Absolute value fractional inequalities

[khavar]

I want to better understand the principals of absolute values. I have been reading about it on the internet, however, it is still not quite clear. The book I am working from is not clear enough for me either.

My confusion begins with proper method.

For example, |(2x-1)/(x-2)| ≥ 1.

Method 1 based on |x|² = x²:
|2x-1|≥|x-2|
(2x-1)² ≥ (x-2)²
4X²-4x+1 ≥ X²-4x+4
3x² ≥ 3
X² ≥ 1
x ≥ ±1, testing intervals reveals that x ≤ -1 and x ≥ 1
Something is missing, this is only one half of the answer.

Method 2:

|(2x-1)/(x-2)| ≥ 1
(2x-1)/(x-2) ≤ -1 and (2x-1)/(x-2) ≥ 1
((2x-1)/(x-2)) + 1 ≤ 0 and ((2x-1)/(x-2)) - 1 ≥ 0
((2x-1)/(x-2)) + ((x-2)/(x-2)) ≤ 0 and ((2x-1)/(x-2)) - ((x-2)/(x-2)) ≥ 0
(2x-1+x-2)/(x-2) ≤ 0 and (2x-1-x+2)/(x-2) ≥ 0
(3x-3)/(x-2) ≤ 0 and (x+1)/(x-2) ≥ 0
3x ≤ 3, x ≤ 2 and x ≥ -1, x ≥ 2
X cannot equal 2.
Testing intervals reveals:
x ≤ -1, x > 2, x < 2, x ≥ 1
x ≤ -1, 1 ≤ x < 2, x > 2

I can mechanically find the answer by following a process, however I don't fully understand the principals taking place within fractional inequalities as well as when combined with absolute values. I won't be able to memorize this lesson unless I understand it completely.

 

[khavar]

Method 1 based on |x|² = x²:
|2x-1|≥|x-2|
(2x-1)² ≥ (x-2)²
4X²-4x+1 ≥ X²-4x+4
3x² ≥ 3
X² ≥ 1
x ≥ ±1, testing intervals reveals that x ≤ -1 and x ≥ 1
Something is missing, this is only one half of the answer.

Perhaps one problem here is that I did not take into consideration that the denominator is not distinctly positive or negative, so I must account for both possibilities with a properly reflected equation.

 

[HallsofIvy]

I want to better understand the principals of absolute values. I have been reading about it on the internet, however, it is still not quite clear. The book I am working from is not clear enough for me either.

My confusion begins with proper method.

For example, |(2x-1)/(x-2)| ≥ 1.

Method 1 based on |x|² = x²:
|2x-1|≥|x-2|
(2x-1)² ≥ (x-2)²
4X²-4x+1 ≥ X²-4x+4
3x² ≥ 3
X² ≥ 1
x ≥ ±1, testing intervals reveals that x ≤ -1 and x ≥ 1
Something is missing, this is only one half of the answer.

Other than adding \(\displaystyle x\ne 2\), which you lost when you multiplied both sides by positive number |x- 2|, that is perfectly good. I don't know why you say "only one half of the answer".

Method 2:

|(2x-1)/(x-2)| ≥ 1
(2x-1)/(x-2) ≤ -1 and (2x-1)/(x-2) ≥ 1
((2x-1)/(x-2)) + 1 ≤ 0 and ((2x-1)/(x-2)) - 1 ≥ 0
((2x-1)/(x-2)) + ((x-2)/(x-2)) ≤ 0 and ((2x-1)/(x-2)) - ((x-2)/(x-2)) ≥ 0
(2x-1+x-2)/(x-2) ≤ 0 and (2x-1-x+2)/(x-2) ≥ 0
(3x-3)/(x-2) ≤ 0 and (x+1)/(x-2) ≥ 0
3x ≤ 3, x ≤ 2 and x ≥ -1, x ≥ 2
X cannot equal 2.
Testing intervals reveals:
x ≤ -1, x > 2, x < 2, x ≥ 1
x ≤ -1, 1 ≤ x < 2, x > 2

All of this reduces to x, \(\displaystyle x\le -1\), \(\displaystyle 1\le x< 2\), \(\displaystyle 2<x\) exactly what you had before.

I can mechanically find the answer by following a process, however I don't fully understand the principals taking place within fractional inequalities as well as when combined with absolute values. I won't be able to memorize this lesson unless I understand it completely.

I'm not sure what "principles" you are talking about. The most important "principle" for inequalities is that if you multiply or divide both sides by a negative number you change the direction of the inequality. You didn't multiply both sides by a negative number here, but you had to be careful about the inequalities.

