Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=6-9x-2x^(2)`

Answer

To determine the intervals in which the function \( f(x) = 6 - 9x - 2x^2 \) is strictly increasing or strictly decreasing, we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(6 - 9x - 2x^2) \]

The derivative of a constant is 0, the derivative of \( -9x \) is \( -9 \), and the derivative of \( -2x^2 \) is \( -4x \). Therefore, we have:

\[ f'(x) = -9 - 4x \]

### Step 2: Set the derivative equal to zero to find critical points
Next, we need to find the critical points by setting \( f'(x) = 0 \):

\[ -9 - 4x = 0 \]

Solving for \( x \):

\[ -4x = 9 \\ x = -\frac{9}{4} \]

### Step 3: Determine the sign of the derivative in the intervals
Now, we will analyze the sign of \( f'(x) \) in the intervals determined by the critical point \( x = -\frac{9}{4} \). We will check the intervals \( (-\infty, -\frac{9}{4}) \) and \( (-\frac{9}{4}, \infty) \).

1. **For the interval \( (-\infty, -\frac{9}{4}) \)**:
Choose a test point, for example, \( x = -3 \):
\[ f'(-3) = -9 - 4(-3) = -9 + 12 = 3 \quad (\text{positive}) \]

Thus, \( f'(x) > 0 \) in this interval, indicating that the function is **increasing**.

2. **For the interval \( (-\frac{9}{4}, \infty) \)**:
Choose a test point, for example, \( x = 0 \):
\[ f'(0) = -9 - 4(0) = -9 \quad (\text{negative}) \]

Thus, \( f'(x) < 0 \) in this interval, indicating that the function is **decreasing**.

### Step 4: Summarize the intervals
From our analysis, we can summarize:

- The function \( f(x) \) is **strictly increasing** on the interval \( (-\infty, -\frac{9}{4}) \).
- The function \( f(x) \) is **strictly decreasing** on the interval \( (-\frac{9}{4}, \infty) \).

### Final Answer
- **Increasing Interval**: \( (-\infty, -\frac{9}{4}) \)
- **Decreasing Interval**: \( (-\frac{9}{4}, \infty) \)