To solve the problem, we need to analyze the quadratic expression \( f(x) = x^2 - 6x + 5 \) and determine the sign of this expression over the specified intervals.
### Step-by-step Solution:
1. **Identify the Quadratic Expression**:
We have the quadratic expression:
\[
f(x) = x^2 - 6x + 5
\]
2. **Find the Roots of the Quadratic**:
To find the roots, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -6 \), and \( c = 5 \).
Plugging in these values:
\[
x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}
\]
\[
= \frac{6 \pm \sqrt{36 - 20}}{2}
\]
\[
= \frac{6 \pm \sqrt{16}}{2}
\]
\[
= \frac{6 \pm 4}{2}
\]
This gives us two roots:
\[
x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{2}{2} = 1
\]
3. **Factor the Quadratic**:
The quadratic can be factored as:
\[
f(x) = (x - 1)(x - 5)
\]
4. **Determine the Sign of \( f(x) \)**:
The roots divide the number line into intervals. We will check the sign of \( f(x) \) in the intervals:
- \( (-\infty, 1) \)
- \( (1, 5) \)
- \( (5, \infty) \)
- **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \):
\[
f(0) = (0 - 1)(0 - 5) = 1 \cdot 5 = 5 \quad (\text{positive})
\]
- **Interval \( (1, 5) \)**: Choose \( x = 3 \):
\[
f(3) = (3 - 1)(3 - 5) = 2 \cdot (-2) = -4 \quad (\text{negative})
\]
- **Interval \( (5, \infty) \)**: Choose \( x = 6 \):
\[
f(6) = (6 - 1)(6 - 5) = 5 \cdot 1 = 5 \quad (\text{positive})
\]
5. **Analyze the Given Statements**:
- **Statement I**: For \( x \in (2, 4) \), \( f(x) \) is negative.
- From our analysis, \( f(x) \) is negative in the interval \( (1, 5) \), which includes \( (2, 4) \). Therefore, **Statement I is true**.
- **Statement II**: For \( x \in (-\infty, 2) \cup (4, \infty) \), \( f(x) \) is positive.
- From our analysis, \( f(x) \) is positive in the intervals \( (-\infty, 1) \) and \( (5, \infty) \), and also positive in the interval \( (2, 4) \) does not hold true. Therefore, **Statement II is false**.
### Conclusion:
- **Only Statement I is true**.
