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  8. #8
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    HallsofIvy's answer is indeed correct. To demonstrate why (-\infty,5)\cup(5,\infty) is wrong, try substituting x=6 into the inequality and see what you get. Actually, any x>5 gives a positive result, in contradiction to this answer.
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  9. #9
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    Smile

    hello
    if we take the first case of what "Hallsofivy" posted.
    we'll have -2\leq x< 5 as a solution.
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  10. #10
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    Quote Originally Posted by Raoh View Post
    hello
    if we take the first case of what "Hallsofivy" posted.
    we'll have -2\leq x< 5 as a solution.
    However, IF the original question was "what is the domain of the function \frac{x+2}{x- 5}, THEN the answer is (-\infty, 5)\cup(5, \infty). Prove It said that in his first response!

    However, I don't see what the "<=" has to do with it then. You find domains of functions, not inequalities.
    Last edited by mr fantastic; November 16th 2009 at 03:49 AM. Reason: Fixed some latex tags
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