Determine for which values of x, the following functions are increasing or decreasing :
`f(x)=x^(3)-12x`

Answer

To determine the intervals where the function \( f(x) = x^3 - 12x \) is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to find the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 - 12x) = 3x^2 - 12 \]

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

\[ 3x^2 - 12 = 0 \]

Dividing the entire equation by 3 gives:

\[ x^2 - 4 = 0 \]

Factoring the equation:

\[ (x - 2)(x + 2) = 0 \]

Thus, the critical points are:

\[ x = 2 \quad \text{and} \quad x = -2 \]

### Step 3: Determine the sign of the derivative in the intervals
We will now test the sign of \( f'(x) \) in the intervals defined by the critical points: \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, -2) \)**:
Choose \( x = -3 \):
\[ f'(-3) = 3(-3)^2 - 12 = 27 - 12 = 15 \quad (\text{positive}) \]

2. **Interval \( (-2, 2) \)**:
Choose \( x = 0 \):
\[ f'(0) = 3(0)^2 - 12 = 0 - 12 = -12 \quad (\text{negative}) \]

3. **Interval \( (2, \infty) \)**:
Choose \( x = 3 \):
\[ f'(3) = 3(3)^2 - 12 = 27 - 12 = 15 \quad (\text{positive}) \]

### Step 4: Conclusion based on the sign of the derivative
From our tests, we can conclude:

- \( f'(x) > 0 \) in the intervals \( (-\infty, -2) \) and \( (2, \infty) \): The function is **increasing** in these intervals.
- \( f'(x) < 0 \) in the interval \( (-2, 2) \): The function is **decreasing** in this interval.

### Final Result
- **Increasing** on \( (-\infty, -2) \) and \( (2, \infty) \)
- **Decreasing** on \( (-2, 2) \)