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Stupid question for calculus

  1. Sep 2, 2004 #1
    we just started our calc class, and our teacher gave us this question
    im sure it is really easy, and i dont know why i just cant get it analytically
    the question is solve

    1
    - < 3 solve for positive and negative case.
    x

    the positive value is easy, X> 1/3 . How would you solve it for negative vaulues analytically?
     
  2. jcsd
  3. Sep 2, 2004 #2

    HallsofIvy

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    Multiplying an inequality by a positive number does not change the direction of the inequality. That's why you can say: For x positive, if 1/x< 3 then 1< 3x and 1/3< x.
    (I first multiplied by x and then by 1/3, both positive numbers.)

    Multiplying an inequality by a negative number reverses the direction of the inequality: For x negative, if 1/x< 3, then 1> 3x (multiply by the negative x and reverse the inequality. Now multiply by the positive number 1/3 and get 1/3> x.
    BUT this was assuming x< 0. Since any x< 0 is necessarily less than 1/3, the solution set is: All x> 1/3 and all x< 0.
     
  4. Sep 2, 2004 #3
    hmm that clears up some stuff i guess but im not quite sure yet

    so the problem is 1/x<3. multiply -1 to both sides, and you get -1/x > -3 (signs are switched) then to solve for X... -1>-3x then divide by -3 (signs are switched?) so x>1/3.... i dont get how to get it algebraically. am i just supposed to accept that when u multiply by a negative number, u just swtich the signs and solve?
     
  5. Sep 2, 2004 #4
    I think you're making it too complicated. Look at it like this:

    Given: 1/x < 3, with x < 0.

    Now, multiply both sides by x, remembering that when you multiply an inequality by a negative quantity, you change the sense of the inequality. So, you have:

    1 > 3x

    Now, divide by the positive 3 and you get: (you don't have to change the sense of the inequality now remember)

    1/3 > x

    You just need to keep track of whether you are multiplying/dividing by a negative or positive entity. Other than that, just solve like a regular algebraic equation.
     
  6. Sep 3, 2004 #5

    HallsofIvy

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    No, I didn't say anything about multiplying by -1. I said multiply by x, in order to get it out of the denominator. To do that you have to consider whether x is positive or negative, just like your teacher told you.

    And, no, you are not "just supposed to accept that when u multiply by a negative number, u just swtich the signs and solve?". You are supposed to have learned that when you learned arithmetic:

    3< 5. How do -3 and -5 compare?
     
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