Mean Value Theorem Examples With Solutions

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Apr 25, 2025 · 6 min read

Mean Value Theorem Examples With Solutions
Mean Value Theorem Examples With Solutions

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    Mean Value Theorem Examples with Solutions: A Comprehensive Guide

    The Mean Value Theorem (MVT) is a cornerstone of calculus, bridging the gap between the instantaneous rate of change (derivative) and the average rate of change over an interval. Understanding the MVT is crucial for many advanced calculus concepts and applications. This comprehensive guide provides numerous examples with detailed solutions, helping you grasp the theorem and its implications.

    Understanding the Mean Value Theorem

    Before diving into examples, let's formally state the Mean Value Theorem:

    If a function f(x) 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 the open interval (a, b) such that:

    f'(c) = [f(b) - f(a)] / (b - a)

    In simpler terms: The instantaneous rate of change at some point c within the interval equals the average rate of change over the entire interval. Geometrically, this means there's at least one point on the curve where the tangent line is parallel to the secant line connecting the endpoints (a, f(a)) and (b, f(b)).

    Key Conditions for the Mean Value Theorem

    It's crucial to remember that the MVT has two essential prerequisites:

    • Continuity: The function f(x) must be continuous on the closed interval [a, b]. This means the function has no breaks or jumps within the interval.
    • Differentiability: The function f(x) must be differentiable on the open interval (a, b). This means the function has a derivative at every point within the interval (no sharp corners or vertical tangents).

    Examples and Solutions: A Step-by-Step Approach

    Let's tackle a range of examples, progressing from simple to more complex scenarios. Each example will follow a consistent problem-solving structure:

    1. Verify conditions: Check if the function satisfies the continuity and differentiability requirements.
    2. Find the average rate of change: Calculate [f(b) - f(a)] / (b - a).
    3. Find the derivative: Calculate f'(x).
    4. Solve for c: Set f'(c) equal to the average rate of change and solve for c.
    5. Verify c is in (a, b): Ensure the value of c falls within the open interval (a, b).

    Example 1: A Simple Polynomial

    Let f(x) = x² + 2x on the interval [0, 1].

    1. Conditions: f(x) is a polynomial, so it's continuous and differentiable everywhere, including on [0, 1].

    2. Average rate of change: [(1)² + 2(1) - (0)² - 2(0)] / (1 - 0) = 3

    3. Derivative: f'(x) = 2x + 2

    4. Solve for c: 2c + 2 = 3 => 2c = 1 => c = 1/2

    5. Verification: c = 1/2 is in (0, 1).

    Example 2: A Trigonometric Function

    Let f(x) = sin(x) on the interval [0, π].

    1. Conditions: sin(x) is continuous and differentiable everywhere.

    2. Average rate of change: [sin(π) - sin(0)] / (π - 0) = 0

    3. Derivative: f'(x) = cos(x)

    4. Solve for c: cos(c) = 0. This occurs at c = π/2.

    5. Verification: c = π/2 is in (0, π).

    Example 3: A Function with a Potential Issue

    Let f(x) = √x on the interval [0, 4].

    1. Conditions: √x is continuous on [0, 4], but its derivative, 1/(2√x), is undefined at x = 0. However, it's differentiable on (0, 4).

    2. Average rate of change: [√4 - √0] / (4 - 0) = 1/2

    3. Derivative: f'(x) = 1/(2√x)

    4. Solve for c: 1/(2√c) = 1/2 => √c = 1 => c = 1

    5. Verification: c = 1 is in (0, 4).

    Example 4: A More Complex Function

    Let f(x) = x³ - 3x + 2 on the interval [-2, 2].

    1. Conditions: f(x) is a polynomial, continuous and differentiable everywhere.

    2. Average rate of change: [(2)³ - 3(2) + 2 - ((-2)³ - 3(-2) + 2)] / (2 - (-2)) = 0

    3. Derivative: f'(x) = 3x² - 3

    4. Solve for c: 3c² - 3 = 0 => c² = 1 => c = ±1

    5. Verification: c = 1 and c = -1 are both in (-2, 2). The MVT guarantees at least one value of c; in this case, there are two.

    Example 5: A Rational Function

    Let f(x) = 1/x on the interval [1, 3].

    1. Conditions: 1/x is continuous and differentiable on (1,3). Note that it's undefined at x = 0, but this is outside the interval of interest.

    2. Average rate of change: (1/3 - 1/1) / (3 - 1) = -1/3

    3. Derivative: f'(x) = -1/x²

    4. Solve for c: -1/c² = -1/3 => c² = 3 => c = ±√3. We only consider the positive root since it must lie in (1,3).

    5. Verification: c = √3 is in (1, 3).

    Example 6: A Piecewise Function (Illustrating Limitations)

    Let f(x) = |x| on the interval [-1, 1].

    1. Conditions: |x| is continuous on [-1, 1], but it's not differentiable at x = 0. Therefore, the MVT does not apply.

    These examples demonstrate the application of the Mean Value Theorem to various function types. Remember that the MVT provides an existence theorem—it guarantees the existence of at least one c, but it doesn't provide a method to find all such c. Furthermore, always rigorously check the continuity and differentiability conditions before applying the theorem.

    Applications of the Mean Value Theorem

    The Mean Value Theorem isn't just a theoretical concept; it has numerous practical applications in various fields:

    • Physics: Determining the instantaneous velocity of an object given its average velocity.
    • Engineering: Analyzing the rate of change in various processes.
    • Economics: Studying the marginal cost or revenue.
    • Computer Science: Approximating functions and solving numerical problems.

    Advanced Concepts and Extensions

    The Mean Value Theorem serves as a foundation for more advanced theorems, including:

    • Cauchy's Mean Value Theorem: A generalization of the MVT that deals with the ratio of the derivatives of two functions.
    • L'Hôpital's Rule: Used to evaluate indeterminate forms (0/0 or ∞/∞) of limits. The proof of L'Hôpital's Rule relies heavily on the MVT.

    Conclusion

    Mastering the Mean Value Theorem is essential for any serious student of calculus. By understanding its conditions and applications, you can unlock deeper insights into the behavior of functions and solve a wide variety of problems. This comprehensive guide, with its diverse examples and step-by-step solutions, provides a solid foundation for tackling more complex problems and exploring advanced calculus concepts. Remember to practice regularly and work through various problems to solidify your understanding. The more examples you work through, the more comfortable and confident you'll become in applying the Mean Value Theorem.

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