MHB Absolute value within absolute value

AI Thread Summary
The discussion revolves around solving the inequality |(2|x-3|-5|x+4|)| > 4, with participants analyzing different methods to determine the solution intervals. They identify critical points where sign changes occur, specifically at x = -4 and x = 3, and discuss how to combine intervals derived from these points. There is a debate about whether to use strict or weak inequalities when defining the intervals, with clarification that the function is continuous, allowing for certain combinations. Ultimately, the participants agree on the correct intervals and the reasoning behind merging them. The conversation highlights the importance of understanding absolute value inequalities and the nuances in defining intervals.
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I did this question two ways, each of which yielded a different answer. I'll post method one first.
$$\left| (2|x-3|-5|x+4|) \right|$$

Sign changes occur when $x<-4$, $-4<x<3$, and $x>3$

If $x<-4$, then $|3x+26|>4$, and so $x>-22/3$ and $x<-10$. Clearly since $x$ much be less than $-4$, only $x<-10$ is a solution to this case.

If $-4<x<3$, then $|x+2|>(4/7)$, and so $x>(-10/7)$ and $x<(-18/7)$, clearly we have the solutions $-4<x<(-18/7)$ and $(-10/7)<x<3$.

Before I continue with the final case, is my logic right?
 
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Were you given an equation?
 
Sorry, left out one part.

$\left| (2|x-3|-5|x+4|) \right|>4$
 
Rido12 said:
I did this question two ways, each of which yielded a different answer. I'll post method one first.
$$\left| (2|x-3|-5|x+4|) \right|$$

Sign changes occur when $x<-4$, $-4<x<3$, and $x>3$

If $x<-4$, then $|3x+26|>4$, and so $x>-22/3$ and $x<-10$. Clearly since $x$ much be less than $-4$, only $x<-10$ is a solution to this case.

If $-4<x<3$, then $|x+2|>(4/7)$, and so $x>(-10/7)$ and $x<(-18/7)$, clearly we have the solutions $-4<x<(-18/7)$ and $(-10/7)<x<3$.

Before I continue with the final case, is my logic right?

You're partially right, note that in the intervals $\left(-\dfrac{22}{3},\,-4 \right]$ and $\left[-4,\,-\dfrac{18}{7} \right)$, you have to combine these so that the second interval that agrees to the given inequality would be $\left(-\dfrac{22}{3},\,-\dfrac{18}{7}\right)$.

Your reasoning for the rest is correct though, good job, Rido12!
 
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Hi Anemone! (Wave)

Right...I left out $\left(-\dfrac{22}{3},\,-4 \right)$.
But why are we allowed to combine the intervals? If both intervals never reach $-4$, how come we can fill in the "hole"? (Wondering)
 
Rido12 said:
Hi Anemone! (Wave)

Right...I left out $\left(-\dfrac{22}{3},\,-4 \right)$.
But why are we allowed to combine the intervals? If both intervals never reach $-4$, how come we can fill in the "hole"? (Wondering)

We need to combine the intervals, because the function is continuous on the entire domain.

Perhaps you want draw a real number line so that you can see the whole situation more clearly? :)
 
I have two open circles on my number line at $x=-4$? (Wondering)
 
Rido12 said:
Sorry, left out one part.

