- #1

brotherbobby

- 699

- 163

- 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##

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

**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.**