Find the interval in which `2x^(3)+9x^(2)+12x-1` is strictly increasing.

Answer

To find the interval in which the function \( f(x) = 2x^3 + 9x^2 + 12x - 1 \) is strictly increasing, we need to follow these steps:

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

\[ f'(x) = \frac{d}{dx}(2x^3 + 9x^2 + 12x - 1) \]

Using the power rule for differentiation:

\[ f'(x) = 6x^2 + 18x + 12 \]

### Step 2: Factor the derivative
Next, we can factor the derivative to find critical points.

\[ f'(x) = 6(x^2 + 3x + 2) \]

Now, we can factor the quadratic:

\[ f'(x) = 6(x + 1)(x + 2) \]

### Step 3: Find critical points
To find the critical points, we set the derivative equal to zero:

\[ 6(x + 1)(x + 2) = 0 \]

This gives us:

\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 4: Test intervals around critical points
Now we will test the intervals determined by the critical points \( x = -2 \) and \( x = -1 \):

1. **Interval \( (-\infty, -2) \)**:
- Choose \( x = -3 \):
\[ f'(-3) = 6(-3 + 1)(-3 + 2) = 6(-2)(-1) = 12 \quad (\text{positive}) \]

2. **Interval \( (-2, -1) \)**:
- Choose \( x = -1.5 \):
\[ f'(-1.5) = 6(-1.5 + 1)(-1.5 + 2) = 6(-0.5)(0.5) = -1.5 \quad (\text{negative}) \]

3. **Interval \( (-1, \infty) \)**:
- Choose \( x = 0 \):
\[ f'(0) = 6(0 + 1)(0 + 2) = 6(1)(2) = 12 \quad (\text{positive}) \]

### Step 5: Conclusion
From our tests, we find that \( f'(x) > 0 \) in the intervals \( (-\infty, -2) \) and \( (-1, \infty) \). Therefore, the function \( f(x) \) is strictly increasing in these intervals.

### Final Answer
The function \( f(x) = 2x^3 + 9x^2 + 12x - 1 \) is strictly increasing in the intervals:

\[ (-\infty, -2) \cup (-1, \infty) \]