Solving an inequality involving absolute values

[brotherbobby]

Homework Statement

Solve $\boldsymbol{\left|\dfrac{x-3}{x+1}\right|\le 1}$

Relevant Equations

Given $|f(x)|\le a\Rightarrow -a\le f(x)\le +a$

Problem Statement : I copy and paste the problem as it appeared in the text to the right.

Attempt (mine) : I copy and paste my attempt using Autodesk Sketchbook$^{\circledR}$ below. I hope the writing is legible.

My answer : I have three answers and confused as to which of them hold.
Crucially, when we say that $|f(x)|\le a\Rightarrow -a\le f(x)\le a$, do we mean that $f(x) \ge -a\; \textbf{(and?)}\; f(x)\le a$ or do we mean that $f(x) \ge -a\; \textbf{(or?)}\; f(x)\le a$?.

I suspect it is the answer OR. (The text disagrees, as you will see in the text solutions soon.)

My three solutions : $x>-1\; \textbf{OR}\; x\ge 1\; \textbf{OR}\; x<-1$

These solutions overlap (giving the condition OR) to yield the solution that $\boxed{x\in (-\infty,+\infty)}-\{-1\}$. In other words, $x$ can be any real number but $-1$.

Check (my answer) : I am aware that my answer is wrong. Given $\boldsymbol{\left|\frac{x-3}{x+1}\right|\le 1}$. To take one number from my solution, $x = -5$. But clearly, $\left|\frac{-5-3}{-5+1}\right|=2\nleq 1$!

Text's Solution : I copy and paste the solution in the text. I must admit that, despite my failure to understand why it took the crucial condition above to be AND, the answer matches the inequality given in the problem upon taking a few representative examples.

A hint or a suggestion would be welcome.

 

[Steve4Physics]

A hint or a suggestion would be welcome.

I’m not sure I fully understand the difficulty but see if this helps.

Consider the number line:
… -3 -2 -1 0 +1 +2 +3 …

When we say (for example) $|y| \le 1$, this is equivalent to saying that y is between -1 and +1 (inclusive) on the number line.

This means two conditions have to be met at the same time:
$y \ge -1$ AND $y \le 1$

If meeting only one condition were sufficient, then OR would be used. But that’s not the case here.

 

[brotherbobby]

If meeting only one condition were sufficient, then OR would be used. But that’s not the case here.

I did not understand this. $|y|\le 1$ implies both conditions : $y\ge -1$ and $y\le 1$. Where is the scope for only one of these conditions to apply?

 

[Steve4Physics]

I did not understand this. $|y|\le 1$ implies both conditions : $y\ge -1$ and $y\le 1$. Where is the scope for only one of these conditions to apply?

We are looking for all allowed values of $y$ which make $|y| \le 1$ true.

Condition 1: $y \ge -1$
Condition 2: $y \le +1$

Condition 1 by itself allows values such as y = +100 (for example).
Condition 2 by itself allows values such as y = -100 (for example).

There is no possibility for only one condition, by itself, to always be true and still satisfy $|y| \le 1$.

Both conditions must simultaneously be met in order to satisfy $|y| \le 1$. Which is why we must ’AND’ the two conditions.

Sorry, I can’t think of another way to explain it.

EDIT. Some signs and tyop's corrected.

 

[brotherbobby]

Condition 1: y≥−1

Condition 2: y≤+1

Condition 1 by itself allows values such as y = -100 (for example).
Condition 2 by itself allows values such as y = +100 (for example).

I suppose you meant that Condition 2 by itself allows y = -100 but Condition 1 will forbid it. Likewise, condition 1 by itself will allow y = 100 but condition 2 will forbid.

Thank you.

May I ask you what would you say if it was given that $|f(x)|\ge 1$? Surely, there are two conditions here too : (1) $f(x)\ge 1$ and $f(x)\le -1$. Is the word and correct here? Should it be replaced by or?

 

[PeroK]

May I ask you what would you say if it was given that $|f(x)|\ge 1$? Surely, there are two conditions here too : (1) $f(x)\ge 1$ and $f(x)\le -1$. Is the word and correct here? Should it be replaced by or?

Note that $|x| \ge 1$ implies $x \le -1$ or $x \ge +1$

Whereas $|x| \le 1$ implies $-1 \le x \le 1$, which means $-1 \le x$ and $x \le 1$.

 

[brotherbobby]

Note that $|x| \ge 1$ implies $x \le -1$ or $x \ge +1$

Whereas $|x| \le 1$ implies $-1 \le x \le 1$, which means $-1 \le x$ and $x \le 1$.

Thank you @PeroK. I hope you are not dismissive of my query above in post# 5 on this matter, though it might appear to be trivial. I think it is an important point, recognition of which would save the student several heartaches when he proceeds to solve problems. Not to mention that I have not seen a textbook where the point is made. The text I am using has only made use of it in its examples, without once mentioning it.

To recapitulate :

1. If $|f(x)|\le a$, we have the finction $f(x)\le a$ and $f(x)\ge -a$. [This may be summed up by saying that $-a\le f(x)\le a$, long as one understands that both conditions must be obeyed simultaneously].

2. If $|f(x)|\ge a$, the function $f(x)\ge a$ or $f(x)\le -a$.

 

[Steve4Physics]

May I ask you what would you say if it was given that $|f(x)|\ge 1$? Surely, there are two conditions here too : (1) $f(x)\ge 1$ and $f(x)\le -1$. Is the word and correct here? Should it be replaced by or?

Yes. In that case 'and' is wrong and 'or' is needed.

I’ll use y rather than f(x), for neatness/clarity.

So now we are looking for all allowed values of $y$ such that $|y| \ge 1$.

Our new conditions are:
$y \le -1$ This is (-∞, -1] on the number line.
$y \ge +1$ This is [+1,∞) on the number line.

Note the two conditions correspond to two non-overlapping sections of the number line. y could be on either section.

So in this case, we require either (not both) of the conditions to apply. We now require $y \le -1$ OR $y \ge +1$.

Compare this to the original question with $|y| \le 1$ [Edit: typo'corrected] . This had two overlapping sections of the number line and $y$ had to be on the single region of overlap. That’s where the AND came from.

 

[brotherbobby]

Compare this to the original question with |y|≥1.

You mean $|y|\le 1$.

 

[Steve4Physics]

You mean $|y|\le 1$.

Yes. I just corrected it but you beat me to it!

 

[pasmith]

Simpler approach: $$\begin{split} |x - 3| &\leq |x + 1| \\ (x - 3)^2 &\leq (x+1)^2 \\ x^2 - 6x + 9 &\leq x^2 + 2x + 1 \\ 8 &\leq 8x \\ 1 &\leq x.\end{split}$$

 

[WWGD]

Maybe you can consider a more geometric approach: |x-1| can be seen as the distance from the point x to 1. Same for |x+3|.
Now we look for the set of points whose distance to 3 are less than the distance to -1.