Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=-3log(1+x)+4log(2+x)-(4)/(2+x)`

Answer

To determine the intervals in which the function \( f(x) = -3\log(1+x) + 4\log(2+x) - \frac{4}{2+x} \) is strictly increasing or strictly decreasing, we need to follow these steps:

### Step 1: Find the derivative \( f'(x) \)

To analyze the behavior of the function, we first differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( -3\log(1+x) + 4\log(2+x) - \frac{4}{2+x} \right) \]

Using the derivative of logarithmic functions and the quotient rule, we have:

\[ f'(x) = -3 \cdot \frac{1}{1+x} \cdot (1) + 4 \cdot \frac{1}{2+x} \cdot (1) - 4 \cdot \left(-\frac{1}{(2+x)^2}\right) \]

This simplifies to:

\[ f'(x) = -\frac{3}{1+x} + \frac{4}{2+x} + \frac{4}{(2+x)^2} \]

### Step 2: Set the derivative to zero to find critical points

To find the critical points where the function changes from increasing to decreasing or vice versa, we set \( f'(x) = 0 \):

\[ -\frac{3}{1+x} + \frac{4}{2+x} + \frac{4}{(2+x)^2} = 0 \]

### Step 3: Solve for \( x \)

To solve this equation, we can multiply through by \( (1+x)(2+x)^2 \) to eliminate the denominators:

\[ -3(2+x)^2 + 4(1+x)(2+x) + 4(1+x) = 0 \]

Expanding and simplifying gives:

\[ -3(4 + 4x + x^2) + 4(2 + 3x + x^2) + 4(1+x) = 0 \]

This leads to:

\[ -12 - 12x - 3x^2 + 8 + 12x + 4x^2 + 4 + 4x = 0 \]

Combining like terms results in:

\[ x^2 - 12 = 0 \]

Thus, we find:

\[ x^2 = 12 \implies x = \pm 2\sqrt{3} \]

### Step 4: Determine intervals of increase and decrease

Now we need to test the sign of \( f'(x) \) in the intervals determined by the critical points \( x = -2\sqrt{3} \) and \( x = 0 \). We will check the sign of \( f'(x) \) in the intervals:

1. \( (-\infty, -2\sqrt{3}) \)
2. \( (-2\sqrt{3}, 0) \)
3. \( (0, \infty) \)

### Step 5: Test points in each interval

1. **For \( x < -2\sqrt{3} \)** (e.g., \( x = -5 \)):
- Calculate \( f'(-5) \) to check the sign.

2. **For \( -2\sqrt{3} < x < 0 \)** (e.g., \( x = -1 \)):
- Calculate \( f'(-1) \) to check the sign.

3. **For \( x > 0 \)** (e.g., \( x = 1 \)):
- Calculate \( f'(1) \) to check the sign.

### Step 6: Conclusion

From the sign tests, we can conclude:

- The function is **strictly increasing** where \( f'(x) > 0 \).
- The function is **strictly decreasing** where \( f'(x) < 0 \).

Thus, the intervals are:

- **Strictly increasing**: \( (-\infty, -2\sqrt{3}) \) and \( (0, \infty) \)
- **Strictly decreasing**: \( (-2\sqrt{3}, 0) \)