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 find 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 the function
We start by calculating the first derivative \( f'(x) \) of the function \( f(x) \).

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

Using the power rule, we differentiate each term:

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

### Step 2: Set the derivative less than or equal to zero
To find where the function is decreasing, we need to solve the inequality:

\[ f'(x) < 0 \]

This translates to:

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

### Step 3: Simplify the inequality
We can simplify the inequality by dividing all terms by 6:

\[ x^2 + 3x + 2 < 0 \]

### Step 4: Factor the quadratic expression
Next, we factor the quadratic expression:

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

So, we rewrite the inequality:

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

### Step 5: Find the critical points
The critical points are found by setting the factors equal to zero:

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

### Step 6: Test intervals around the critical points
We will test the intervals determined by the critical points \( x = -2 \) and \( x = -1 \):

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

2. **Interval \( (-2, -1) \)**: Choose \( x = -1.5 \)
\[ (-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 < 0 \]

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

### Step 7: Conclusion
The function \( f(x) \) is decreasing where \( f'(x) < 0 \), which occurs in the interval:

\[ (-2, -1) \]

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

### Final Answer
The interval on which \( f(x) \) is decreasing is \( (-2, -1) \).

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