The function `f(x)=x(x+3)e^(-(1/2)x)` satisfies the conditions of Rolle's theorem in (-3,0). The value of c, is

Options

A. 0
B. -1
C. -2
D. -3

Explanation

To solve the problem, we need to find the value of \( c \) in the function \( f(x) = x(x+3)e^{-\frac{1}{2}x} \) that satisfies the conditions of Rolle's theorem in the interval \((-3, 0)\).

### Step 1: Verify the conditions of Rolle's theorem
Rolle's theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Check continuity and differentiability**:
- The function \( f(x) = x(x+3)e^{-\frac{1}{2}x} \) is a product of polynomial and exponential functions. Both polynomials and exponentials are continuous and differentiable everywhere. Therefore, \( f(x) \) is continuous on \([-3, 0]\) and differentiable on \((-3, 0)\).

2. **Check the endpoints**:
- Calculate \( f(-3) \) and \( f(0) \):
\[ f(-3) = (-3)(-3 + 3)e^{-\frac{1}{2}(-3)} = (-3)(0)e^{\frac{3}{2}} = 0 \]
\[ f(0) = (0)(0 + 3)e^{-\frac{1}{2}(0)} = 0 \]
- Since \( f(-3) = f(0) = 0 \), the conditions of Rolle's theorem are satisfied.

### Step 2: Differentiate the function
Now we need to find \( f'(x) \) and set it to zero to find \( c \).

1. **Differentiate using the product rule**:
\[ f'(x) = \frac{d}{dx}[x(x+3)] \cdot e^{-\frac{1}{2}x} + x(x+3) \cdot \frac{d}{dx}[e^{-\frac{1}{2}x}] \]
- The derivative of \( x(x+3) \) is \( 2x + 3 \).
- The derivative of \( e^{-\frac{1}{2}x} \) is \( -\frac{1}{2} e^{-\frac{1}{2}x} \).

Thus,
\[ f'(x) = (2x + 3)e^{-\frac{1}{2}x} + x(x+3)\left(-\frac{1}{2} e^{-\frac{1}{2}x}\right) \]
\[ = e^{-\frac{1}{2}x}\left((2x + 3) - \frac{1}{2}x(x + 3)\right) \]

### Step 3: Set the derivative to zero
Now, we set \( f'(x) = 0 \):
\[ (2x + 3) - \frac{1}{2}x(x + 3) = 0 \]
Multiply through by 2 to eliminate the fraction:
\[ 2(2x + 3) - x(x + 3) = 0 \]
\[ 4x + 6 - x^2 - 3x = 0 \]
\[ -x^2 + x + 6 = 0 \]
Rearranging gives:
\[ x^2 - x - 6 = 0 \]

### Step 4: Factor the quadratic equation
Now we factor the quadratic:
\[ (x - 3)(x + 2) = 0 \]
Thus, the solutions are:
\[ x = 3 \quad \text{or} \quad x = -2 \]

### Step 5: Determine the valid solution for \( c \)
Since we are looking for \( c \) in the interval \((-3, 0)\), we discard \( c = 3 \) and keep:
\[ c = -2 \]

### Final Answer
The value of \( c \) is \( -2 \).

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