Problem 45 - Calculus of a Single Variable

Question

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). If the Mean Value Theorem cannot be applied, explain why not. \(f(x)=\sin x, \quad[0, \pi]\)

Step-by-Step Solution

Answer

The Mean Value Theorem can be applied for the function \(f(x) = \sin x\) on the closed interval \([0, \pi]\). The value of \(c\) that satisfies the condition of the Mean Value Theorem for this function and interval is \(c = \frac{\pi}{2}\).

Step 1: Check the Application Conditions of MVT

The Mean Value Theorem is applicable if the function is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). The function \(f(x) = \sin x\) is continuous and differentiable for all real numbers, thus it's continuous on \([0, \pi]\) and differentiable on \((0, \pi)\). Therefore, the Mean Value Theorem can be applied.

Step 2: Apply the Mean Value Theorem

According to the Mean Value Theorem, there exists a number \(c\) in the open interval \((0, \pi)\) where the derivative of the function equals the average rate of change of the function on the interval. That is, \(f'(c) = \frac{f(\pi) - f(0)}{\pi - 0}\). Substitute \(f(x) = \sin x\), we have, \(f'(c) = \cos(c) = \frac{\sin(\pi) - \sin(0)}{\pi - 0} = 0.\) So, we need to solve the equation \(\cos(c) = 0\) for \(c\) in the interval \((0, \pi)\).

Step 3: Solve for the Value of \(c\)

The equation \(\cos(c) = 0\) has solutions \(c = \frac{\pi}{2} + n\pi\) for \(n\) being any integer. However, we need to find the solution in the open interval \((0, \pi)\). The only value satisfying this is \(c = \frac{\pi}{2}\). Thus, \(c = \frac{\pi}{2}\) is the solution.

Key Concepts

The Importance of Continuity

Continuity is a fundamental concept in calculus that plays a key role in the Mean Value Theorem (MVT). A function is continuous on a closed interval \[a, b\] if there are no breaks, jumps, or holes in the graph of the function within this interval. For MVT to be applicable, the function must be continuous on \[a, b\].
This ensures that the function behaves predictably and smoothly throughout the interval, allowing us to calculate meaningful averages and derivatives.
When dealing with trigonometric functions like \((x) = \ sin x\), continuity is generally assured because trigonometric functions are continuous over their entire domains. This means functions such as sine and cosine have no breaks and fit the requirement for applying MVT.

• Verify continuity by checking for breaks or discontinuities in the interval.
• If continuous, one MVT condition is satisfied.

Understanding Derivability

Derivability refers to the ability of a function to have a derivative. To apply the Mean Value Theorem, a function must be differentiable on an open interval \(a, b\). This means we should be able to calculate the derivative at every point within this interval.
For \(f(x) = \sin x\), it is not only continuous everywhere but also differentiable. This trigonometric function has a derivative of \(f'(x) = \cos x\), which can be calculated smoothly and constantly across any interval including \(0, \pi\).
If both continuity and differentiability are satisfied, the MVT can be applied. Differentiability ensures that there are no sharp turns or cusps, which makes the math straightforward.

• Check the differentiability at each point in the open interval.
• No sharp turns or undefined points mean MVT can be applied.

Exploring Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in mathematics, particularly in calculus.
The function \(f(x) = \sin x\) exhibits smooth, wave-like behavior and is periodic, meaning it repeats its values in regular intervals.
This periodic behavior contributes to both its continuity and derivability across any real number, enhancing its reliability for calculus applications like MVT.
These functions are known for their endless continuity and simple derivatives:

• Derivatives: \(f'(x) = \cos x\).
• Periodic and continuous across all real numbers.

Understanding these properties helps in tackling problems involving intervals and derivatives easily. This ensures that any results or solutions drawn are mathematically sound.

Interval Analysis in Calculus

Working with intervals is crucial when applying the Mean Value Theorem.
An interval defines the section of the function where we're analyzing behavior. For the function \(f(x) = \sin x\) over \([0, \pi]\), we look at how the function behaves between these end-points.
Interval analysis involves:

• Closed interval \[a, b\]: where the function must be continuous.
• Open interval \(a, b\): where the function must be differentiable.

For the given problem, \(c = \frac{\pi}{2}\) is the point within \(0, \pi\) where the derivative equals the average rate of change. This point is important because it connects the instantaneous rate (calculated by the derivative) to the average rate, illustrating MVT.
Knowing how intervals work and how to analyze them helps understand where certain mathematical theorems, like MVT, apply and yield meaningful results.