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

Options

A. `[-1, infty)`
B. `(-infty, - 2]`
C. `[-2, - 1]`
D. `[-1,1]`

Answer

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

### Step 1: Find the derivative of the function
To find where the function is decreasing, we first need to compute its derivative \( f'(x) \).

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

Calculating the derivative:

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

### Step 2: Set the derivative less than zero
Next, we need to find the intervals where the function is decreasing, which occurs when the derivative is less than zero:

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

### Step 3: Factor the derivative
We can factor out the common term from the derivative:

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

Now, we can factor the quadratic expression:

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

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 intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).

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

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

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

### Step 6: Conclusion
The function \( f(x) \) is decreasing in the interval where \( f'(x) < 0 \), which is:

\[ (-2, -1) \]

Thus, the interval on which the function \( f(x) \) is decreasing is \( (-2, -1) \).