For what values of x, the function `f(x)=(x + 1)^3(x - 3)^3` is increasing on R.

Options

A. `(oo,1)`
B. `(1,oo)`
C. `(-1,1)`
D. (0,1)

Answer

To determine the values of \( x \) for which the function \( f(x) = (x + 1)^3 (x - 3)^3 \) is increasing, we need to find the derivative \( f'(x) \) and analyze where it is greater than zero.

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

Using the product rule, 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 the derivatives:
\[ \frac{d}{dx}[(x + 1)^3] = 3(x + 1)^2 \]
\[ \frac{d}{dx}[(x - 3)^3] = 3(x - 3)^2 \]

Substituting back into the derivative:
\[ f'(x) = 3(x + 1)^2 (x - 3)^3 + (x + 1)^3 \cdot 3(x - 3)^2 \]

### Step 2: Factor out common terms

We can factor out \( 3(x + 1)^2 (x - 3)^2 \):
\[ f'(x) = 3(x + 1)^2 (x - 3)^2 \left[ (x - 3) + (x + 1) \right] \]
\[ = 3(x + 1)^2 (x - 3)^2 (2x - 2) \]
\[ = 6(x + 1)^2 (x - 3)^2 (x - 1) \]

### Step 3: Determine where \( f'(x) > 0 \)

The function \( f(x) \) is increasing when \( f'(x) > 0 \):
\[ 6(x + 1)^2 (x - 3)^2 (x - 1) > 0 \]

### Step 4: Analyze the factors

1. \( 6 \) is always positive.
2. \( (x + 1)^2 \) is always non-negative and is zero at \( x = -1 \).
3. \( (x - 3)^2 \) is always non-negative and is zero at \( x = 3 \).
4. \( (x - 1) \) is positive when \( x > 1 \).

### Step 5: Find intervals

- \( (x + 1)^2 = 0 \) at \( x = -1 \)
- \( (x - 3)^2 = 0 \) at \( x = 3 \)
- \( (x - 1) = 0 \) at \( x = 1 \)

We check the sign of \( f'(x) \) in the intervals determined by these critical points: \( (-\infty, -1) \), \( (-1, 1) \), \( (1, 3) \), and \( (3, \infty) \).

1. For \( x < -1 \): All factors are positive, \( f'(x) > 0 \).
2. For \( -1 < x < 1 \): \( (x + 1)^2 \) is positive, \( (x - 3)^2 \) is positive, \( (x - 1) < 0 \) → \( f'(x) < 0 \).
3. For \( 1 < x < 3 \): \( (x + 1)^2 \) is positive, \( (x - 3)^2 \) is positive, \( (x - 1) > 0 \) → \( f'(x) > 0 \).
4. For \( x > 3 \): All factors are positive, \( f'(x) > 0 \).

### Conclusion

The function \( f(x) \) is increasing in the intervals:
- \( (-\infty, -1) \)
- \( (1, 3) \)
- \( (3, \infty) \)

Thus, the function is increasing for \( x > 1 \).

### Final Answer
The function \( f(x) \) is increasing for \( x > 1 \).