Absolute value within absolute value

[Dethrone]

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?

 

[MarkFL]

Were you given an equation?

 

[Dethrone]

Sorry, left out one part.

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

 

[anemone]

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!

 

[Dethrone]

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)

 

[anemone]

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? :)

 

[Dethrone]

I have two open circles on my number line at $x=-4$? (Wondering)

 

[Prove It]

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.

 

[anemone]

...

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 $x=-4$, $\left| 3x+26 \right|=\left| 3(-4)+26 \right|=11>4$.

 

[Dethrone]

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

 

[anemone]

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 $\left(-\dfrac{22}{3},\,-\dfrac{18}{7}\right)$.

 

[Dethrone]

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?

 

[anemone]

...

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)

 

[Dethrone]

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 $x-3 <0$?

 

[anemone]

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!

 

[Dethrone]

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 $x\ge 3$?

 

[anemone]

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.

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! :)