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Math Help - [SOLVED] Inequalities

  1. #1
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    [SOLVED] Inequalities

    This is a little embarassing, but it seems like I do not understand why the answers are what they are for these inequalities.

    1.  3x^2 - 2x-1 \geq 0

    First, I factor the left-hand side.

     \therefore (3x+1)(x-1) \geq 0

    So,  x - 1 \geq 0 if we divide both sides by '3x+1'.

    Thus,  x \geq 1

    Here's the part I don't understand. Naturally, I think that if we divide 'x-1' from both sides. We would get

     3x+1 \geq 0

    which yields

     3x \geq -1 \therefore x \geq -\frac{1}{3} (*)

    But this can not be right. By counterexample, let x = 0; this inequality is false.

    It should be  x \leq -\frac{1}{3} .

    I believe the reason for this is since for (x-1), if x is a negative number, then (x-1) < 0. So in order for the inequality to be equal then '3x+1' must also be negative (since a negative times a negative equals a positive) which explains why it should be

     3x + 1 \leq 0 and not  3x + 1 \geq 0

    Hence we solve, we should get  x \leq 1/3 . What am I doing wrong at (*)?

    =============================
    2. Another confusing inequality.

     \frac{1}{3-x} < -2

    I multiply both sides by '3-x'
     1 < -2(3-x)
     1 < -6+2x
     7 < 2x
     \frac{7}{2} < x (*)

    This answer makes no sense. If x = 4, then the inequality is false.

    Then, I tried a different way. I add both sides by 2

     \frac{1}{3-x} + 2 < 0
     \frac{2(3-x)+1}{3-x} < 0
     \frac{6-2x+1}{3-x} < 0
     7 - 2x < 0

    So,  x < \frac{7}{2} . Since, x can not be 3, or else the denominator is zero. I conclude that  3 < x < \frac{7}{2}

    What am I doing wrong at (*)?

    Thank you.
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  2. #2
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    The problem is when you divide both sides by a variable factor! you are losing information about the ineqality.

    After you factor you get

    (3x+1)(x-1) \ge 0 from here take your two zero's (this is where the inequality can change signs)

    so we get x=-\frac{1}{3} \mbox{ or } x=1

    Now we have three different parts of the number line

    \left(-\infty,-\frac{1}{3} \right)

    \left(-\frac{1}{3},1 \right) and

    (1,\infty)

    Now we check a point from each interval to see if it is a solution

    \left(-\infty,-\frac{1}{3} \right) check x=-1

    In the original and we get (3(-1)+1)(-1-1)=4 \ge 0 so the above set is a solution.

    Now we check \left(-\frac{1}{3},1 \right) with x=0 and we get (3(0)+1)(0-1)=-1 \le 0 so it is false and is not a solution.

    finally we check (1,\infty) with x=2 (3(2)+1)(2-1)=7 \ge 0 so this is a solution.

    So the solution to the inequality is

    \left(-\infty,-\frac{1}{3} \right]  \cup [1,\infty)

    For the rational ineqality get a common denomiantor and get the zero's from both the numerator and denomiator and do the same thing.

    Good luck.
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  3. #3
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    Thank you TheEmptySet. I was reviewing through my Calculus book and did not know/forgotten how to approach these problems. It was driving me crazy for a while. Your explanation makes more sense. So, I just choose arbitrary numbers within these intervals and check whether the inequality holds... ok. Thanks again. Cheers.
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