Problem 173 - Calculus Volume 1

Question

Determine whether the Mean Value Theorem applies for the functions over the given interval \([a, b] .\) Justify your answer. $$ f(x)=\tan (2 \pi x) \text { over }[0,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because the function is discontinuous and not differentiable on the interval \\( [0, 2] \\\).

Step 1: Check Continuity on the Closed Interval

To apply the Mean Value Theorem, the function must be continuous on the closed interval \( [0, 2] \). The function \( f(x) = \tan(2 \pi x) \) is discontinuous whenever its argument \( 2\pi x \) is an odd multiple of \( \frac{\pi}{2} \). Solving \( 2\pi x = \frac{\pi}{2} + k\pi \), for integers \( k \), gives points of discontinuity at \( x = \frac{1}{4} + \frac{k}{2} \). The point \( x = \frac{1}{2} \) is within interval \( [0, 2] \). Thus, \( f(x) \) is discontinuous in \( [0, 2] \).

Step 2: Check Differentiability on the Open Interval

Another condition of the Mean Value Theorem is that the function must be differentiable on the open interval \( (0, 2) \). Differentiability requires the function to be continuous, which fails at \( x = \frac{1}{2} \). Consequently, \( f(x) = \tan(2 \pi x) \) is not differentiable over \( (0, 2) \).

Step 3: Conclusion

Since \( f(x) = \tan(2 \pi x) \) is neither continuous over \( [0, 2] \) nor differentiable over \( (0, 2) \), it does not satisfy the requirements of the Mean Value Theorem. Hence, the theorem cannot be applied.

Key Concepts

Continuity

To determine if the Mean Value Theorem applies to a function, we first check for continuity on a closed interval. A function is continuous at a point if the following conditions are met:

• The function is defined at that point.
• The limit of the function as it approaches the point from both sides exists.
• The limit equals the function's value at the point.

When a function is continuous over an entire interval, it means the above conditions hold for all points in that interval.
In the given exercise, we have the function \(f(x) = \tan(2 \pi x)\), which is periodic because the tangent function repeats its values in intervals of \(\pi\). Continuity issues arise particularly where the tangent function approaches vertical asymptotes, i.e., where \(2\pi x\) equals an odd multiple of \(\frac{\pi}{2}\).
The equation \(2 \pi x = \frac{\pi}{2} + k\pi\) helps identify these points. Solving gives \(x = \frac{1}{4} + \frac{k}{2}\). The critical point at \(x = \frac{1}{2}\) falls within the interval \([0, 2]\). Therefore, the function is not continuous on that interval, which means the Mean Value Theorem can't be applied.

Differentiability

Differentiability is another core condition for applying the Mean Value Theorem. A function is differentiable at a point if it has a defined derivative at that point. This means the function not only has to be continuous at that point, but also its graph should have a tangent line that is not vertical.
For differentiability, the derivative needs to be defined throughout the open interval. However, if there is any point in the interval where the function is not continuous, it automatically fails to be differentiable at that location.
For \(f(x) = \tan(2 \pi x)\), the discontinuity at \(x = \frac{1}{2}\) means it fails the differentiability test in the interval \((0, 2)\). As differentiability requires continuity throughout the interval, the function is not suitable for applying the Mean Value Theorem.

Intervals

Intervals play a crucial role in the application of the Mean Value Theorem. They define the domain over which the theorem's criteria must hold. An interval can be closed, open, or half-open (closed on one end and open on the other).

• Closed interval \([a, b]\): Includes all points from \(a\) to \(b\), and both \(a\) and \(b\) are part of the interval.
• Open interval \((a, b)\): Includes all points between \(a\) and \(b\), but excludes \(a\) and \(b\) themselves.
• Half-open interval \([a, b)\) or \((a, b]\): Include all points in between but only one of the boundary points.

In the original problem, the closed interval \([0, 2]\) is given. The Mean Value Theorem requires continuity over this entire interval and differentiability over the open interval \((0, 2)\). The discontinuity and lack of differentiability at \(x = \frac{1}{2}\) within these intervals prevent the theorem from being applied effectively.