The interval on which `f(x)=2x^(3)+9x^(2)+12x-1` is decreasing in

Options

A. `(-1,oo)`
B. `(-2,-1)`
C. `(-oo,-2)`
D. `(-1,1)`

Answer

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

### Step 1: Find the derivative of \( f(x) \)
The first step is to differentiate the function \( f(x) \).

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

Using the power rule, we differentiate each term:

- The derivative of \( 2x^3 \) is \( 6x^2 \).
- The derivative of \( 9x^2 \) is \( 18x \).
- The derivative of \( 12x \) is \( 12 \).
- The derivative of a constant (-1) is \( 0 \).

Thus, we have:

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

### Step 2: Set the derivative less than zero
To find the intervals where \( f(x) \) is decreasing, we need to find where \( f'(x) < 0 \).

\[ 6x^2 + 18x + 12 < 0 \]

### Step 3: Factor the derivative
First, we can factor out the common factor of 6:

\[ 6(x^2 + 3x + 2) < 0 \]

Now, we need to factor the quadratic \( x^2 + 3x + 2 \):

\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]

Thus, we rewrite the inequality:

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

### Step 4: Find the critical points
The critical points occur when the expression equals zero:

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

This gives us the critical points:

\[ x = -1 \quad \text{and} \quad x = -2 \]

### Step 5: Test intervals around the critical points
We will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).

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

2. **Interval \( (-2, -1) \)**: Choose \( x = -1.5 \)
\[ f'(-1.5) = 6(-1.5 + 1)(-1.5 + 2) = 6(-0.5)(0.5) = -1.5 < 0 \]

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

### Step 6: Conclusion
From our tests, we find that \( f'(x) < 0 \) in the interval \( (-2, -1) \). Therefore, the function \( f(x) \) is decreasing on the interval:

\[ \boxed{(-2, -1)} \]