Question

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

Short answer

Rolle's Theorem does not apply to the function \(f(x)=\cot \frac{x}{2}\) on the interval \([\pi, 3\pi]\) because it isn't continuous on the whole interval. Despite the function being differentiable and \(f(a)=f(b)\), one failure is enough to make Rolle's Theorem inapplicable.

Step-by-step solution

Check the Continuity of the Function

First, verify if \(f(x)=\cot \frac{x}{2}\) is continuous in the interval. The cotangent function is undefined at \(x=n\pi\). Within the interval \([\pi, 3 \pi]\), there is \(x=2\pi\) where the function is discontinuous. Hence, the function \(f(x)=\cot \frac{x}{2}\) is not continuous on \([\pi, 3 \pi]\).

Check for Differentiability

Even though the function is not continuous in the entire interval (thus failing Rolle's Theorem), we can continue and check its differentiability. The function \(f(x)=\cot \frac{x}{2}\) is differentiable everywhere it is defined. Given that the derivative of \(\cot x\) is \(-\csc^2(x)\), it can be concluded that the given function is differentiable in its domain.

Check for Equal Values at the Endpoint

The third condition to check is whether \(f(a) = f(b)\). This is clearly given in the task where there are points \(a\) and \(b\) in the interval where \(f(a)=f(b)\). Thus this condition is fulfilled.

Conclusion

Rolle's theorem fails to apply to \(f(x)=\cot \frac{x}{2}\) in the given interval due to the function not being continuous on the entire interval. Despite meeting the two other conditions (differentiable between the intervals and \(f(a)=f(b)\)), Rolle's theorem cannot be applied if one condition fails.

Key concepts

Continuity in Calculus

In calculus, continuity of a function is a fundamental concept that ensures the smoothness of the function's graph. A function is said to be continuous on an interval if it does not have any sudden jumps, breaks, or holes in that interval. For a function to satisfy the conditions of Rolle's Theorem, it must be continuous on the closed interval \[a, b\].

When looking at the function in question, \(f(x)=\cot \frac{x}{2}\), it is necessary to recognize that the cotangent function can have points of discontinuity. Specifically, cotangent is undefined whenever its argument is an integer multiple of \(\pi\), which translates to points on the graph where the function will have vertical asymptotes. Within the interval \[\pi, 3\pi\], the function \(\cot \frac{x}{2}\) indeed hits a discontinuity at \(x=2\pi\), since \(\frac{2\pi}{2}=\pi\), corresponding to an undefined point for the cotangent function.

To check continuity, one must ensure there are no such points of discontinuity within the interval. Since the given function has a point of discontinuity, it violates the first condition of Rolle's Theorem, highlighting the importance of continuous behavior for its application.

Differentiability

Differentiability is another key concept in calculus, closely related to the notion of continuity. For a function to be differentiable at a point, it must be continuous at that point, and we must be able to compute its derivative there. The derivative of a function at any point represents the slope of the tangent line to the function's graph at that point. Rolle's Theorem requires that the function is not only continuous on the closed interval \[a, b\], but also that it must be differentiable on the open interval \(a, b\).

In the exercise, we encountered the function \(f(x)=\cot \frac{x}{2}\), and while it is discontinuous at \(x=2\pi\) within the interval \[\pi, 3\pi\], it is differentiable wherever it is continuous. The derivative here is found using the chain rule, taking into account that the derivative of \(\cot x\) is \( -\csc^2(x)\). Therefore, the differentiability of the function is intact where the function is continuous but does not compensate for the absence of continuity at \(x=2\pi\). Since differentiability depends on continuity, and continuity is already broken, Rolle's Theorem cannot be applied.

Cotangent Function

The cotangent function, denoted as \(\cot x\), is one of the trigonometric functions that arises from the reciprocal of the tangent function, meaning \(\cot x = \frac{1}{\tan x}\). It describes the ratio of the adjacent side to the opposite side of a right-angled triangle. The cotangent function exhibits periodic behavior with periods of \(\pi\), where it goes through its entire set of values from negative to positive infinity and is undefined at integer multiples of \(\pi\).

While the cotangent function may present continuity and differentiability within certain bounds, it has inherent discontinuities at points where the argument is an integer multiple of \(\pi\). As such, any analysis of functions involving cotangent for the purposes of applying calculus theorems like Rolle's must thoroughly investigate the behavior of the function at these critical points.