$\left| (2|x-3|-5|x+4|) \right|>4$

$\displaystyle \begin{align*} \left| 2 \left| x - 3 \right| - 5 \left| x + 4 \right| \right| &> 4 \\ \left( 2 \left| x - 3 \right| - 5 \left| x + 4 \right| \right) ^2 &> 4^2 \\ \left( 2 \left| x- 3 \right| \right) ^2 - 20 \left| x - 3 \right| \left| x + 4 \right| + \left( -5 \left| x + 4 \right| \right) ^2 &> 16 \\ 4 \left(x - 3 \right) ^2 - 20 \left| x - 3 \right| \left| x + 4 \right| + 25 \left( x + 4 \right) ^2 &> 16 \\ 4 \left( x - 3 \right) ^2 + 25 \left( x + 4 \right) ^2 - 16 &> 20 \left| x - 3 \right| \left| x + 4 \right| \\ 4\left( x^2 - 6x + 9 \right) + 25 \left( x^2 + 8x + 16 \right) - 16 &> 20 \left| x - 3 \right| \left| x + 4 \right| \\ 4x^2 - 24x + 36 + 25x^2 + 200x + 400 - 16 &> 20 \left| x - 3 \right| \left| x + 4 \right| \\ 29x^2 + 176x + 420 &> 20 \left| x - 3 \right| \left| x + 4 \right| \\ \left( 29x^2 + 176x + 420 \right) ^2 &> \left( 20 \left| x - 3 \right| \left| x + 4 \right| \right) ^2 \\ 841x^4 + 10\,208x^3 + 55\,336x^2 + 147\,840x + 176\,400 &> 400 \left( x - 3 \right) ^2 \left( x + 4 \right) ^2 \\ 841x^4 + 10\,208x^3 + 55\,336x^2 + 147\,840x + 176\,400 &> 400x^4 + 800x^3 - 9200 x^2 - 9600x + 57\,600 \\ 441x^4 + 9408x^3 + 64\,536x^2 + 157\,440x + 118\,800 &> 0 \\ 3 \left( x + 10 \right) \left( 3x + 22 \right) \left( 7x + 10 \right) \left( 7x + 18 \right) &> 0 \end{align*}$

Now solving the factored equation will tell you the x intercepts. Substituting points in between the x intercepts will tell you where the function is positive and negative.
 
Rido12 said:
...

Sign changes occur when $x<-4$, $-4<x<3$, and $x>3$...

Hi Rido12 again,

I'm sorry, I re-read your first post and noticed I haven't pointed out that sign changes occur when we have the weak inequalities over the domain as you have cited, i.e. $x\le -4$, $-4\le x \le3$, and $x\ge 3$, note that when $x=-4$, $\left| 3x+26 \right|=\left| 3(-4)+26 \right|=11>4$.
 
  • #10
So to edit my statements:Sign changes occur when $x\le -4$, $-4\le x\le 3$, and $x\le 3$

If $x\le -4$, then $|3x+26|>4$, and so $x>-22/3$ and $x<-10$. Clearly since $x$ much be $\le -4$, so $x<-10$ and $(-22/3)<x\le 4$

If $-4\le x\le 3$, then $|x+2|>(4/7)$, and so $x>(-10/7)$ and $x<(-18/7)$, clearly we have the solutions $-4\le x<(-18/7)$ and $(-10/7)<x\le 3$.

Now, since both $(-22/3)<x\le 4$ and $-4\le x<(-18/7)$ have a strong inequality at $-4$, we can amalgamate them?

@Prove it, I have tried that method as a third method, but it's quite tedious as a quiz question. But nevertheless, a quite solid method when you don't want to think...! :D
 
  • #11
Rido12 said:
So to edit my statements:Sign changes occur when $x\le -4$, $-4\le x\le 3$, and $x\le 3$

If $x\le -4$, then $|3x+26|>4$, and so $x>-22/3$ and $x<-10$. Clearly since $x$ much be $\le -4$, so $x<-10$ and $(-22/3)<x\le 4$

If $-4\le x\le 3$, then $|x+2|>(4/7)$, and so $x>(-10/7)$ and $x<(-18/7)$, clearly we have the solutions $-4\le x<(-18/7)$ and $(-10/7)<x\le 3$.

Now, since both $(-22/3)<x\le 4$ and $-4\le x<(-18/7)$ have a strong inequality at $-4$, we can amalgamate them?

Correct! Since both sets of interval are defined when $x=-4$, we can then merge them that yields $\left(-\dfrac{22}{3},\,-\dfrac{18}{7}\right)$.
 
