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Rolle's Theorem applies to \( f(x) = \sin(5x) \) on \( \left[\frac{\pi}{5}, \frac{2\pi}{5}\right] \) because the function is continuous and differentiable on the interval, and \( f\left(\frac{\pi}{5}\right) = f\left(\frac{2\pi}{5}\right) = 0 \). Therefore, there exists at least one \( c \) in \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \) where \( f'(c) = 0 \).
Solución
To determine whether **Rolle's Theorem** applies to the function \( f(x) = \sin(5x) \) on the interval \( \left[\frac{\pi}{5}, \frac{2\pi}{5}\right] \), we'll verify the three necessary conditions of the theorem:
### **Rolle's Theorem Conditions**
1. **Continuity on \([a, b]\):**
- The function \( f(x) = \sin(5x) \) is a sine function, which is continuous for all real numbers.
- **Conclusion:** \( f(x) \) is continuous on \( \left[\frac{\pi}{5}, \frac{2\pi}{5}\right] \).
2. **Differentiability on \((a, b)\):**
- The function \( f(x) = \sin(5x) \) is differentiable everywhere since it's a smooth trigonometric function.
- **Conclusion:** \( f(x) \) is differentiable on \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \).
3. **Equal Function Values at the Endpoints:**
- Evaluate \( f \) at the endpoints:
\[
f\left(\frac{\pi}{5}\right) = \sin\left(5 \cdot \frac{\pi}{5}\right) = \sin(\pi) = 0
\]
\[
f\left(\frac{2\pi}{5}\right) = \sin\left(5 \cdot \frac{2\pi}{5}\right) = \sin(2\pi) = 0
\]
- **Conclusion:** \( f\left(\frac{\pi}{5}\right) = f\left(\frac{2\pi}{5}\right) = 0 \).
### **Applying Rolle's Theorem**
Since all three conditions are satisfied, **Rolle's Theorem** guarantees that there exists at least one number \( c \) in \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \) such that:
\[
f'(c) = 0
\]
### **Finding \( c \)**
1. **Compute the derivative of \( f(x) \):**
\[
f'(x) = \frac{d}{dx} \sin(5x) = 5\cos(5x)
\]
2. **Set the derivative equal to zero and solve for \( c \):**
\[
5\cos(5c) = 0 \implies \cos(5c) = 0
\]
\[
\cos(5c) = 0 \implies 5c = \frac{\pi}{2} + k\pi \quad \text{for some integer } k
\]
\[
c = \frac{\pi}{10} + \frac{k\pi}{5}
\]
3. **Determine \( c \) within the interval \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \):**
- Let's find \( k \) such that \( c \) falls within \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \):
\[
\frac{\pi}{5} < \frac{\pi}{10} + \frac{k\pi}{5} < \frac{2\pi}{5}
\]
- Solving for \( k \), we find \( k = 1 \):
\[
c = \frac{\pi}{10} + \frac{\pi}{5} = \frac{3\pi}{10}
\]
- **Verification:**
\[
\frac{\pi}{5} = \frac{2\pi}{10} < \frac{3\pi}{10} < \frac{4\pi}{10} = \frac{2\pi}{5}
\]
- \( c = \frac{3\pi}{10} \) lies within the interval \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \).
### **Conclusion**
**Rolle's Theorem** **does apply** to the function \( f(x) = \sin(5x) \) on the interval \( \left[\frac{\pi}{5}, \frac{2\pi}{5}\right] \). Specifically, there exists a point \( c = \frac{3\pi}{10} \) in the open interval \( \left(\frac{\pi}{5}, \frac{2\pi}{5}\right) \) where the derivative \( f'(c) = 0 \).
Revisado y aprobado por el equipo de tutoría de UpStudy
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