Find the domain of the function : `f(x)=1/(sqrt((log)_(1/2)(x^2-7x+13)))`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{\log_{1/2}(x^2 - 7x + 13)}} \), we need to ensure that the expression inside the square root is positive, since the square root function is defined only for non-negative values, and the logarithm function must also be defined.

### Step 1: Set the condition for the logarithm
The logarithm \( \log_{1/2}(x^2 - 7x + 13) \) must be greater than 0:
\[ \log_{1/2}(x^2 - 7x + 13) > 0 \]

### Step 2: Convert the logarithm inequality
Since the base of the logarithm is \( \frac{1}{2} \), which is less than 1, the inequality reverses when we convert it to an exponential form:
\[ x^2 - 7x + 13 < 1 \]

### Step 3: Rearrange the inequality
Rearranging the inequality gives:
\[ x^2 - 7x + 12 < 0 \]

### Step 4: Factor the quadratic expression
Now, we factor the quadratic expression:
\[ (x - 3)(x - 4) < 0 \]

### Step 5: Determine the intervals
To solve the inequality \( (x - 3)(x - 4) < 0 \), we find the critical points where the expression equals zero:
- The critical points are \( x = 3 \) and \( x = 4 \).

Next, we test the intervals defined by these points:
1. For \( x < 3 \) (e.g., \( x = 2 \)): \( (2 - 3)(2 - 4) = (-1)(-2) = 2 > 0 \)
2. For \( 3 < x < 4 \) (e.g., \( x = 3.5 \)): \( (3.5 - 3)(3.5 - 4) = (0.5)(-0.5) = -0.25 < 0 \)
3. For \( x > 4 \) (e.g., \( x = 5 \)): \( (5 - 3)(5 - 4) = (2)(1) = 2 > 0 \)

### Step 6: Conclusion about the intervals
The expression \( (x - 3)(x - 4) < 0 \) is satisfied in the interval \( (3, 4) \).

### Final Domain
Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain} = (3, 4) \]