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Solve the inequality x^3 < x

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve for x

    2. Relevant equations

    x^3 < x


    3. The attempt at a solution


    x^3 < x
    x^3 - x < 0
    x(x^2 - 1) < 0
    x(x+1)(x-1) < 0

    For the expression on the left to be less than zero, it has to be two positives + negative or three negatives right? I've tried setting the different terms less than x, and greater than x but there's so many combinations and I get intervals with infinities which I can't make sense of.

    What is a systematic/formal way of finding the solution set (and checking that it's right) without having to graph or doing it by trial and error?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 17, 2011 #2
    I would start differently... I would divide both sides by x to get x^2 < 1. Try that :p
     
  4. Aug 17, 2011 #3

    PeterO

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    Look at the three factors, (x-1) , x , and (x+1)

    What value(s) of x will make (x-1) negative
    What value(s) of x will make x negative
    What value(s) of x will make (x+1) negative

    From that you can establish value(s) of x when all three are negative, or when just one on the factors is negative. That is what you said you wanted.

    Of course the correct way to solve this problem is to draw a sketch graph and look at it!!!!!
     
  5. Aug 17, 2011 #4

    PeterO

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    CAUTION!!! you don't know what to do with the inequality sign !!

    if x is negative, so you are dividing by a negative, the < sign becomes >
    if x is positive it is OK
    if x = 0 you can't divide by x anyway!!
     
  6. Aug 17, 2011 #5
    But we don't know if x is negative or not so the sign may or may not flip

    I got
    (x-1) < 0
    (-∞, 1)

    x < 0
    (-∞, 0)

    (x+1) < 0
    (-∞, -1)
     
  7. Aug 17, 2011 #6

    PeterO

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    all correct.

    Now if the third factor is negative - so the range (-∞, -1) , are the other factors positive or negative?
     
  8. Aug 18, 2011 #7

    HallsofIvy

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    The simplest way to do this is to note that x- a is negative as long as x< a and positive for x> a. Also, the three numbers, -1, 0, and 1, divide the real numbers into four intervals.
    If x< -1, then all three of x+ 1= x- (-1), x= x- 0, and x- 1 are negative.

    If -1< x< 0, then x+ 1 is positive while x and x-1 are still negative.

    If 0< x< 1, x+ 1 and x are positive while x- 1 is still negative.

    If 1< x, all three are positive.

    Notice that as we "pass" one of those values exactly one factor changes sign.
     
  9. Aug 18, 2011 #8

    gb7nash

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    This is the way I usually solve these:

    Trying to solve x^3 < x is equivalent to asking where the polynomial x^3 - x is negative. To do this, set x^3 - x = 0, solve for the zeros.

    Once you have the zeros, test a point in each interval and from this, you'll obtain the answer (this is similar to HallsofIvy's approach)
     
  10. Aug 18, 2011 #9

    HallsofIvy

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    That's also works on non-polynomial inequalities while the method I used does not.
     
  11. Aug 24, 2011 #10
    I've seen this method before but I didn't understand how you'd know to divide it into 4 intervals...just logic?
     
  12. Aug 24, 2011 #11

    Mark44

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    Assuming you're talking about the original problem, which is the inequality x3 < x, the reason for the four intervals is that the equation x3 - x = 0 has three roots: x = -1, x = 0, and x = 1. These three roots divide the number line into four regions. Since the only places where x3 - x can equal zero are at the three roots, the expression x3 - x must be either positive or negative on each of the four regions. So it suffices to pick a value in each of the four regions. If x3 - x is positive at that value, it has to be positive at each other point in that region. The expression x3 - x can't change sign in any of the four regions, because to do so, it must be zero at some point in the region. We have already established that the only places where x3 - x is zero are at x = -1, x = 0, and x = 1.
     
  13. Aug 26, 2011 #12
    ^Ah, that makes sense, but I'm missing the part where you treat the inequality sign as an equal sign
     
  14. Aug 26, 2011 #13

    Mark44

    Staff: Mentor

    You look at the equation formed from the inequality so that you can find the roots (zeroes) of the polynomial. Since these points are the only values where the polynomial is zero, and are the only places where it could possibly change sign, then at each interval between two roots the polynomial must be either positive or negative.
     
  15. Aug 26, 2011 #14

    PeterO

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    The question was an inequality with a < sign.

    The best way to fully understand the function is to also have a look at what happens when the sign is > , and what happens when the sign is =

    You don't want either of them in your answer, but knowing exactly where those solutions are makes it easier to avoid them.
     
  16. Aug 26, 2011 #15

    vela

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    I've attached a picture to illustrate what Mark explained above. From the plot of f(x)=x^3-x, you can see that for all points in intervals II and IV, the graph is above the axis, and for all points in I and III, the graph is below the axis. To get from one region to another (at the red dots), the function has to cross the axis, so the endpoints of the intervals correspond to where x^3 - x = 0.
     

    Attached Files:

    Last edited: Aug 26, 2011
  17. Aug 28, 2011 #16
    That makes sense

    Thanks, the graph does make things all the working here clearer

    Hmm...I think I recall being told to solve inequalities with equal signs and then putting them back in once everything was....dunno, ha.


    So after going back to HallsOfIvy's post, I got (-∞, -1) and (0, 1) as my answer, which agrees with Grapher...I assume that's all good then, right?

    Many thanks to everyone who contributed to this thread...this is the kind of stuff I can't find in textbooks
     
  18. Aug 28, 2011 #17

    HallsofIvy

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    The point is that a continuous (and polynomials are continuous) can't "jump" over 0. The only places a polynomial can change from "< 0" to "> 0" and vice-versa, is where it is equal to 0.


    Hey, if you agree with me, you must be right!:approve:
     
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