To determine whether $f(x) = |x|$ satisfies Rolle's Theorem on the interval $[-1, 1]$, we need to verify the following three conditions:

Rolle's Theorem Conditions:

1. $f(x)$ is continuous on the closed interval $[a, b]$:

   • The function $f(x) = |x|$ is continuous for all real numbers, including on $[-1, 1]$. Therefore, this condition is satisfied.

2. $f(x)$ is differentiable on the open interval $(a, b)$:

   • The derivative of $f(x) = |x|$ is: \begin{cases} 1 & \text{if } x > 0, \\ -1 & \text{if } x < 0. \end{cases}$$ However, $$f'(x)$$ is undefined at $$x = 0$$. Since $$x = 0$$ lies within the open interval $$(-1, 1)$$, $$f(x)$$ is not differentiable on the entire open interval. Thus, this condition is **not satisfied**.

3. $f(a) = f(b)$:

   • At the endpoints of the interval $[-1, 1]$, $f(-1) = |{-1}| = 1$ and $f(1) = |1| = 1$. Therefore, $f(-1) = f(1)$, and this condition is satisfied.

Conclusion:

Although $f(x) = |x|$ satisfies the conditions of continuity on $[-1, 1]$ and equal function values at the endpoints, it does not satisfy the differentiability condition on $(-1, 1)$. Therefore, Rolle's Theorem does not apply to $f(x) = |x|$ on the interval $[-1, 1]$.

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Would you like to explore similar examples or delve deeper into this topic? Here are some related questions to consider:

1. What happens if we modify the interval for $f(x) = |x|$?
2. Can a function fail Rolle's Theorem due to endpoints instead of differentiability?
3. How does differentiability impact the application of Rolle's Theorem?
4. What is an example of a function that does satisfy Rolle's conditions on a specific interval?
5. How does the absolute value function behave near sharp points or corners?

Tip: Rolle’s Theorem often hinges on differentiability, so check for sharp turns or cusps in the graph of a function!