To find the domain of the function \( f(x) = \sqrt{\frac{(x-1)(x+2)}{(x-3)(x-4)}} \), we need to ensure that the expression under the square root is non-negative and that the denominator is not zero.
### Step 1: Identify the conditions for the square root
The expression inside the square root must be greater than or equal to zero:
\[
\frac{(x-1)(x+2)}{(x-3)(x-4)} \geq 0
\]
### Step 2: Determine the critical points
The critical points occur when the numerator or denominator is zero:
- From the numerator \( (x-1)(x+2) = 0 \):
- \( x - 1 = 0 \) gives \( x = 1 \)
- \( x + 2 = 0 \) gives \( x = -2 \)
- From the denominator \( (x-3)(x-4) = 0 \):
- \( x - 3 = 0 \) gives \( x = 3 \)
- \( x - 4 = 0 \) gives \( x = 4 \)
The critical points are \( x = -2, 1, 3, 4 \).
### Step 3: Test intervals around the critical points
We will test the sign of the expression in the intervals determined by the critical points:
- Intervals: \( (-\infty, -2) \), \( (-2, 1) \), \( (1, 3) \), \( (3, 4) \), \( (4, \infty) \)
1. **Interval \( (-\infty, -2) \)**:
- Choose \( x = -3 \):
\[
\frac{(-3-1)(-3+2)}{(-3-3)(-3-4)} = \frac{(-4)(-1)}{(-6)(-7)} = \frac{4}{42} > 0
\]
2. **Interval \( (-2, 1) \)**:
- Choose \( x = 0 \):
\[
\frac{(0-1)(0+2)}{(0-3)(0-4)} = \frac{(-1)(2)}{(-3)(-4)} = \frac{-2}{12} < 0
\]
3. **Interval \( (1, 3) \)**:
- Choose \( x = 2 \):
\[
\frac{(2-1)(2+2)}{(2-3)(2-4)} = \frac{(1)(4)}{(-1)(-2)} = \frac{4}{2} > 0
\]
4. **Interval \( (3, 4) \)**:
- Choose \( x = 3.5 \):
\[
\frac{(3.5-1)(3.5+2)}{(3.5-3)(3.5-4)} = \frac{(2.5)(5.5)}{(0.5)(-0.5)} = \frac{13.75}{-0.25} < 0
\]
5. **Interval \( (4, \infty) \)**:
- Choose \( x = 5 \):
\[
\frac{(5-1)(5+2)}{(5-3)(5-4)} = \frac{(4)(7)}{(2)(1)} = \frac{28}{2} > 0
\]
### Step 4: Compile results
From the tests, we find:
- The expression is non-negative in the intervals \( (-\infty, -2) \), \( (1, 3) \), and \( (4, \infty) \).
- The points \( x = 3 \) and \( x = 4 \) are excluded since they make the denominator zero.
### Step 5: Write the domain
Thus, the domain of the function \( f(x) \) is:
\[
\text{Domain} = (-\infty, -2) \cup [1, 3) \cup (4, \infty)
\]