Your "Method 2" seems unecessarily complicated to me. Instead of doing all of the inequalities you did, you can look at the corresponding equation. That is, start by solving \(\displaystyle \left|\frac{2x- 1}{x- 2}\right|= 1\). Now that is the same as |2x- 1|= |x- 2| which is the same as 2x- 1= x- 2 or 2x- 1= -(x- 2)= 2- x. The first equation is the same as x= -1 and the second is the same as x= 1. Now, the "principle" is that a function can change from "> 1" to "< 1" and vice-versa only where it is equal to 1 or where it is not continuous. We have just seen that this function is 0 at x= -1 and 1, and, of course, it is not continuous at x= 2.

So we look at the intervals x< -1, -1< x< 1, 1< x< 2, and x>2. x= -2 is in x< -1 and then \(\displaystyle \left|\frac{2x-1}{x-2}\right|\)\(\displaystyle = \left|\frac{-4-1}{-2-2}\right|\)\(\displaystyle = \frac{5}{4}> 1\) so all number less than -1 satisfy this inequality.

0 lies between -1 and 1 and \(\displaystyle \left|\frac{2x-1}{x- 2}\right|= \left|\frac{-1}{-2}\right|= \frac{1}{2}< 1\) so that all numbers between -1 and 1 do NOT satisfy the inequality.

3/2 lies between 1 and 2 and \(\displaystyle \left|\frac{2x-1}{x-2}\right|= \left|\frac{2}{-1/2}\right|= 4> 1\) so all numbers between 1 and 2 satisfy the inequality.

3 is larger than 2 and \(\displaystyle \left|\frac{2x-1}{x-2}\right|= \left|\frac{5}{1}\right|= 5> 1\) so every number larger than 3 safisfies the inequality. Putting those together gives the same result as before.

 

[JeffM]

I want to better understand the principals of absolute values. I have been reading about it on the internet, however, it is still not quite clear. The book I am working from is not clear enough for me either.

My confusion begins with proper method.

For example, |(2x-1)/(x-2)| ≥ 1.

Method 1 based on |x|² = x²:
|2x-1|≥|x-2| Fine
(2x-1)² ≥ (x-2)² Good
4X²-4x+1 ≥ X²-4x+4 Do not change between upper case and lower case for the same variable.
3x² ≥ 3 Going strong
X² ≥ 1 Super
x ≥ ±1, This is technically wrong: \(\displaystyle x^2 \ge 1 \implies x \ge 1\ or\ x \le - 1.\) It is false that \(\displaystyle x^2 \ge 1 \implies x \ge - 1.\)
testing intervals reveals that x ≤ -1 and x ≥ 1
Something is missing, this is only one half of the answer.

Method 2:

|(2x-1)/(x-2)| ≥ 1
(2x-1)/(x-2) ≤ -1 and (2x-1)/(x-2) ≥ 1 Or, not and
((2x-1)/(x-2)) + 1 ≤ 0 and ((2x-1)/(x-2)) - 1 ≥ 0
((2x-1)/(x-2)) + ((x-2)/(x-2)) ≤ 0 and ((2x-1)/(x-2)) - ((x-2)/(x-2)) ≥ 0
(2x-1+x-2)/(x-2) ≤ 0 and (2x-1-x+2)/(x-2) ≥ 0
(3x-3)/(x-2) ≤ 0 and (x+1)/(x-2) ≥ 0
3x ≤ 3, x ≤ 2 and x ≥ -1, x ≥ 2 This is wrong. You have assumed that x - 2 > 0, but it may be true that x - 2 < 0.
X cannot equal 2. This is certainly correct.
Testing intervals reveals:
x ≤ -1, x > 2, x < 2, x ≥ 1
x ≤ -1, 1 ≤ x < 2, x > 2

I can mechanically find the answer by following a process, however I don't fully understand the principals taking place within fractional inequalities as well as when combined with absolute values. I won't be able to memorize this lesson unless I understand it completely.

First, whatever method you use, \(\displaystyle x \ne 2.\)

Second, your method 1 has a subtlety implied in it.

