Problem 162 - Calculus Volume 1

Question

Use the Mean Value Theorem and find all points \(0 < c < 2\) such that \(f(2)-f(0)=f^{\prime}(c)(2-0)\) $$ f(x)=\sin (\pi x) $$

Step-by-Step Solution

Answer

The values of \( c \) are \( \frac{1}{2} \) and \( \frac{3}{2} \).

Step 1: Recall the Mean Value Theorem

The Mean Value Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that:\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Step 2: Calculate \( f(2) \) and \( f(0) \)

Compute the values of the function \( f(x) = \sin(\pi x) \) at \( x = 2 \) and \( x = 0 \):\[ f(2) = \sin(2\pi) = 0 \]\[ f(0) = \sin(0) = 0 \]

Step 3: Compute the Difference \( f(2) - f(0) \)

Subtract the function values:\[ f(2) - f(0) = 0 - 0 = 0 \]

Step 4: Find the Derivative \( f'(x) \)

Differentiate \( f(x) = \sin(\pi x) \):\[ f'(x) = \pi \cos(\pi x) \]

Step 5: Solve \( f'(c)(2-0) = 0 \)

Set up the equation using the Mean Value Theorem:\[ f'(c)(2 - 0) = 0 \]This simplifies to:\[ 2\pi \cos(\pi c) = 0 \]Divide both sides by \( 2\pi \):\[ \cos(\pi c) = 0 \]

Step 6: Solve \( \cos(\pi c) = 0 \) for \( 0 < c < 2 \)

The cosine function is zero at odd multiples of \( \frac{\pi}{2} \). Set \( \pi c = \frac{(2n + 1)\pi}{2} \) for integer \( n \):\[ c = \frac{2n + 1}{2} \]Find values of \( c \) within \( 0 < c < 2 \):- For \( n = 0 \), \( c = \frac{1}{2} \)- For \( n = 1 \), \( c = \frac{3}{2} \)These are the possible values of \( c \).

Step 7: Verify Values in Interval

Check that the values \( c = \frac{1}{2} \) and \( c = \frac{3}{2} \) fit within the interval \( 0 < c < 2 \), which they do.

Key Concepts

Understanding the Derivative of Sinusoidal Functions

When dealing with sinusoidal functions like sine and cosine, it's important to grasp how their derivatives are derived. Consider the function \( f(x) = \sin(\pi x) \). The derivative tells us the rate at which the function's value is changing with respect to \( x \).

• The derivative of \( \sin(\pi x) \) involves the chain rule, which combines the derivative of sine and the inner function \( \pi x \).
• First, calculate the derivative of the outer function \( \sin(u) \), where \( u = \pi x \). The derivative of \( \sin(u) \) is \( \cos(u) \).
• Next, multiply by the derivative of \( \pi x \), which is simply \( \pi \).

Thus, \( f'(x) = \pi \cos(\pi x) \). This tells us how the sine wave oscillates between its extrema and zeros as \( x \) changes.

Continuity and Differentiability in the Mean Value Theorem

The Mean Value Theorem (MVT) is a pivotal concept in calculus. For it to apply, a function must be both continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). This is crucial because:

• Continuity means that the function has no breaks, jumps, or holes within \([a, b]\).
• Differentiability indicates that the function has a defined derivative at every point within the interval \((a, b)\).

For the function \( f(x) = \sin(\pi x) \), it's continuous and differentiable on any interval on the real number line because trigonometric functions like sine are inherently smooth and differentiable everywhere. Therefore, the MVT guarantees that there is at least one point \( c \) in \((0, 2)\) where the instantaneous rate of change equals the average rate between the endpoints.

Approaching Calculus Problem Solving

Solving calculus problems using the Mean Value Theorem requires a structured approach. This involves:

• First, checking if the prerequisites for the theorem are satisfied: continuity and differentiability.
• Identifying the points on the interval and evaluating the function at those points.
• Compute the difference \( f(b) - f(a) \) and apply the Mean Value Theorem to find the derivative at some \( c \).
• Solve any equations that arise, as in the case with \( \cos(\pi c) = 0 \), to find suitable \( c \)-values.
• Finally, verify that these \( c \)-values lie within the specified interval.

This approach ensures that you systematically check all necessary conditions, compute correctly, and utilize calculus concepts effectively to arrive at the solution.