`|x^(2)-4x+4|ge 1`

Answer

To solve the inequality \( |x^2 - 4x + 4| \geq 1 \), we will follow these steps:

### Step 1: Rewrite the expression inside the modulus
The expression \( x^2 - 4x + 4 \) can be factored:
\[ x^2 - 4x + 4 = (x - 2)^2 \]
Thus, the inequality becomes:
\[ |(x - 2)^2| \geq 1 \]

### Step 2: Remove the modulus
Since \( (x - 2)^2 \) is always non-negative, we can drop the modulus:
\[ (x - 2)^2 \geq 1 \]

### Step 3: Solve the inequality
This inequality can be split into two cases:
1. \( (x - 2)^2 \geq 1 \)
2. \( (x - 2)^2 \leq -1 \) (This case is not possible since a square cannot be negative)

So we only need to solve:
\[ (x - 2)^2 \geq 1 \]

Taking the square root of both sides (considering both the positive and negative roots), we have:
\[ x - 2 \geq 1 \quad \text{or} \quad x - 2 \leq -1 \]

### Step 4: Solve each case
**Case 1:**
\[ x - 2 \geq 1 \]
Adding 2 to both sides:
\[ x \geq 3 \]

**Case 2:**
\[ x - 2 \leq -1 \]
Adding 2 to both sides:
\[ x \leq 1 \]

### Step 5: Combine the solutions
The solutions from both cases give us:
\[ x \leq 1 \quad \text{or} \quad x \geq 3 \]

### Step 6: Write the final answer in interval notation
The solution can be expressed in interval notation as:
\[ (-\infty, 1] \cup [3, \infty) \]

### Final Answer:
\[ x \in (-\infty, 1] \cup [3, \infty) \]

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