The value of c in Rolle's theorem for the function `f (x) =(x(x+1))/e^x` defined on `[-1,0]` is

Options

A. `0.5`
B. `(1+sqrt(5))/(2)`
C. `(1-sqrt(5))/(2)`
D. `-0.5`

Answer

To find the value of \( c \) in Rolle's theorem for the function \( f(x) = \frac{x(x+1)}{e^x} \) defined on the interval \([-1, 0]\), we will follow these steps:

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

1. **Check continuity and differentiability**:
- The function \( f(x) = \frac{x(x+1)}{e^x} \) is a quotient of two functions where \( e^x \) is never zero. Therefore, \( f(x) \) is continuous and differentiable on \([-1, 0]\).

2. **Calculate \( f(-1) \) and \( f(0) \)**:
- \( f(-1) = \frac{(-1)(-1 + 1)}{e^{-1}} = \frac{0}{\frac{1}{e}} = 0 \)
- \( f(0) = \frac{0(0 + 1)}{e^0} = \frac{0}{1} = 0 \)

Since \( f(-1) = f(0) \), the conditions of Rolle's theorem are satisfied.

### Step 2: Find the derivative \( f'(x) \)
To find \( c \), we need to compute the derivative \( f'(x) \).

Using the quotient rule:
\[ f'(x) = \frac{(e^x)(2x + 1) - (x^2 + x)(e^x)}{(e^x)^2} \]
This simplifies to:
\[ f'(x) = \frac{(2x + 1)e^x - (x^2 + x)e^x}{(e^x)^2} \]
Factoring out \( e^x \):
\[ f'(x) = \frac{e^x((2x + 1) - (x^2 + x))}{(e^x)^2} = \frac{(2x + 1 - x^2 - x)e^x}{e^{2x}} = \frac{(1 + x - x^2)}{e^x} \]

### Step 3: Set the derivative equal to zero
To find \( c \), we set the numerator of \( f'(x) \) equal to zero:
\[ 1 + x - x^2 = 0 \]
Rearranging gives:
\[ -x^2 + x + 1 = 0 \quad \text{or} \quad x^2 - x - 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
This gives us two potential values for \( c \):
\[ c_1 = \frac{1 + \sqrt{5}}{2}, \quad c_2 = \frac{1 - \sqrt{5}}{2} \]

### Step 5: Determine which value is in the interval \((-1, 0)\)
- \( c_1 = \frac{1 + \sqrt{5}}{2} \) is greater than 1, hence not in the interval \([-1, 0]\).
- \( c_2 = \frac{1 - \sqrt{5}}{2} \) is approximately \(-0.618\), which lies in the interval \([-1, 0]\).

### Conclusion
The value of \( c \) in Rolle's theorem for the function \( f(x) = \frac{x(x+1)}{e^x} \) defined on \([-1, 0]\) is:
\[ c = \frac{1 - \sqrt{5}}{2} \]