If `f(x)=sin x/ e^(x)` in `[0,pi]` then `f(x)`

Options

A. satisfies Rolle's theorem and `c=(pi)/4` so that `f'((pi)/4)=4`
B. does not satisfy Rolle's theorem but `f'((pi)/4)gt0`
C. satisfies Rolle's theorem and `f'((pi)/4)=0`
D. satisfies Lagrange's Mean Value theorem but `f'((pi)/4)!=0`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{\sin x}{e^x} \) on the interval \([0, \pi]\) and apply Rolle's Theorem.

### Step-by-Step Solution:

1. **Check the Conditions of Rolle's Theorem:**
- **Function Definition:** The function \( f(x) = \frac{\sin x}{e^x} \) is defined for all \( x \) in \([0, \pi]\).
- **Continuity:** Since \( \sin x \) is continuous on \([0, \pi]\) and \( e^x \) is also continuous and never zero, \( f(x) \) is continuous on \([0, \pi]\).
- **Differentiability:** The function \( f(x) \) is differentiable on the open interval \( (0, \pi) \) because both \( \sin x \) and \( e^x \) are differentiable.

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

3. **Apply Rolle's Theorem:**
- According to Rolle's Theorem, since \( f(0) = f(\pi) \), there exists at least one point \( c \) in the interval \( (0, \pi) \) such that \( f'(c) = 0 \).

4. **Find the Derivative \( f'(x) \):**
- Use the quotient rule to find \( f'(x) \):
\[ f'(x) = \frac{(e^x)(\cos x) - (\sin x)(e^x)}{(e^x)^2} \]
Simplifying this gives:
\[ f'(x) = \frac{e^x \cos x - e^x \sin x}{e^{2x}} = \frac{\cos x - \sin x}{e^x} \]

5. **Set the Derivative to Zero:**
- Set \( f'(x) = 0 \):
\[ \cos x - \sin x = 0 \]
- This simplifies to:
\[ \tan x = 1 \]
- The solution to this equation in the interval \( (0, \pi) \) is:
\[ x = \frac{\pi}{4} \]

6. **Conclusion:**
- Therefore, according to Rolle's Theorem, there exists at least one point \( c = \frac{\pi}{4} \) in the interval \( (0, \pi) \) such that \( f'(c) = 0 \).

### Final Answer:
The point \( c \) where \( f'(c) = 0 \) is \( c = \frac{\pi}{4} \).