The function `f(x)=x^(3)-3x ` is

Options

A. Increasing in `(-oo,-1) cup (1,oo) and ` decreasing in (-1,1)
B. Decreasing in `(-oo,-1) cup (1,oo) and ` increasing in (-1,1)
C. Increasing in `(0,oo)` and decreasing in `(-oo,0)`
D. Decreasing in `(0,oo)` and increasing in `(-oo,0)`

Answer

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

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

\[ f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3 \]

### Step 2: Set the derivative equal to zero to find critical points
Next, we set the derivative equal to zero to find the critical points.

\[ 3x^2 - 3 = 0 \]

Dividing both sides by 3:

\[ x^2 - 1 = 0 \]

Factoring the equation:

\[ (x - 1)(x + 1) = 0 \]

Thus, the critical points are:

\[ x = -1 \quad \text{and} \quad x = 1 \]

### Step 3: Determine the sign of the derivative in each interval
We will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).

1. **Interval \( (-\infty, -1) \)**:
- Choose \( x = -2 \):
\[ f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9 \quad (\text{positive}) \]

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

3. **Interval \( (1, \infty) \)**:
- Choose \( x = 2 \):
\[ f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \quad (\text{positive}) \]

### Step 4: Summarize the intervals of increase and decrease
From our analysis of the sign of \( f'(x) \):

- The function is **increasing** on the intervals \( (-\infty, -1) \) and \( (1, \infty) \).
- The function is **decreasing** on the interval \( (-1, 1) \).

### Final Answer
- **Increasing**: \( (-\infty, -1) \cup (1, \infty) \)
- **Decreasing**: \( (-1, 1) \)