To find the domain of the function \( f(x) = \sqrt{\log\left(\frac{5x - x^2}{6}\right)} \), we need to ensure that the expression inside the square root is non-negative, and that the logarithm is defined and positive. Let's break this down step by step.
### Step 1: Ensure the argument of the logarithm is positive
The logarithm function is defined only for positive arguments. Therefore, we need:
\[
\frac{5x - x^2}{6} > 0
\]
This simplifies to:
\[
5x - x^2 > 0
\]
### Step 2: Rearranging the inequality
Rearranging the inequality gives:
\[
-x^2 + 5x > 0
\]
Factoring out \( -1 \):
\[
-x(x - 5) > 0
\]
### Step 3: Analyzing the inequality
To solve the inequality \( -x(x - 5) > 0 \), we can find the critical points by setting the expression equal to zero:
\[
-x(x - 5) = 0 \implies x = 0 \quad \text{or} \quad x = 5
\]
Now we test the intervals determined by these critical points: \( (-\infty, 0) \), \( (0, 5) \), and \( (5, \infty) \).
- For \( x < 0 \): Choose \( x = -1 \), \( -(-1)(-1 - 5) = -1 \cdot (-6) = 6 > 0 \) (not valid for our function).
- For \( 0 < x < 5 \): Choose \( x = 1 \), \( -1(1 - 5) = -1 \cdot (-4) = 4 > 0 \) (valid).
- For \( x > 5 \): Choose \( x = 6 \), \( -6(6 - 5) = -6 \cdot 1 = -6 < 0 \) (not valid).
Thus, the solution to the inequality \( 5x - x^2 > 0 \) is:
\[
0 < x < 5
\]
### Step 4: Ensure the logarithm is greater than or equal to 1
Next, we need the logarithm to be non-negative:
\[
\log\left(\frac{5x - x^2}{6}\right) \geq 0
\]
This implies:
\[
\frac{5x - x^2}{6} \geq 1
\]
Multiplying both sides by 6 (since 6 is positive):
\[
5x - x^2 \geq 6
\]
### Step 5: Rearranging this inequality
Rearranging gives:
\[
-x^2 + 5x - 6 \geq 0
\]
Factoring:
\[
-(x^2 - 5x + 6) \geq 0 \implies x^2 - 5x + 6 \leq 0
\]
### Step 6: Finding roots of the quadratic
Finding the roots of \( x^2 - 5x + 6 = 0 \):
\[
(x - 2)(x - 3) = 0 \implies x = 2 \quad \text{or} \quad x = 3
\]
### Step 7: Analyzing the quadratic inequality
To solve \( x^2 - 5x + 6 \leq 0 \), we test the intervals determined by the roots \( 2 \) and \( 3 \):
- For \( x < 2 \): Choose \( x = 1 \), \( 1^2 - 5(1) + 6 = 2 > 0 \) (not valid).
- For \( 2 \leq x \leq 3 \): Choose \( x = 2.5 \), \( (2.5)^2 - 5(2.5) + 6 = -0.25 \leq 0 \) (valid).
- For \( x > 3 \): Choose \( x = 4 \), \( 4^2 - 5(4) + 6 = 2 > 0 \) (not valid).
Thus, the solution to \( x^2 - 5x + 6 \leq 0 \) is:
\[
2 \leq x \leq 3
\]
### Step 8: Finding the intersection of the two conditions
Now we combine the two conditions:
1. From \( 5x - x^2 > 0 \): \( 0 < x < 5 \)
2. From \( x^2 - 5x + 6 \leq 0 \): \( 2 \leq x \leq 3 \)
The intersection of these two intervals is:
\[
[2, 3]
\]
### Final Domain
Thus, the domain of the function \( f(x) \) is:
\[
\boxed{[2, 3]}
\]