  • #12
Thanks everyone! Here is my solution: :D

Case 1: ($x\le -4$)

$x<-10$, $-\frac{22}{3}<x\le -4$

Case 2: ($-4 \le x \le 3$)
$-4\le x < -\frac{18}{7}$, $-\frac{10}{7}<x\le 3$

Case 3: ($x\ge 3$)

$x<-10$ => untrue statement, inadmissible
$x>-\frac{22}{3}$ => also not true

Therefore:
$(-\infty,-10) \cup (-\frac{22}{3},-\frac{18}{7}) \cup(-\frac{10}{7}, \infty)$

Correct :D?
 
  • #13
Rido12 said:
...

Therefore:
$(-\infty,-10) \cup (-\frac{22}{3},-\frac{18}{7}) \cup(-\frac{10}{7}, \infty)$

Correct :D?

Yeap! You got them all correct! Well done!(Yes)
 
  • #14
My prof deals with inequalities differently. Where you said :$x\le -4$, $-4\le x\le 3$, and $x\le 3$, he does it differently.

For example, case 2 is when we have $-4\le x\le 3$, where each are separate by a strict inequality. My prof likes to go back to the definition of the absolute value, namely, when $|x-a|>b$, we have when $x\ge a$ or when $x<a$. (Notice, he doesn't say $x\le a$. So in the context of our problem, he would say $x-3<0$ and $x+4\ge 0$. I have worked it out his way, and while you merged $(-22/3, -4]$ with $[-4, -18/7)$, he would have $(-22/3, -4)$ and $[-4, -18/7)$. So the strict inequality is missing from one of the -4's. Will these methods always be the same?

And we are allowed to say "$x\le -4$, $-4\le x\le 3$, and $x\le 3$" because we're not redefining the "function" because clearly those intervals would fail the vertical line test...we are only separating the critical numbers?

TL;DR:

If we are given an inequality, say $|x-3|>b$
Does it matter whether we separate the domain for when $x-3\ge 0$ and when $x-3\le 0$ versus when we separate the domain as $x-3 \ge 0$ and $x-3 <0$?
 
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  • #15
Rido12 said:
My prof deals with inequalities differently. Where you said :$x\le -4$, $-4\le x\le 3$, and $x\le 3$, he does it differently, in the context of our problem, he would say $x-3<0$ and $x+4\ge 0$.

Hi Rido12,

Ops...I'm sorry...your prof is absolutely right, I saw that my previous reply(#9) has some error in it, I should have explained it in the following manner for you, so that I would end up making error in defining the weak/strict inequality!:(It's like when we have $f(x)=2\left| x-3 \right|$, we can clear the absolute value sign by rewriting it piece-wisely:

$f(x)=\begin{cases}2(x-3), & x\ge 3, & \\[3pt] \\[3pt] -2(x-3), & x<3 \\ \end{cases}$

Also, when we have $g(x)=5\left| x+4 \right|$, it can be redefined as:

$g(x)=\begin{cases}5(x+4), & x\ge -4, & \\[3pt] \\[3pt] -5(x+4), & x<-4 \\ \end{cases}$

Let's say for now, our purpose is to study the difference of these two functions in the interval $x<3$ AND $x<-4$, which the resulting combined interval would be $x<-4$ and not $x\le -4$, like what I previously have told you...sorry!
 
  • #16
Alright, thanks Anemone!

That makes sense :D But the answer is still the same, right?
Interval $(a,b)$ with $[b, c)$ merges to $(a,c)$? Since while one has an "open circle", the other closes that "circle"?

So would $x\le -4$, $-4\le x \le3$, and $x\ge 3$ be rewritten like: $x< -4$, $-4\le x <3$, and $x\ge 3$?
 
  • #17
Rido12 said:
Alright, thanks Anemone!

You're welcome, Rido12!

I'm so glad that you gave me chance to fix things right when I bungled the interval of the inequality. :o

Rido12 said:
That makes sense :D But the answer is still the same, right?
Interval $(a,b)$ with $[b, c)$ merges to $(a,c)$? Since while one has an "open circle", the other closes that "circle"?

So would $x\le -4$, $-4\le x \le3$, and $x\ge 3$ be rewritten like: $x< -4$, $-4\le x <3$, and $x\ge 3$?

Yes... yes and yes! :)
 
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