To find the domain of the function \( f(x) = \sqrt{\frac{2x + 1}{x^3 - 3x^2 + 2x}} \), we need to ensure that the expression inside the square root is non-negative, and the denominator is not zero.
### Step-by-step Solution:
1. **Set the condition for the square root:**
\[
\frac{2x + 1}{x^3 - 3x^2 + 2x} \geq 0
\]
2. **Factor the denominator:**
First, we factor the denominator \( x^3 - 3x^2 + 2x \):
\[
x^3 - 3x^2 + 2x = x(x^2 - 3x + 2)
\]
Now, we can factor \( x^2 - 3x + 2 \):
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]
Thus, the denominator can be written as:
\[
x^3 - 3x^2 + 2x = x(x - 1)(x - 2)
\]
3. **Rewrite the inequality:**
Now we rewrite the inequality:
\[
\frac{2x + 1}{x(x - 1)(x - 2)} \geq 0
\]
4. **Find the critical points:**
The critical points are found by setting the numerator and denominator to zero:
- Numerator: \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \)
- Denominator: \( x(x - 1)(x - 2) = 0 \) gives \( x = 0, 1, 2 \)
Thus, the critical points are \( x = -\frac{1}{2}, 0, 1, 2 \).
5. **Test intervals around the critical points:**
We will test the sign of the expression in the intervals defined by the critical points:
- Interval \( (-\infty, -\frac{1}{2}) \)
- Interval \( (-\frac{1}{2}, 0) \)
- Interval \( (0, 1) \)
- Interval \( (1, 2) \)
- Interval \( (2, \infty) \)
We can choose test points from each interval:
- For \( x = -1 \) in \( (-\infty, -\frac{1}{2}) \):
\[
\frac{2(-1) + 1}{(-1)(-1 - 1)(-1 - 2)} = \frac{-2 + 1}{-1 \cdot -2 \cdot -3} = \frac{-1}{6} < 0
\]
- For \( x = -\frac{1}{4} \) in \( (-\frac{1}{2}, 0) \):
\[
\frac{2(-\frac{1}{4}) + 1}{(-\frac{1}{4})(-\frac{1}{4} - 1)(-\frac{1}{4} - 2)} = \frac{-\frac{1}{2} + 1}{(-\frac{1}{4})(-\frac{5}{4})(-\frac{9}{4})} = \frac{\frac{1}{2}}{\frac{45}{64}} > 0
\]
- For \( x = \frac{1}{2} \) in \( (0, 1) \):
\[
\frac{2(\frac{1}{2}) + 1}{(\frac{1}{2})(\frac{1}{2} - 1)(\frac{1}{2} - 2)} = \frac{1 + 1}{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})} > 0
\]
- For \( x = 1.5 \) in \( (1, 2) \):
\[
\frac{2(1.5) + 1}{(1.5)(1.5 - 1)(1.5 - 2)} = \frac{3 + 1}{(1.5)(0.5)(-0.5)} < 0
\]
- For \( x = 3 \) in \( (2, \infty) \):
\[
\frac{2(3) + 1}{(3)(3 - 1)(3 - 2)} = \frac{6 + 1}{3 \cdot 2 \cdot 1} > 0
\]
6. **Combine the results:**
The intervals where the expression is non-negative are:
- \( (-\frac{1}{2}, 0) \)
- \( (0, 1) \)
- \( (2, \infty) \)
7. **Include critical points:**
We must also check the critical points:
- At \( x = -\frac{1}{2} \), \( f(x) = 0 \) (included)
- At \( x = 0 \), the function is undefined (not included)
- At \( x = 1 \), the function is undefined (not included)
- At \( x = 2 \), the function is undefined (not included)
### Final Domain:
Thus, the domain of the function \( f(x) \) is:
\[
\boxed{[-\frac{1}{2}, 0) \cup (0, 1) \cup (2, \infty)}
\]
