Problem 35 - Essential Calculus

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 \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). $$ f(x)=\sin x,[0, \pi] $$

Step-by-Step Solution

Answer

Yes, the Mean Value Theorem can be applied to \(f(x)=\sin x\) on the interval \([0, \pi]\). The value of 'c' which satisfies the theorem is \(c=\pi/2\).

Step 1: Validate the Conditions for Mean Value Theorem

First, we need to confirm if our function \(f(x)=\sin x\) on the interval \([0, \pi]\) meets the Mean Value Theorem requirements. The function \(f(x)=\sin x\) is continuous and differentiable for all real numbers. Hence, it's continuous on the closed interval [0, \(\pi\)] and differentiable on the open interval (0, \(\pi\)). This validates the conditions for the Mean Value Theorem.

Step 2: Apply the Mean Value Theorem

Now, we apply the Mean Value Theorem which states that there is at least one 'c' in the open interval (0, \(\pi\)) such that \(f'(c)=(f(b)-f(a))/(b-a)\). Let's calculate \(f'(x)=\cos x\) and \(f(b)-f(a)=\sin\pi - \sin0 = 0\). Substituting these values in the equation we get \(\cos c = 0\).

Step 3: Solve for 'c'

Solving the equation \(\cos c = 0\) for \(c\) within the interval, we find the solutions \(c=\pi/2\) and \(c=3\pi/2\). However, \(3\pi/2\) is outside (0, \(\pi\)), so our only valid value is \(c=\pi/2\).

Key Concepts

Differentiability

To determine whether the Mean Value Theorem (MVT) applies, we first consider the concept of differentiability. Differentiability is about whether the derivative of a function exists at every point in its domain. For a function to be differentiable on an interval, it must have a derivative at each point inside that interval.

For the function \( f(x) = \sin x \) on the open interval \((0, \pi)\), we can find its derivative as \( f'(x) = \cos x \). The cosine function is defined and continuous everywhere, thus making \( \sin x \) differentiable on \((0, \pi)\).

In this step, verifying differentiability ensures that the function behaves smoothly. Being differentiable guarantees that there are no sharp corners or discontinuities in the interval, an essential criterion for applying the MVT.

Continuity

Continuity is another cornerstone concept needed when applying the Mean Value Theorem. A function is continuous on an interval if there are no breaks, jumps, or holes in its graph over that interval.

The sine function, \( f(x) = \sin x \), is continuous over all real numbers, meaning it is uninterrupted and flowing smoothly everywhere, including on our closed interval [0, \(\pi\)]. This property allows us to confidently state that \( \sin x \) is continuous on [0, \(\pi\)], satisfying one of the Mean Value Theorem’s requirements.

Validating this condition assures that the behavior of the function will be predictable and stable, allowing us to pursue further analysis using the MVT.

Sin Function

The sine function is essential in both trigonometry and calculus. Its smooth and periodic nature makes it perfect for illustrating certain calculus principles. Sine of \( x \), written \( \sin x \), oscillates between -1 and 1 and completes a full cycle every \( 2\pi \) units.

On the interval [0, \(\pi\)], the sine function starts at 0, rises to 1 at \(\pi/2\), and falls back to 0 at \(\pi\). This behavior is crucial when applying the Mean Value Theorem because we need to analyze these changes in function value. By understanding how \( \sin x \) behaves on this interval, we can accurately apply calculus techniques like differentiation and integration.

Sine’s differentiability and continuity over all real numbers make it a stepping stone in deeper mathematical concepts like the MVT.

Trigonometric Differentiation

In the context of the Mean Value Theorem, understanding trigonometric differentiation is key. It involves finding the derivative of trigonometric functions like sine, cosine, and tangent.

For \( f(x) = \sin x \), the derivative is found using basic rules as \( f'(x) = \cos x \). This derivative reveals rate changes, important for understanding how quickly things like angles or cycles change over time.

In the given problem, after finding \( f'(x) = \cos x \), the next step was to substitute into the MVT equation \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Solving \( \cos c = 0 \) yields critical values within the specified interval. Since \( \cos x \) equals zero at \( x = \pi/2 \), this value satisfies the MVT’s requirements.

This process highlights how essential it is to understand not just the behaviors of functions but also their rates of change, fitting perfectly into concepts of analysis and theoretical applications.