F(x) = tan x can you apply rolle theorem on [0,pi]

To determine whether Rolle's Theorem can be applied to the function \( f(x) = \tan(x) \) on the interval \([0, \pi]\), let's check the conditions of Rolle's Theorem:

### Rolle’s Theorem Conditions:
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 analyze each condition for \( f(x) = \tan(x) \) on the interval \([0, \pi]\):

1. **Continuity on \([0, \pi]\):**
   - \( \tan(x) \) is undefined at \( x = \frac{\pi}{2} \) because \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), and \( \cos\left(\frac{\pi}{2}\right) = 0 \), leading to a vertical asymptote at \( x = \frac{\pi}{2} \). Therefore, \( \tan(x) \) is not continuous on the entire interval \([0, \pi]\).

2. **Differentiability on \((0, \pi)\):**
   - Since \( \tan(x) \) is not continuous at \( x = \frac{\pi}{2} \), it also fails to be differentiable at this point.

3. **Equality at the endpoints:**
   - We check \( f(0) = \tan(0) = 0 \) and \( f(\pi) = \tan(\pi) = 0 \). So, \( f(0) = f(\pi) \), which satisfies this condition.

### Conclusion:
Rolle's Theorem cannot be applied to \( f(x) = \tan(x) \) on \([0, \pi]\) because the function is not continuous or differentiable on the entire interval due to the vertical asymptote at \( x = \frac{\pi}{2} \).

Would you like further details on Rolle's Theorem, or do you have any questions?

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Here are five related questions:
1. What is the formal statement of Rolle’s Theorem?
2. Can Rolle’s Theorem be applied to polynomials?
3. How does the Intermediate Value Theorem relate to Rolle's Theorem?
4. What are some real-life applications of Rolle’s Theorem?
5. Can we apply the Mean Value Theorem to \( f(x) = \tan(x) \) on \([0, \pi]\)?

**Tip:** To apply Rolle’s Theorem, always verify the continuity and differentiability of the function over the given interval.

Explore whether Rolle's Theorem can be applied to f(x) = tan(x) on the interval [0, π]. This solution checks the conditions of continuity, differentiability, and endpoint equality, and explains why the function does not meet all the requirements due to a vertical asymptote at x = π/2. Suitable for undergraduate calculus students.

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