`f(x)=6-12x-18x^(2)`

Options

A. `I_(1)=(-infty, -3),I_(2)=(-3, infty)`
B. `I_(1)=(- infty, -(1)/(3)), I_(2)=(-(1)/(3), infty)`
C. `I_(1)=(-2,1),I_(2)=(1,3)`
D. none of these

Answer

To solve the problem of determining where the function \( f(x) = 6 - 12x - 18x^2 \) is increasing or decreasing, 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 \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(6 - 12x - 18x^2) \]

Calculating the derivative:

- The derivative of a constant (6) is 0.
- The derivative of \(-12x\) is \(-12\).
- The derivative of \(-18x^2\) is \(-36x\).

Thus, we have:

\[ f'(x) = -12 - 36x \]

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

\[ -12 - 36x = 0 \]

Solving for \( x \):

\[ -36x = 12 \\ x = -\frac{12}{36} = -\frac{1}{3} \]

### Step 3: Determine the sign of the derivative
Now we will determine where the function is increasing or decreasing by testing intervals around the critical point \( x = -\frac{1}{3} \).

We will test the intervals:
1. \( (-\infty, -\frac{1}{3}) \)
2. \( (-\frac{1}{3}, \infty) \)

#### Interval 1: \( x < -\frac{1}{3} \)
Choose a test point, for example, \( x = -1 \):

\[ f'(-1) = -12 - 36(-1) = -12 + 36 = 24 > 0 \]

This means \( f(x) \) is increasing in the interval \( (-\infty, -\frac{1}{3}) \).

#### Interval 2: \( x > -\frac{1}{3} \)
Choose a test point, for example, \( x = 0 \):

\[ f'(0) = -12 - 36(0) = -12 < 0 \]

This means \( f(x) \) is decreasing in the interval \( (-\frac{1}{3}, \infty) \).

### Step 4: Conclusion
From our analysis, we conclude that:
- The function \( f(x) \) is increasing on the interval \( (-\infty, -\frac{1}{3}) \).
- The function \( f(x) \) is decreasing on the interval \( (-\frac{1}{3}, \infty) \).

### Summary of Results
- **Increasing**: \( (-\infty, -\frac{1}{3}) \)
- **Decreasing**: \( (-\frac{1}{3}, \infty) \)