91Ó°ÊÓ

In calculus, continuity is one of the fundamental building blocks. A function is continuous over an interval if there are no breaks, jumps, or holes in the graph of the function over that range. This means you could draw the graph without lifting your pencil. For a function to be continuous at a point, the following three conditions must be met:
- The function must be defined at that point.
- The limit of the function must exist as it approaches that point.
- The limit of the function as it approaches the point must be equal to the function value at that point.
In the context of the function\( f(x) = \sqrt{(2-x^{2/3})^3} \), it becomes apparent that it is not continuous over the interval \([-1, 1]\). The function is undefined at \(x = 1\) because it makes the term under the square root negative, causing a break in the graph. Thus, it fails the condition of continuity, which is required for Rolle's Theorem.

Differentiability is a property that tells us whether a function has a derivative at each point in its domain. If a function is differentiable at a point, it means the graph of the function has a well-defined tangent line, and hence a derivative, at that point. One crucial aspect to remember is that differentiability requires continuity.
If a function is not continuous at a point, it cannot be differentiable there either. This is because the discontinuity creates an undefined or infinite rate of change at the point, which prevents the function from having a smooth tangent line.
For \(f(x)=\sqrt{(2-x^{2/3})^3} \), since the function is not continuous in the interval \([-1,1]\), it cannot be differentiable as well. This further supports why Rolle's Theorem does not apply here, since all aspects of differentiability derive from the foundation of continuity.

Rolle's Theorem is a significant result in calculus that helps us understand the behavior of differentiable functions within closed intervals. The theorem states that for a function \(f\), if:
- The function is continuous on the closed interval \([a,b]\).
- The function is differentiable on the open interval \(a, b\).
- The function values at the endpoints are equal, i.e., \(f(a) = f(b)\).
Then, there exists at least one point \(c\) in the interval \(a, b\) such that the derivative \(f'(c) = 0\). This means the tangent to the graph at this point is horizontal.
In the case of our function \(f(x)=\sqrt{(2-x^{2/3})^3}\), even though \(f(-1) = f(1)\), the function is not continuous on the interval \([-1,1]\). Therefore, it does not satisfy the first condition of Rolle's Theorem, making the theorem inapplicable. Understanding the conditions of Rolle's Theorem is crucial as any violation can prevent the theorem from being used.