Let `f (x) =x ^(3) + 6x ^(2) + ax +2, if (-3, -1)` is the largest possible interval for which `f (x)` is decreasing function, then `a=`

Options

A. 3
B. 9
C. `-2`
D. 1

Answer

To find the value of \( a \) such that the function \( f(x) = x^3 + 6x^2 + ax + 2 \) is decreasing on the interval \((-3, -1)\), we need to follow these steps:

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

\[ f'(x) = \frac{d}{dx}(x^3 + 6x^2 + ax + 2) = 3x^2 + 12x + a \]

### Step 2: Set the derivative less than or equal to zero
For the function to be decreasing on the interval \((-3, -1)\), the derivative must be negative in that interval:

\[ f'(x) < 0 \quad \text{for } x \in (-3, -1) \]

### Step 3: Analyze the quadratic \( 3x^2 + 12x + a \)
The expression \( 3x^2 + 12x + a \) is a quadratic function. Since the coefficient of \( x^2 \) is positive, the parabola opens upwards. For the quadratic to be negative in the interval \((-3, -1)\), it must have roots at \( x = -3 \) and \( x = -1 \).

### Step 4: Find the roots of the quadratic
The roots of the quadratic can be expressed using the factored form:

\[ f'(x) = 3(x + 3)(x + 1) \]

Expanding this gives:

\[ f'(x) = 3(x^2 + 4x + 3) = 3x^2 + 12x + 9 \]

### Step 5: Compare coefficients
Now we can compare this with our expression for \( f'(x) \):

\[ 3x^2 + 12x + a = 3x^2 + 12x + 9 \]

From this, we can see that:

\[ a = 9 \]

### Conclusion
Thus, the value of \( a \) such that \( f(x) \) is decreasing on the interval \((-3, -1)\) is:

\[ \boxed{9} \]