Solve the inequality x^3 < x

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Homework Statement



Solve for x

Homework Equations



x^3 < x


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?

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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I would start differently... I would divide both sides by x to get x^2 < 1. Try that :p
 
  • #3
PeterO
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Homework Statement



Solve for x

Homework Equations



x^3 < x


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?

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


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!!!!!
 
  • #4
PeterO
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I would start differently... I would divide both sides by x to get x^2 < 1. Try that :p

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!!
 
  • #5
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I would start differently... I would divide both sides by x to get x^2 < 1. Try that :p

But we don't know if x is negative or not so the sign may or may not flip

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!!!!!

I got
(x-1) < 0
(-∞, 1)

x < 0
(-∞, 0)

(x+1) < 0
(-∞, -1)
 
  • #6
PeterO
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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)

all correct.

Now if the third factor is negative - so the range (-∞, -1) , are the other factors positive or negative?
 
  • #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.
 
  • #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)
 
  • #9
HallsofIvy
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That's also works on non-polynomial inequalities while the method I used does not.
 
  • #10
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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.

I've seen this method before but I didn't understand how you'd know to divide it into 4 intervals...just logic?
 
  • #11
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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.
 
  • #12
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^Ah, that makes sense, but I'm missing the part where you treat the inequality sign as an equal sign
 
  • #13
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^Ah, that makes sense, but I'm missing the part where you treat the inequality sign as an equal sign
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.
 
  • #14
PeterO
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^Ah, that makes sense, but I'm missing the part where you treat the inequality sign as an equal sign

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.
 
  • #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.
 

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  • #16
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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.

That makes sense

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.

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

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.

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
 
  • #17
HallsofIvy
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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.
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.


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

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