To find the values of \( x \) for which \( f(x) \) is positive, we start with the function given:
\[
f(x) = \frac{x^2 - 6x + 5}{x^2 - 5x + 6}
\]
### Step 1: Factor the numerator and denominator
**Numerator:**
The numerator \( x^2 - 6x + 5 \) can be factored as follows:
\[
x^2 - 6x + 5 = (x - 1)(x - 5)
\]
**Denominator:**
The denominator \( x^2 - 5x + 6 \) can be factored as:
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]
Thus, we can rewrite \( f(x) \) as:
\[
f(x) = \frac{(x - 1)(x - 5)}{(x - 2)(x - 3)}
\]
### Step 2: Determine the critical points
The critical points occur when the numerator is zero or the denominator is zero:
- **Numerator is zero:**
\((x - 1)(x - 5) = 0\) gives \( x = 1 \) and \( x = 5 \).
- **Denominator is zero:**
\((x - 2)(x - 3) = 0\) gives \( x = 2 \) and \( x = 3 \).
Thus, the critical points are \( x = 1, 2, 3, 5 \).
### Step 3: Create a sign chart
We will analyze the sign of \( f(x) \) in the intervals defined by the critical points:
- \( (-\infty, 1) \)
- \( (1, 2) \)
- \( (2, 3) \)
- \( (3, 5) \)
- \( (5, \infty) \)
### Step 4: Test the intervals
1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \)
\[
f(0) = \frac{(0 - 1)(0 - 5)}{(0 - 2)(0 - 3)} = \frac{(1)(5)}{(2)(3)} = \frac{5}{6} > 0
\]
2. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \)
\[
f(1.5) = \frac{(1.5 - 1)(1.5 - 5)}{(1.5 - 2)(1.5 - 3)} = \frac{(0.5)(-3.5)}{(-0.5)(-1.5)} = \frac{-1.75}{0.75} < 0
\]
3. **Interval \( (2, 3) \)**: Choose \( x = 2.5 \)
\[
f(2.5) = \frac{(2.5 - 1)(2.5 - 5)}{(2.5 - 2)(2.5 - 3)} = \frac{(1.5)(-2.5)}{(0.5)(-0.5)} = \frac{-3.75}{-0.25} > 0
\]
4. **Interval \( (3, 5) \)**: Choose \( x = 4 \)
\[
f(4) = \frac{(4 - 1)(4 - 5)}{(4 - 2)(4 - 3)} = \frac{(3)(-1)}{(2)(1)} = \frac{-3}{2} < 0
\]
5. **Interval \( (5, \infty) \)**: Choose \( x = 6 \)
\[
f(6) = \frac{(6 - 1)(6 - 5)}{(6 - 2)(6 - 3)} = \frac{(5)(1)}{(4)(3)} = \frac{5}{12} > 0
\]
### Step 5: Compile the results
From the sign analysis, we find that \( f(x) > 0 \) in the intervals:
- \( (-\infty, 1) \)
- \( (2, 3) \)
- \( (5, \infty) \)
### Final Answer
Thus, the values of \( x \) for which \( f(x) \) is positive are:
\[
x \in (-\infty, 1) \cup (2, 3) \cup (5, \infty)
\]
