If `f(x)=x^3+4x^2+ax+5` is a monotonically decreasing function of x in the largest possible interval `(-2,-2//3), then the value of a is

Options

A. 4
B. 2
C. -1
D. none of these

Explanation

To determine the value of \( a \) such that the function \( f(x) = x^3 + 4x^2 + ax + 5 \) is monotonically decreasing on the interval \( (-2, -\frac{2}{3}) \), we need to follow these steps:

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

\[ f'(x) = \frac{d}{dx}(x^3 + 4x^2 + ax + 5) = 3x^2 + 8x + a \]

### Step 2: Set the derivative to zero at the endpoints
Since the function is monotonically decreasing in the interval \( (-2, -\frac{2}{3}) \), we need to ensure that the derivative \( f'(x) \) is zero at the endpoints of this interval.

1. **At \( x = -2 \)**:
\[ f'(-2) = 3(-2)^2 + 8(-2) + a = 3(4) - 16 + a = 12 - 16 + a = a - 4 \]
Setting this equal to zero gives:
\[ a - 4 = 0 \implies a = 4 \]

2. **At \( x = -\frac{2}{3} \)**:
Next, we check the derivative at the other endpoint:
\[ f'\left(-\frac{2}{3}\right) = 3\left(-\frac{2}{3}\right)^2 + 8\left(-\frac{2}{3}\right) + a \]
Simplifying this:
\[ = 3 \cdot \frac{4}{9} - \frac{16}{3} + a = \frac{4}{3} - \frac{16}{3} + a = -\frac{12}{3} + a = -4 + a \]
Setting this equal to zero gives:
\[ -4 + a = 0 \implies a = 4 \]

### Step 3: Conclusion
Since both conditions at the endpoints yield the same value for \( a \), we conclude that the value of \( a \) for which the function \( f(x) \) is monotonically decreasing in the interval \( (-2, -\frac{2}{3}) \) is:

\[ \boxed{4} \]