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Math Help - More polynomial inequalities question

  1. #1
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    More polynomial inequalities question

    I had another question for the problem (x(x-8))/(x^2-9x-52)<=0

    my work is:
    x(x-8)/(x+4)(x-13)<=0

    i then did a number line with the numbers -4,0,8,13 and did the sign chart underneath. The answer I ended getting is (-4,0)U(8,13) I retried the problem several times and still got the same answer.

    I also did the problem (x-4)/(x+13)=>0 with that i got the answer (-infinity,-13]U[4,infinity)

    both answers were incorrect also.
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  2. #2
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    The answer for the first one you have is almost correct. It is \left( { - 4,0} \right]\bigcup {\left[ {8,13} \right)} rather than \left( { - 4,0} \right)\bigcup {\left( {8,13} \right)}.

    Same deal with the second. It is \left( { - \infty ,13} \right)\bigcup {\left[ {4,\infty } \right)} not \left( { - \infty ,13} \right]\bigcup {\left[ {4,\infty } \right)}.

    Those intervals can be tricky at times. If it is undefined at the endpoint, you should use the open interval. Also, always use an open interval for \infty, as you seem to know.

    Edit: forgot to add, the interval is closed in the first problem at 0 and 8 because those values make the function 0, not undefined. Since the inequality wants 0 included, the interval is closed. Hopefully that clarifies it a bit.
    Last edited by xifentoozlerix; March 28th 2008 at 02:00 PM.
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  3. #3
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    2. (-infinity,-13)U[4,infinity)
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  4. #4
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    The key idea to solving a problem like \frac{x(x-8)}{(x+4)(x-13)} < 0 is to use the fact that the numerator and denominator must have different signs. So either

    x(x-8) > 0 AND (x+4)(x-13) < 0

    OR

    x(x-8) < 0 AND (x+4)(x-13) > 0 .

    Let's look at the first case:

    Question: When is x(x+8) > 0?
    Answer: When x and x+8 are both positive OR both negative.

    x > 0 AND x-8 > 0 \Rightarrow x > 8. \text{   } (8,\infty)
    OR
    x < 0 and  x - 8 < 0 \Rightarrow x < 0. \text{   }(-\infty,0)

    Conclusion 1:
    Putting this together we say the numerator is positive when x is in (-\infty,0) \cup (8,\infty)

    Question: When is (x+4)(x-13) < 0?
    Answer: When x+4 and x-13 have opposite signs.

    This happens when:
    x < -4 AND x > 13 (this is not possible)
    OR
    x > -4 AND x < 13. \text{   } (-4,13)

    Conclusion 2:
    So we conclude that the denominator is negative when x is in (-4,13).

    Finally, we require that the numerator is positive AND the denominator is negative meaning both the first and second conclusions are met. This happens only when x is in both intervals (their intersection) yielding as you pointed out (-4,0) \cup (8,13). Now check when \frac{x(x-8)}{(x+4)(x-13)} = 0 and also check the other case: x(x-8) < 0 AND (x+4)(x-13) > 0 .

    Whenever you have an OR you are talking union:

    x is in A OR B \iff x is in A \cup B.

    Whenever you have an AND you are talking intersection:

    x is in A AND B \iff x is in A \cap B.
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