The function defined as f(x) = `2x^(3) -6x +3` is

Options

A. strictly increasing in `(-oo ,-1) cup (1,oo) ` and strictly decreasing in (-1,1)
B. strictly increasing in (-1,1) and strictly decreasing in`(-oo ,-1) cup (1,oo) ` .
C. strictly increasing in `(-oo , -1) cup [ 1,oo) ` and strictly decreasing in [-1,1].
D. strictly increasing in [-1,1] and strictly decreasing in `(-oo ,-1) cup [1,oo)`

Answer

To determine the intervals of increase and decrease for the function \( f(x) = 2x^3 - 6x + 3 \), we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 6x + 3) \]

Using the power rule for differentiation:

\[ f'(x) = 3 \cdot 2x^{3-1} - 6 \cdot 1 = 6x^2 - 6 \]

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

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

Factoring out the common term:

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

This gives us:

\[ x^2 - 1 = 0 \]

### Step 3: Solve for \( x \)
Solving the equation \( x^2 - 1 = 0 \):

\[ x^2 = 1 \implies x = 1 \quad \text{or} \quad x = -1 \]

### Step 4: Determine intervals for increasing and decreasing
Now we will test the intervals determined by the critical points \( x = -1 \) and \( x = 1 \). The intervals are:

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

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

- **Interval 1: \( (-\infty, -1) \)** (choose \( x = -2 \)):
\[ f'(-2) = 6(-2)^2 - 6 = 6(4) - 6 = 24 - 6 = 18 \quad (\text{positive}) \]

- **Interval 2: \( (-1, 1) \)** (choose \( x = 0 \)):
\[ f'(0) = 6(0)^2 - 6 = 0 - 6 = -6 \quad (\text{negative}) \]

- **Interval 3: \( (1, \infty) \)** (choose \( x = 2 \)):
\[ f'(2) = 6(2)^2 - 6 = 6(4) - 6 = 24 - 6 = 18 \quad (\text{positive}) \]

### Step 5: Conclusion on intervals of increase and decrease
From our analysis:

- \( f'(x) > 0 \) in the intervals \( (-\infty, -1) \) and \( (1, \infty) \) (function is strictly increasing).
- \( f'(x) < 0 \) in the interval \( (-1, 1) \) (function is strictly decreasing).

### Final Answer:
- The function \( f(x) \) is strictly increasing on the intervals \( (-\infty, -1) \) and \( (1, \infty) \).
- The function \( f(x) \) is strictly decreasing on the interval \( (-1, 1) \).