Determine for which values of x, the following functions are increasing or decreasing :
`f(x)=2x^(3)-24x+107.`

Answer

To determine the intervals where the function \( f(x) = 2x^3 - 24x + 107 \) is increasing or decreasing, we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(2x^3 - 24x + 107) \]

Using the power rule, we find:

\[ f'(x) = 6x^2 - 24 \]

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

\[ 6x^2 - 24 = 0 \]

### Step 3: Solve for \( x \)
Now, we solve the equation:

\[ 6x^2 = 24 \]
\[ x^2 = 4 \]
\[ x = \pm 2 \]

The critical points are \( x = -2 \) and \( x = 2 \).

### Step 4: Test intervals around the critical points
We will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ f'(-3) = 6(-3)^2 - 24 = 54 - 24 = 30 \quad (> 0) \]
So, \( f(x) \) is increasing in this interval.

2. **Interval \( (-2, 2) \)**: Choose \( x = 0 \)
\[ f'(0) = 6(0)^2 - 24 = -24 \quad (< 0) \]
So, \( f(x) \) is decreasing in this interval.

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ f'(3) = 6(3)^2 - 24 = 54 - 24 = 30 \quad (> 0) \]
So, \( f(x) \) is increasing in this interval.

### Step 5: Summarize the results
- The function \( f(x) \) is **increasing** on the intervals \( (-\infty, -2) \) and \( (2, \infty) \).
- The function \( f(x) \) is **decreasing** on the interval \( (-2, 2) \).

### Final Answer
- **Increasing**: \( (-\infty, -2) \) and \( (2, \infty) \)
- **Decreasing**: \( (-2, 2) \)

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