The set of real values of `x` satisfying `log_(1/2) (x^2-6x+12)>=-2`

Options

A. `(-oo, 2]`
B. `[2, 4]`
C. `[4, oo)`
D. none of these

Explanation

To solve the inequality \( \log_{\frac{1}{2}} (x^2 - 6x + 12) \geq -2 \), we will follow these steps:

### Step 1: Convert the logarithmic inequality to exponential form
Using the property of logarithms, if \( \log_b(a) = c \), then \( a = b^c \). Here, we have:
\[ x^2 - 6x + 12 \geq \left(\frac{1}{2}\right)^{-2} \]
Calculating \( \left(\frac{1}{2}\right)^{-2} \):
\[ \left(\frac{1}{2}\right)^{-2} = 2^2 = 4 \]
Thus, the inequality becomes:
\[ x^2 - 6x + 12 \geq 4 \]

### Step 2: Rearrange the inequality
Now, we rearrange the inequality:
\[ x^2 - 6x + 12 - 4 \geq 0 \]
This simplifies to:
\[ x^2 - 6x + 8 \geq 0 \]

### Step 3: Factor the quadratic expression
Next, we factor the quadratic expression:
\[ x^2 - 6x + 8 = (x - 2)(x - 4) \]
So, we rewrite the inequality as:
\[ (x - 2)(x - 4) \geq 0 \]

### Step 4: Find the critical points
The critical points of the expression are \( x = 2 \) and \( x = 4 \). These points will help us determine the intervals to test.

### Step 5: Test the intervals
We will test the intervals determined by the critical points \( x = 2 \) and \( x = 4 \):
1. **Interval \( (-\infty, 2) \)**: Choose \( x = 0 \)
\[ (0 - 2)(0 - 4) = ( -2)( -4) = 8 \quad (\text{positive}) \]
2. **Interval \( (2, 4) \)**: Choose \( x = 3 \)
\[ (3 - 2)(3 - 4) = (1)(-1) = -1 \quad (\text{negative}) \]
3. **Interval \( (4, \infty) \)**: Choose \( x = 5 \)
\[ (5 - 2)(5 - 4) = (3)(1) = 3 \quad (\text{positive}) \]

### Step 6: Determine the solution set
From our tests, we find:
- The expression is non-negative in the intervals \( (-\infty, 2] \) and \( [4, \infty) \).

### Final Solution
Thus, the solution set for the inequality \( \log_{\frac{1}{2}} (x^2 - 6x + 12) \geq -2 \) is:
\[ x \in (-\infty, 2] \cup [4, \infty) \]