Problem 2 - Calculus of a Single Variable

Question

Explain why Rolle's Theorem does not apply to the function even though there exist \(a\) and \(b\) such that \(f(a)=f(b) .\) \(f(x)=\cot \frac{x}{2}, \quad[\pi, 3 \pi]\)

Step-by-Step Solution

Answer

Rolle's Theorem does not apply to the function \(f(x)=\cot \frac{x}{2}\) in the interval \([\pi, 3 \pi]\) because the function is not continuous and differentiable at \(x=2\pi\), which falls in the interval \([\pi, 3 \pi]\).

Step 1: Analyze Continuity

The function \(f(x)=\cot \frac{x}{2}\) is discontinuous at \(x=2\pi k\) for any integer \(k\). Within the interval \([\pi, 3 \pi]\), \(x=2\pi\) is a point of discontinuity.

Step 2: Analyze Differentiability

Even if the function \(f(x)=\cot \frac{x}{2}\) were to be continuous on the whole interval, it is still not differentiable at \(x=2\pi k\). At these points, the derivative of \(f(x)\) is not defined.

Step 3: Conclusion

Rolle's Theorem does not apply to the function \(f(x)=\cot \frac{x}{2}\) in the interval \([\pi, 3 \pi]\) because this function is not continuous and differentiable in the whole interval.

Key Concepts

Understanding Continuity

In calculus, continuity is a critical concept that helps describe how a function behaves on a given interval. A function is continuous on an interval if there are no breaks, jumps, or gaps within that interval. Continuity ensures that as you move along the graph of the function, there is a smooth transition from point to point.

For a function to be continuous on an interval \[a, b\], it must meet the following criteria:

• The function is defined at every point within the interval, including the endpoints.
• The limit of the function as it approaches a point from either side exists and is equal to the function's value at that point.

In the case of the function \(f(x)=\cot \frac{x}{2}\), it is important to note that it has discontinuities at points where \(x\) makes the denominator zero, specifically at \(x = 2\pi k\) for any integer \(k\). Within the interval \[\pi, 3\pi\], this happens at \(x=2\pi\), causing a break in the function. Therefore, \(f(x)\) is not continuous over the entire interval.

Exploring Differentiability

Differentiability is the property of a function that describes how smoothly it changes and whether a tangent line can be drawn at every point within an interval. A function is differentiable on an interval if a derivative exists at every point in that interval.

For a function to be differentiable on an interval \[a, b\], it must satisfy:

• The function is continuous on the interval \[a, b\].
• The derivative of the function exists at all points within \[a, b\].

Even if a function is continuous everywhere, it doesn't mean it is differentiable everywhere. Sharp points and vertical tangents are instances where a function might be continuous but not differentiable. In the function \(f(x)=\cot \frac{x}{2}\), discontinuities at \(x=2\pi k\) also mean that the derivative does not exist at these points.

In our specific interval \[\pi, 3\pi\], \(x=2\pi\) becomes problematic, proving that \(f(x)\) is not differentiable throughout.

Discontinuous Functions and Their Traits

A discontinuous function is one that has breaks, jumps, or points where it is not defined within an interval. These characteristics prevent the function from being smooth and continuous over the interval.

Several types of discontinuity exist:

• Removable Discontinuity: A point discontinuity where a function could be made continuous if a single point is defined or redefined.
• Jump Discontinuity: The function makes a 'jump' from one value to another abruptly.
• Infinite Discontinuity: Occurs when a function approaches infinity at a point within the interval.

\(f(x)=\cot \frac{x}{2}\) specifically features an \('undefined'\) discontinuity at points like \(x=2\pi\) within \[\pi, 3\pi\]. This is due to its tangent nature, as the cotangent function tends to infinity when its argument equals a multiple of \(\pi\).

These discontinuities play crucial roles when considering Rolle's Theorem, which requires the function to be continuous over the entire closed interval.