To determine which of the given functions do not satisfy the conditions of Rolle's Theorem, we need to analyze each function based on the three conditions of the theorem:
1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).
3. The function must satisfy f(a) = f(b).
Let's evaluate each option step by step.
### Step 1: Analyze Option 1 - \( f(x) = e^x \sin x \) on the interval [0, \( \frac{\pi}{2} \)]
- **Continuity**: The function \( e^x \) is continuous for all \( x \) and \( \sin x \) is also continuous for all \( x \). Therefore, \( f(x) \) is continuous on [0, \( \frac{\pi}{2} \)].
- **Differentiability**: Since both \( e^x \) and \( \sin x \) are differentiable everywhere, \( f(x) \) is differentiable on (0, \( \frac{\pi}{2} \)).
- **Check f(0) and f(\( \frac{\pi}{2} \))**:
- \( f(0) = e^0 \sin 0 = 0 \)
- \( f(\frac{\pi}{2}) = e^{\frac{\pi}{2}} \sin(\frac{\pi}{2}) = e^{\frac{\pi}{2}} \)
Since \( f(0) \neq f(\frac{\pi}{2}) \), the third condition is not satisfied.
### Conclusion for Option 1:
**Does not satisfy Rolle's Theorem.**
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### Step 2: Analyze Option 2 - \( f(x) = (x + 1)^2 (2x - 3)^5 \) on the interval [-1, \( \frac{3}{2} \)]
- **Continuity**: This is a polynomial function, which is continuous everywhere, hence continuous on [-1, \( \frac{3}{2} \)].
- **Differentiability**: Polynomial functions are also differentiable everywhere, thus differentiable on (-1, \( \frac{3}{2} \)).
- **Check f(-1) and f(\( \frac{3}{2} \))**:
- \( f(-1) = (0)^2 (-5)^5 = 0 \)
- \( f(\frac{3}{2}) = (2.5)^2 (0)^5 = 0 \)
Since \( f(-1) = f(\frac{3}{2}) \), all three conditions are satisfied.
### Conclusion for Option 2:
**Satisfies Rolle's Theorem.**
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### Step 3: Analyze Option 3 - \( f(x) = \sin |x| \) on the interval [\( \pi \), \( 2\pi \)]
- **Continuity**: The function \( \sin |x| \) is continuous everywhere, hence continuous on [\( \pi \), \( 2\pi \)].
- **Differentiability**: The function is differentiable on the interval [\( \pi \), \( 2\pi \)] since there are no points where it is not differentiable.
- **Check f(\( \pi \)) and f(\( 2\pi \))**:
- \( f(\pi) = \sin(\pi) = 0 \)
- \( f(2\pi) = \sin(2\pi) = 0 \)
Since \( f(\pi) = f(2\pi) \), all three conditions are satisfied.
### Conclusion for Option 3:
**Satisfies Rolle's Theorem.**
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### Step 4: Analyze Option 4 - \( f(x) = \sin(\frac{1}{x}) \) on the interval [-\( \frac{\pi}{2} \), \( \frac{\pi}{2} \)]
- **Continuity**: The function \( \sin(\frac{1}{x}) \) is not defined at \( x = 0 \), which lies within the interval [-\( \frac{\pi}{2} \), \( \frac{\pi}{2} \)]. Therefore, it is not continuous on this interval.
- **Differentiability**: Since the function is not continuous at \( x = 0 \), it cannot be differentiable there.
- **Check f(-\( \frac{\pi}{2} \)) and f(\( \frac{\pi}{2} \))**: Since the function is not defined at \( x = 0 \), we cannot check this condition.
### Conclusion for Option 4:
**Does not satisfy Rolle's Theorem.**
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### Final Conclusion:
The functions that do not satisfy the conditions of Rolle's Theorem are:
- **Option 1: \( f(x) = e^x \sin x \)**
- **Option 4: \( f(x) = \sin(\frac{1}{x}) \)**
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