To find the domain of the function \( f(x) = \frac{1}{\sqrt{x^3 - 3x^2 + 2x}} \), we need to ensure that the expression inside the square root is positive, as the square root of a non-positive number is not defined in the real number system.
### Step-by-Step Solution:
1. **Set up the inequality**:
We need to solve the inequality:
\[
x^3 - 3x^2 + 2x > 0
\]
2. **Factor the expression**:
First, we can factor out \( x \) from the expression:
\[
x(x^2 - 3x + 2) > 0
\]
Now, we need to factor the quadratic \( x^2 - 3x + 2 \).
3. **Factor the quadratic**:
The quadratic factors as:
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]
Therefore, we can rewrite our inequality as:
\[
x(x - 1)(x - 2) > 0
\]
4. **Identify critical points**:
The critical points from the factors are:
\[
x = 0, \quad x = 1, \quad x = 2
\]
5. **Test intervals**:
We will test the sign of the expression \( x(x - 1)(x - 2) \) in the intervals defined by the critical points:
- Interval \( (-\infty, 0) \)
- Interval \( (0, 1) \)
- Interval \( (1, 2) \)
- Interval \( (2, \infty) \)
- For \( x < 0 \) (e.g., \( x = -1 \)):
\[
(-1)(-1 - 1)(-1 - 2) = (-1)(-2)(-3) < 0
\]
- For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)):
\[
(0.5)(0.5 - 1)(0.5 - 2) = (0.5)(-0.5)(-1.5) > 0
\]
- For \( 1 < x < 2 \) (e.g., \( x = 1.5 \)):
\[
(1.5)(1.5 - 1)(1.5 - 2) = (1.5)(0.5)(-0.5) < 0
\]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[
(3)(3 - 1)(3 - 2) = (3)(2)(1) > 0
\]
6. **Combine results**:
The expression \( x(x - 1)(x - 2) > 0 \) is positive in the intervals \( (0, 1) \) and \( (2, \infty) \).
7. **Write the domain**:
Therefore, the domain of the function \( f(x) \) is:
\[
(0, 1) \cup (2, \infty)
\]
### Final Answer:
The domain of \( f(x) = \frac{1}{\sqrt{x^3 - 3x^2 + 2x}} \) is:
\[
(0, 1) \cup (2, \infty)
\]
