Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=(x+1)^(3)(x-3)^(3).`

Answer

To determine the intervals in which the function \( f(x) = (x + 1)^3 (x - 3)^3 \) is strictly increasing or strictly decreasing, we will follow these steps:

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

Using the product rule of differentiation, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}[(x + 1)^3] \cdot (x - 3)^3 + (x + 1)^3 \cdot \frac{d}{dx}[(x - 3)^3] \]

Calculating each derivative:

1. \( \frac{d}{dx}[(x + 1)^3] = 3(x + 1)^2 \)
2. \( \frac{d}{dx}[(x - 3)^3] = 3(x - 3)^2 \)

Thus, we have:

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

Factoring out the common terms:

\[ f'(x) = 3(x + 1)^2 (x - 3)^2 \left[(x - 3) + (x + 1)\right] \]

Simplifying the expression inside the brackets:

\[ f'(x) = 3(x + 1)^2 (x - 3)^2 (2x - 2) = 6(x + 1)^2 (x - 3)^2 (x - 1) \]

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

To find the critical points, we set \( f'(x) = 0 \):

\[ 6(x + 1)^2 (x - 3)^2 (x - 1) = 0 \]

This gives us the following critical points:

1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x - 3 = 0 \) → \( x = 3 \)
3. \( x - 1 = 0 \) → \( x = 1 \)

### Step 3: Determine the sign of \( f'(x) \) in the intervals defined by the critical points

The critical points divide the number line into the following intervals:

1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, 3) \)
4. \( (3, \infty) \)

We will test a point from each interval to determine the sign of \( f'(x) \):

- **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \)
\[ f'(-2) = 6(-2 + 1)^2 (-2 - 3)^2 (-2 - 1) = 6(1)^2(25)(-3) < 0 \quad \text{(Decreasing)} \]

- **Interval \( (-1, 1) \)**: Choose \( x = 0 \)
\[ f'(0) = 6(0 + 1)^2 (0 - 3)^2 (0 - 1) = 6(1)(9)(-1) < 0 \quad \text{(Decreasing)} \]

- **Interval \( (1, 3) \)**: Choose \( x = 2 \)
\[ f'(2) = 6(2 + 1)^2 (2 - 3)^2 (2 - 1) = 6(3)^2(1)(1) > 0 \quad \text{(Increasing)} \]

- **Interval \( (3, \infty) \)**: Choose \( x = 4 \)
\[ f'(4) = 6(4 + 1)^2 (4 - 3)^2 (4 - 1) = 6(5)^2(1)(3) > 0 \quad \text{(Increasing)} \]

### Step 4: Summarize the intervals

- **Strictly Decreasing**: \( (-\infty, -1) \) and \( (-1, 1) \)
- **Strictly Increasing**: \( (1, 3) \) and \( (3, \infty) \)

### Final Answer

- The function \( f(x) \) is strictly decreasing on the intervals \( (-\infty, -1) \) and \( (-1, 1) \).
- The function \( f(x) \) is strictly increasing on the intervals \( (1, 3) \) and \( (3, \infty) \).