Find domain of `f(x)=sqrt(log_(1/2)((5x-x^2)/4))`

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{1/2}\left(\frac{5x - x^2}{4}\right)} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ \log_{1/2}\left(\frac{5x - x^2}{4}\right) \geq 0 \]

### Step 1: Understand the logarithmic function
The logarithm \( \log_{1/2}(y) \) is non-negative when \( y \leq 1 \) because the base \( \frac{1}{2} \) is less than 1. Therefore, we need to set up the inequality:

\[ \frac{5x - x^2}{4} \leq 1 \]

### Step 2: Solve the inequality
To solve this inequality, we first multiply both sides by 4 (since 4 is positive, the inequality direction remains the same):

\[ 5x - x^2 \leq 4 \]

Rearranging gives:

\[ -x^2 + 5x - 4 \leq 0 \]

This can be rewritten as:

\[ x^2 - 5x + 4 \geq 0 \]

### Step 3: Factor the quadratic
Next, we factor the quadratic expression:

\[ x^2 - 5x + 4 = (x - 1)(x - 4) \]

### Step 4: Determine the intervals
Now we need to find the intervals where this product is non-negative. The roots of the equation \( (x - 1)(x - 4) = 0 \) are \( x = 1 \) and \( x = 4 \). We will test the intervals \( (-\infty, 1) \), \( (1, 4) \), and \( (4, \infty) \).

1. For \( x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 1)(0 - 4) = (-1)(-4) = 4 \quad (\text{positive}) \]

2. For \( 1 < x < 4 \) (e.g., \( x = 2 \)):
\[ (2 - 1)(2 - 4) = (1)(-2) = -2 \quad (\text{negative}) \]

3. For \( x > 4 \) (e.g., \( x = 5 \)):
\[ (5 - 1)(5 - 4) = (4)(1) = 4 \quad (\text{positive}) \]

### Step 5: Combine the results
The expression \( (x - 1)(x - 4) \geq 0 \) is true for:

\[ x \in (-\infty, 1] \cup [4, \infty) \]

### Step 6: Check the original function
We also need to ensure that the argument of the logarithm \( \frac{5x - x^2}{4} \) is positive:

\[ 5x - x^2 > 0 \implies x(5 - x) > 0 \]

This inequality holds for \( 0 < x < 5 \).

### Step 7: Find the intersection of intervals
Now, we find the intersection of the intervals:

1. From the logarithm condition: \( (-\infty, 1] \cup [4, \infty) \)
2. From the positivity condition: \( (0, 5) \)

The intersection gives us:

\[ (0, 1] \cup [4, 5) \]

### Final Answer
Thus, the domain of the function \( f(x) \) is:

\[ \boxed{(0, 1] \cup [4, 5)} \]