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

Answer

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

### Step-by-step Solution:

1. **Check the conditions of Rolle's Theorem**:
- The function \( f(x) \) must be continuous on the closed interval \([-3, 0]\) and differentiable on the open interval \((-3, 0)\).
- We can see that \( f(x) \) is a product of polynomials and the exponential function, which are both continuous and differentiable everywhere. Therefore, \( f(x) \) satisfies the conditions of Rolle's theorem.

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

3. **Find the derivative \( f'(x) \)**:
- We will use the product rule to differentiate \( f(x) \):
\[ f(x) = x(x + 3)e^{-\frac{1}{2}x} \]
- Let \( u = x(x + 3) \) and \( v = e^{-\frac{1}{2}x} \).
- The derivative using the product rule \( (uv)' = u'v + uv' \):
- Calculate \( u' \):
\[ u' = (x^2 + 3x)' = 2x + 3 \]
- Calculate \( v' \):
\[ v' = e^{-\frac{1}{2}x} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2}e^{-\frac{1}{2}x} \]
- Now apply the product rule:
\[ f'(x) = (2x + 3)e^{-\frac{1}{2}x} + x(x + 3)\left(-\frac{1}{2}e^{-\frac{1}{2}x}\right) \]
- Factor out \( e^{-\frac{1}{2}x} \):
\[ f'(x) = e^{-\frac{1}{2}x} \left( (2x + 3) - \frac{1}{2}x(x + 3) \right) \]

4. **Set \( f'(c) = 0 \)**:
- We need to solve:
\[ (2c + 3) - \frac{1}{2}c(c + 3) = 0 \]
- Multiply through by 2 to eliminate the fraction:
\[ 2(2c + 3) - c(c + 3) = 0 \]
- Simplifying gives:
\[ 4c + 6 - c^2 - 3c = 0 \implies -c^2 + c + 6 = 0 \]
- Rearranging:
\[ c^2 - c - 6 = 0 \]

5. **Solve the quadratic equation**:
- Factor the quadratic:
\[ (c - 3)(c + 2) = 0 \]
- Thus, the solutions are:
\[ c = 3 \quad \text{or} \quad c = -2 \]

6. **Determine which value of \( c \) lies in the interval \((-3, 0)\)**:
- The value \( c = 3 \) is not in the interval \((-3, 0)\).
- The value \( c = -2 \) is in the interval \((-3, 0)\).

### Conclusion:
The value of \( c \) that satisfies the conditions of Rolle's theorem is:
\[ \boxed{-2} \]