The function f(X) = `int((x-2)dx)/(x^(2)-7x+12)`

Options

A. decreases on R
B. increases on `R~ (2,3)`
C. increases on `[2,3) cup (4, oo)`
D. decreases on `(2, oo)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \int \frac{x-2}{x^2 - 7x + 12} \, dx \) and determine where it is increasing or decreasing.

### Step-by-Step Solution:

1. **Identify the Function**:
The function given is:
\[ f(x) = \int \frac{x-2}{x^2 - 7x + 12} \, dx \]

2. **Differentiate the Function**:
To find where the function is increasing or decreasing, we need to find the derivative \( f'(x) \). By the Fundamental Theorem of Calculus, we have:
\[ f'(x) = \frac{x-2}{x^2 - 7x + 12} \]

3. **Factor the Denominator**:
We can factor the denominator:
\[ x^2 - 7x + 12 = (x-3)(x-4) \]
Thus, we can rewrite \( f'(x) \) as:
\[ f'(x) = \frac{x-2}{(x-3)(x-4)} \]

4. **Determine Critical Points**:
The critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined.
- Setting the numerator to zero:
\[ x - 2 = 0 \implies x = 2 \]
- The denominator is undefined at:
\[ (x-3)(x-4) = 0 \implies x = 3 \text{ and } x = 4 \]

5. **Test Intervals**:
We will test the sign of \( f'(x) \) in the intervals determined by the critical points \( x = 2, 3, 4 \):
- Intervals: \( (-\infty, 2) \), \( (2, 3) \), \( (3, 4) \), \( (4, \infty) \)

- **Interval \( (-\infty, 2) \)**: Choose \( x = 0 \):
\[ f'(0) = \frac{0-2}{(0-3)(0-4)} = \frac{-2}{12} < 0 \quad \text{(decreasing)} \]

- **Interval \( (2, 3) \)**: Choose \( x = 2.5 \):
\[ f'(2.5) = \frac{2.5-2}{(2.5-3)(2.5-4)} = \frac{0.5}{(-0.5)(-1.5)} > 0 \quad \text{(increasing)} \]

- **Interval \( (3, 4) \)**: Choose \( x = 3.5 \):
\[ f'(3.5) = \frac{3.5-2}{(3.5-3)(3.5-4)} = \frac{1.5}{(0.5)(-0.5)} < 0 \quad \text{(decreasing)} \]

- **Interval \( (4, \infty) \)**: Choose \( x = 5 \):
\[ f'(5) = \frac{5-2}{(5-3)(5-4)} = \frac{3}{(2)(1)} > 0 \quad \text{(increasing)} \]

6. **Summarize the Results**:
- \( f(x) \) is decreasing on \( (-\infty, 2) \) and \( (3, 4) \).
- \( f(x) \) is increasing on \( (2, 3) \) and \( (4, \infty) \).

### Conclusion:
The function \( f(x) \) is increasing on the intervals \( [2, 3) \) and \( (4, \infty) \), and decreasing on the intervals \( (-\infty, 2) \) and \( (3, 4) \).