Find the interval in which the function f(x) =2 log (x-2)`-x^(2)+4x +1` is increasing .

Answer

To find the interval in which the function \( f(x) = 2 \log(x - 2) - x^2 + 4x + 1 \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}[2 \log(x - 2)] - \frac{d}{dx}[x^2] + \frac{d}{dx}[4x] + \frac{d}{dx}[1] \]

Using the derivative rules, we get:

\[ f'(x) = 2 \cdot \frac{1}{x - 2} - 2x + 4 \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we set the derivative greater than zero:

\[ f'(x) > 0 \]

This gives us:

\[ \frac{2}{x - 2} - 2x + 4 > 0 \]

### Step 3: Combine the terms
To combine the terms, we first find a common denominator, which is \( x - 2 \):

\[ \frac{2 - 2x(x - 2) + 4(x - 2)}{x - 2} > 0 \]

Simplifying the numerator:

\[ 2 - 2x^2 + 4x - 8 = -2x^2 + 4x - 6 \]

Thus, we have:

\[ \frac{-2x^2 + 4x - 6}{x - 2} > 0 \]

### Step 4: Factor the quadratic expression
Now, we factor the quadratic expression in the numerator:

\[ -2(x^2 - 2x + 3) = -2(x - 1)(x - 3) \]

So we rewrite the inequality as:

\[ \frac{-2(x - 1)(x - 3)}{x - 2} > 0 \]

### Step 5: Determine critical points
The critical points from the factors are \( x = 1, x = 2, x = 3 \). We will analyze the sign of the expression in the intervals determined by these points.

### Step 6: Test intervals
We will test the intervals: \( (-\infty, 1) \), \( (1, 2) \), \( (2, 3) \), and \( (3, \infty) \).

1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \)
\[ \frac{-2(0 - 1)(0 - 3)}{0 - 2} = \frac{-2(-1)(-3)}{-2} = \text{Positive} \]

2. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \)
\[ \frac{-2(1.5 - 1)(1.5 - 3)}{1.5 - 2} = \frac{-2(0.5)(-1.5)}{-0.5} = \text{Negative} \]

3. **Interval \( (2, 3) \)**: Choose \( x = 2.5 \)
\[ \frac{-2(2.5 - 1)(2.5 - 3)}{2.5 - 2} = \frac{-2(1.5)(-0.5)}{0.5} = \text{Positive} \]

4. **Interval \( (3, \infty) \)**: Choose \( x = 4 \)
\[ \frac{-2(4 - 1)(4 - 3)}{4 - 2} = \frac{-2(3)(1)}{2} = \text{Negative} \]

### Step 7: Conclusion
The function \( f(x) \) is increasing in the intervals where \( f'(x) > 0 \). From our tests, we find that:

- The function is increasing in the intervals \( (-\infty, 1) \) and \( (2, 3) \).

### Final Answer
Thus, the function \( f(x) \) is increasing in the intervals \( (-\infty, 1) \) and \( (2, 3) \).

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