Let's analyze each option based on Rolle's Theorem. For a function $f$ to satisfy Rolle's theorem on a closed interval $[a, b]$, it must meet three conditions:

1. Continuity on $[a, b]$.
2. Differentiability on the open interval $(a, b)$.
3. Equal values at the endpoints: $f(a) = f(b)$.

Given the function $f(x) = |x|$ on the interval $[-1, 1]$:

1. $f(x)$ is continuous on $[-1, 1]$, since absolute value functions are continuous everywhere.
2. $f(x)$ is not differentiable at $x = 0$, as there is a cusp at $x = 0$.
3. $f(-1) = |{-1}| = 1$ and $f(1) = |1| = 1$, so $f(-1) = f(1)$.

Since the function fails the differentiability condition (it’s not differentiable at $x = 0$), the correct answer is:

The function $f$ does not satisfy the hypotheses of Rolle's theorem on the interval $[-1,1]$, since $f$ is not differentiable at $x = 0$.

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Related Questions

1. How do we determine if a function is differentiable at a point?
2. What are other examples of functions that are continuous but not differentiable?
3. How would the result change if the interval was $[0,1]$ instead of $[-1,1]$?
4. Why is continuity alone not enough to apply Rolle's Theorem?
5. Can Rolle's Theorem be applied to functions with endpoints not included in the interval?

Tip

When checking if Rolle's Theorem applies, always verify differentiability over the entire open interval—not just at the endpoints.