Certainly! Let's solve the inequality [tex]\((|| x-1|-2|-3)(|x-2|-3) \geq 0\)[/tex] step-by-step.
### Step 1: Understand the individual expressions
Let's start by examining each part of the inequality separately:
1. Expression A: [tex]\(||x-1|-2|-3\)[/tex]
2. Expression B: [tex]\(|x-2|-3\)[/tex]
### Step 2: Analyze Expression A
Consider [tex]\(||x-1|-2|-3\)[/tex]:
- Let [tex]\(y = |x - 1|\)[/tex], thus the expression becomes [tex]\(|y - 2| - 3\)[/tex].
- Notice [tex]\(|y - 2|\)[/tex] can be split into two cases:
- When [tex]\(y \geq 2\)[/tex], [tex]\(|y - 2| = y - 2\)[/tex].
- When [tex]\(y < 2\)[/tex], [tex]\(|y - 2| = 2 - y\)[/tex].
For the outer absolute value (with [tex]\(-3\)[/tex]):
- When [tex]\(y - 2 \geq 3\)[/tex], [tex]\(y - 2 - 3 \geq 0 \Rightarrow y \geq 5\)[/tex].
- When [tex]\(y - 2 < 3\)[/tex], splitting further into:
- When [tex]\(2 - y \geq 3\)[/tex], [tex]\(2 - y - 3 \geq 0 \Rightarrow y \leq -1\)[/tex].
- When [tex]\(2 - y < 3\)[/tex], [tex]\(|2 - y - 3| = -(y - 2 - 3) = -y + 5 - 3 = -y + 5\)[/tex].
Thus, the entire expression [tex]\(||x-1|-2|-3\)[/tex] simplifies to three intervals based on the value of [tex]\(y\)[/tex]:
- If [tex]\(y \leq -1\)[/tex], [tex]\(|y - 2| - 3\)[/tex].
- [tex]\( -1 < y < 2\)[/tex], [tex]\(|-y + 5 - 3\)[/tex].
- [tex]\( y \geq 5\)[/tex], [tex]\( y - 2 - 3 \geq y - 5 \)[/tex].
### Step 3: Analyze Expression B
Consider [tex]\(|x - 2| - 3\)[/tex]:
- When [tex]\(x - 2 \geq 3\)[/tex], [tex]\(|x - 2| = x - 2\)[/tex], then [tex]\(x \geq 5\)[/tex].
- When [tex]\(x - 2 < 3\)[/tex], splitting into two cases as well:
- When [tex]\(x - 2 \leq -3\)[/tex], [tex]\(x \leq -1\)[/tex].
- When [tex]\(-3 < x - 2 < 3\)[/tex], reducing directly:
This results in:
- For [tex]\(x \geq 5\)[/tex], [tex]\(x - 2 - 3 \geq 0\)[/tex]
- For [tex]\(x \leq -1\)[/tex], [tex]\(|x-2|-3 = -(x-2)-3\)[/tex]
- For [tex]\(-1 < x < 5)\, no solution.
### Step 4: Combine both expressions
We combine \(|x-2|-3 \geq 0\)[/tex] and check whether it's positive:
- When [tex]\(x \geq 5\)[/tex], [tex]\(|x-2| = x - 5 \)[/tex].
- When [tex]\( x \leq -1\)[/tex].
Multiplying and combining twice:
[tex]\((|| x-1|-2|-3)(|x-2)-3)\)[/tex]:
Result in intervals:
1. [tex]\(-1 \leq x < 5\)[/tex]
2. [tex]\(x \leq 6\)[/tex]
3. Solutions simplified between:
\((-oo, 11), 1<=x<=5, 5<= x<=6 )
Therefore, the final result is:
[tex]\[
((-1 \leq x \leq 5)) | (6 \leq x < \infty) | (-\infty < x \leq -4)
\][/tex]
