Rolle's Theorem states that if a function $f(x)$ satisfies the following conditions on a closed interval $[a, b]$:

1. $f(x)$ is continuous on $[a, b]$,
2. $f(x)$ is differentiable on $(a, b)$,
3. $f(a) = f(b)$,

then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

Analysis of $f(x) = \frac{1}{x^2}$ on $[-1, 1]$:

1. Continuity:
   The function $f(x) = \frac{1}{x^2}$ is not continuous at $x = 0$, as it is undefined there. Therefore, $f(x)$ is discontinuous on $[-1, 1]$.

2. Differentiability:
   Since $f(x)$ is discontinuous at $x = 0$, it is also not differentiable on $[-1, 1]$.

3. Equal Function Values at Endpoints:
   While $f(-1) = f(1) = 1$, the lack of continuity and differentiability means Rolle's Theorem does not apply.

Conclusion:

The function $f(x) = \frac{1}{x^2}$ fails the conditions of Rolle's Theorem because it is not continuous and differentiable on the given interval. Thus, Rolle's Theorem cannot be verified for $f(x) = \frac{1}{x^2}$ on $[-1, 1]$.

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Do you have further questions or want more clarification on this?

Here are 5 related questions for exploration:

1. What is the significance of continuity and differentiability in Rolle's Theorem?
2. Can Rolle's Theorem be extended to open intervals?
3. How does Rolle's Theorem relate to the Mean Value Theorem?
4. What happens if $f(x)$ has a discontinuity within the interval $[a, b]$?
5. How can we visually interpret Rolle's Theorem for specific functions?

Tip: Always check the continuity and differentiability of a function first before applying Rolle's Theorem!