To determine the interval in which the function \( f(x) = \int_{0}^{x} \log_{0.5} \left( \frac{2t - 8}{t - 2} \right) dt \) is increasing, we need to analyze the derivative \( f'(x) \).
### Step 1: Find the derivative \( f'(x) \)
Using the Fundamental Theorem of Calculus, we can differentiate \( f(x) \):
\[
f'(x) = \log_{0.5} \left( \frac{2x - 8}{x - 2} \right)
\]
### Step 2: Set the derivative greater than zero
For \( f(x) \) to be increasing, we need:
\[
f'(x) > 0 \implies \log_{0.5} \left( \frac{2x - 8}{x - 2} \right) > 0
\]
### Step 3: Analyze the logarithmic inequality
Since the base of the logarithm is \( 0.5 \) (which is less than 1), the logarithm is positive when its argument is less than 1:
\[
\frac{2x - 8}{x - 2} < 1
\]
### Step 4: Solve the inequality
Rearranging the inequality:
\[
2x - 8 < x - 2
\]
\[
2x - x < 8 - 2
\]
\[
x < 6
\]
Now, we also need to consider the condition when the argument of the logarithm is defined and positive:
\[
\frac{2x - 8}{x - 2} > 0
\]
### Step 5: Find critical points
The critical points occur when the numerator and denominator are zero:
- Numerator: \( 2x - 8 = 0 \) gives \( x = 4 \)
- Denominator: \( x - 2 = 0 \) gives \( x = 2 \)
### Step 6: Test intervals around critical points
We will test the sign of \( \frac{2x - 8}{x - 2} \) in the intervals \( (-\infty, 2) \), \( (2, 4) \), and \( (4, 6) \):
1. **Interval \( (-\infty, 2) \)**:
- Choose \( x = 0 \):
\[
\frac{2(0) - 8}{0 - 2} = \frac{-8}{-2} = 4 > 0
\]
2. **Interval \( (2, 4) \)**:
- Choose \( x = 3 \):
\[
\frac{2(3) - 8}{3 - 2} = \frac{-2}{1} = -2 < 0
\]
3. **Interval \( (4, 6) \)**:
- Choose \( x = 5 \):
\[
\frac{2(5) - 8}{5 - 2} = \frac{2}{3} > 0
\]
4. **Interval \( (6, \infty) \)**:
- Choose \( x = 7 \):
\[
\frac{2(7) - 8}{7 - 2} = \frac{6}{5} > 0
\]
### Step 7: Combine results
From the analysis:
- \( f'(x) > 0 \) in the intervals \( (-\infty, 2) \) and \( (4, 6) \).
- The function is not defined at \( x = 2 \) and is negative in \( (2, 4) \).
### Conclusion
The intervals in which \( f(x) \) is increasing are:
\[
(-\infty, 2) \cup (4, 6)
\]
Thus, the correct answer is:
**(c) \((-∞, 2) \cup (4, ∞)\)**.
