The function `f (x) = 2x ^(3) + 9x ^(2) + 12 x + 20` is :

Options

A. increasing in `(-oo,-2)` and decreasing in `(-2,-1).`
B. increasing in `(-oo,-2)uu(-1,oo)` and decreasing in `(-2,-1).`
C. increasing in `(-2,-1)` and decreasing in `(-oo,-2)uu(-1,oo).`
D. increasing in `(-oo, -2)nn(-1,oo) and ` decreasing in `(-2,-1).`

Answer

To determine the nature of the function \( f(x) = 2x^3 + 9x^2 + 12x + 20 \), we will analyze its increasing and decreasing intervals by finding its derivative and determining where this derivative is positive or negative.

### Step-by-Step Solution:

1. **Write the function**:
\[ f(x) = 2x^3 + 9x^2 + 12x + 20 \]

**Hint**: Start by clearly stating the function you are working with.

2. **Differentiate the function**:
To find the critical points and analyze the behavior of the function, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3) + \frac{d}{dx}(9x^2) + \frac{d}{dx}(12x) + \frac{d}{dx}(20) \]
Using the power rule, we differentiate each term:
\[ f'(x) = 3 \cdot 2x^{3-1} + 2 \cdot 9x^{2-1} + 12 + 0 = 6x^2 + 18x + 12 \]

**Hint**: Remember to apply the power rule correctly for each term.

3. **Set the derivative to zero**:
To find the critical points, we set the derivative equal to zero:
\[ 6x^2 + 18x + 12 = 0 \]
We can simplify this equation by dividing all terms by 6:
\[ x^2 + 3x + 2 = 0 \]

**Hint**: Simplifying the equation makes it easier to solve.

4. **Factor the quadratic equation**:
We can factor the quadratic:
\[ (x + 1)(x + 2) = 0 \]
This gives us the critical points:
\[ x = -1 \quad \text{and} \quad x = -2 \]

**Hint**: Look for two numbers that multiply to the constant term and add to the coefficient of \( x \).

5. **Determine intervals**:
The critical points divide the number line into intervals. We will test the sign of \( f'(x) \) in the intervals:
- \( (-\infty, -2) \)
- \( (-2, -1) \)
- \( (-1, \infty) \)

**Hint**: Choose test points from each interval to determine the sign of the derivative.

6. **Test the intervals**:
- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ f'(-3) = 6(-3)^2 + 18(-3) + 12 = 54 - 54 + 12 = 12 \quad (\text{positive}) \]
- For \( -2 < x < -1 \) (e.g., \( x = -1.5 \)):
\[ f'(-1.5) = 6(-1.5)^2 + 18(-1.5) + 12 = 13.5 - 27 + 12 = -1.5 \quad (\text{negative}) \]
- For \( x > -1 \) (e.g., \( x = 0 \)):
\[ f'(0) = 6(0)^2 + 18(0) + 12 = 12 \quad (\text{positive}) \]

**Hint**: Calculate the derivative at the test points to determine if the function is increasing or decreasing.

7. **Conclusion**:
- The function is increasing on the intervals \( (-\infty, -2) \) and \( (-1, \infty) \).
- The function is decreasing on the interval \( (-2, -1) \).

**Final Answer**: The function \( f(x) \) is increasing on \( (-\infty, -2) \cup (-1, \infty) \) and decreasing on \( (-2, -1) \).