\(\displaystyle a \le - \sqrt{b} \le 0 \implies a * a \ge a\left(-\sqrt{b}\right)\ and\ \left(-\sqrt{b}\right)a \ge \left(-\sqrt{b}\right)\left(- \sqrt{b}\right) \ge \left(\sqrt{b}\right) * 0 \implies\)

\(\displaystyle a * a \ge \left(- \sqrt{b}\right)\left(- \sqrt{b}\right) \ge 0 \implies a^2 \ge b \ge 0\ if a \le - \sqrt{b} \le 0.\)

\(\displaystyle And\ a \ge \sqrt{b} \ge 0 \implies a^2 \ge b \ge 0\ if\ a \ge \sqrt{b} > 0.\)

\(\displaystyle a^2 \ge b \ge 0 \implies a \le - \sqrt{b} \le 0\ OR\ a \ge \sqrt{b} \ge 0.\)

So the answer is \(\displaystyle \left|\dfrac{2x - 1}{x - 2}\right| \ge 1\ if\ x^2 \ge 1\ and\ x \ne 2 \implies x \le - 1\ or\ 1 \le x < 2\ or\ x > 2.\)

In method 2, you did not explore all the possibilities.

 

[khavar]

Other than adding \(\displaystyle x\ne 2\), which you lost when you multiplied both sides by positive number |x- 2|, that is perfectly good. I don't know why you say "only one half of the answer".


All of this reduces to x, \(\displaystyle x\le -1\), \(\displaystyle 1\le x< 2\), \(\displaystyle 2<x\) exactly what you had before.


I'm not sure what "principles" you are talking about. The most important "principle" for inequalities is that if you multiply or divide both sides by a negative number you change the direction of the inequality. You didn't multiply both sides by a negative number here, but you had to be careful about the inequalities.

Your "Method 2" seems unecessarily complicated to me. Instead of doing all of the inequalities you did, you can look at the corresponding equation. That is, start by solving \(\displaystyle \left|\frac{2x- 1}{x- 2}\right|= 1\). Now that is the same as |2x- 1|= |x- 2| which is the same as 2x- 1= x- 2 or 2x- 1= -(x- 2)= 2- x. The first equation is the same as x= -1 and the second is the same as x= 1. Now, the "principle" is that a function can change from "> 1" to "< 1" and vice-versa only where it is equal to 1 or where it is not continuous. We have just seen that this function is 0 at x= -1 and 1, and, of course, it is not continuous at x= 2.

So we look at the intervals x< -1, -1< x< 1, 1< x< 2, and x>2. x= -2 is in x< -1 and then \(\displaystyle \left|\frac{2x-1}{x-2}\right|\)\(\displaystyle = \left|\frac{-4-1}{-2-2}\right|\)\(\displaystyle = \frac{5}{4}> 1\) so all number less than -1 satisfy this inequality.

0 lies between -1 and 1 and \(\displaystyle \left|\frac{2x-1}{x- 2}\right|= \left|\frac{-1}{-2}\right|= \frac{1}{2}< 1\) so that all numbers between -1 and 1 do NOT satisfy the inequality.

3/2 lies between 1 and 2 and \(\displaystyle \left|\frac{2x-1}{x-2}\right|= \left|\frac{2}{-1/2}\right|= 4> 1\) so all numbers between 1 and 2 satisfy the inequality.

3 is larger than 2 and \(\displaystyle \left|\frac{2x-1}{x-2}\right|= \left|\frac{5}{1}\right|= 5> 1\) so every number larger than 3 safisfies the inequality. Putting those together gives the same result as before.

Thank you, HallsofIvy. "...or where it is not continuous." This part made me realize why I was thinking about it in such a complicated fashion. The answer was right there in front of me.

 

[khavar]

First, whatever method you use, \(\displaystyle x \ne 2.\)

Second, your method 1 has a subtlety implied in it.

\(\displaystyle a \le - \sqrt{b} \le 0 \implies a * a \ge a\left(-\sqrt{b}\right)\ and\ \left(-\sqrt{b}\right)a \ge \left(-\sqrt{b}\right)\left(- \sqrt{b}\right) \ge \left(\sqrt{b}\right) * 0 \implies\)

\(\displaystyle a * a \ge \left(- \sqrt{b}\right)\left(- \sqrt{b}\right) \ge 0 \implies a^2 \ge b \ge 0\ if a \le - \sqrt{b} \le 0.\)

\(\displaystyle And\ a \ge \sqrt{b} \ge 0 \implies a^2 \ge b \ge 0\ if\ a \ge \sqrt{b} > 0.\)

\(\displaystyle a^2 \ge b \ge 0 \implies a \le - \sqrt{b} \le 0\ OR\ a \ge \sqrt{b} \ge 0.\)

So the answer is \(\displaystyle \left|\dfrac{2x - 1}{x - 2}\right| \ge 1\ if\ x^2 \ge 1\ and\ x \ne 2 \implies x \le - 1\ or\ 1 \le x < 2\ or\ x > 2.\)

In method 2, you did not explore all the possibilities.

Thank you, JeffM. I appreciate the corrections; it's good. This morning I'm going to study your notations at the end of your post, I don't yet completely understand what they represent.