Mean Value Theorem Examples With Solutions

Understanding the Mean Value Theorem: Examples and Step-by-Step Solutions

The Mean Value Theorem (MVT) is a cornerstone of differential calculus, bridging the gap between a function's average rate of change over an interval and its instantaneous rate of change at a specific point. In essence, it guarantees that for a function that is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one point c in (a, b) where the instantaneous rate of change (the derivative, f'(c)) equals the average rate of change over [a, b]. This powerful theorem is not just an abstract concept; it has profound implications in physics, engineering, and economics, proving the existence of solutions before we even find them. Mastering it through varied examples is key to unlocking deeper mathematical intuition.

Prerequisites and Geometric Intuition

Before diving into examples, two conditions are non-negotiable for the MVT to apply:

1. 2. Differentiability on (a, b]: The function must have a defined derivative at every point inside the open interval. Continuity on [a, b]: The function must have no breaks, jumps, or holes within the entire closed interval, including the endpoints. Corners, cusps, or vertical tangents within (a, b) invalidate the theorem.

Geometrically, the theorem states that for a smooth curve (satisfying the conditions), there is at least one point where the tangent line is parallel to the secant line connecting the endpoints (a, f(a)) and (b, f(b)). The slope of the secant line is the average rate of change, (f(b) - f(a))/(b - a). The theorem guarantees a point c where f'(c) equals this slope.

Step-by-Step Application Framework

Solving MVT problems follows a reliable pattern:

1. Still, Verify Conditions: Explicitly state that f(x) is continuous on [a, b] and differentiable on (a, b). For polynomials, this is automatic. For other functions, check for domain restrictions or non-differentiable points. In practice, 2. Calculate the Average Rate of Change: Compute (f(b) - f(a))/(b - a). Still, this is your target slope. But 3. Find the Derivative: Compute f'(x).
2. Set Up the Equation: Solve f'(c) = (f(b) - f(a))/(b - a) for c. But 5. Check the Interval: Confirm that your solution(s) for c lie strictly within (a, b). Discard any that are equal to a or b. Plus, 6. Interpret: State the conclusion in the context of the theorem. Also, "Which means, the Mean Value Theorem guarantees that at least one point c = ... exists in `(a, b) where f'(c) equals the average rate of change.

Detailed Worked Examples

Example 1: A Simple Polynomial

Function: f(x) = x³ - 6x² + 9x + 1 on the interval [1, 4].

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Solution:

1. Conditions: f(x) is a polynomial. All polynomials are continuous and differentiable everywhere on ℝ. Conditions are satisfied.
2. Average Rate of Change:
   • f(1) = (1)³ - 6(1)² + 9(1) + 1 = 1 - 6 + 9 + 1 = 5
   • f(4) = (4)³ - 6(4)² + 9(4) + 1 = 64 - 96 + 36 + 1 = 5
   • Average slope = (f(4) - f(1)) / (4 - 1) = (5 - 5) / 3 = 0.
3. Derivative: f'(x) = 3x² - 12x + 9.
4. Solve for c: Set f'(c) = 0. 3c² - 12c + 9 = 0 Divide by 3: c² - 4c + 3 = 0 Factor: (c - 1)(c - 3) = 0 Solutions: c = 1 or c = 3.
5. Check Interval: The interval is (1, 4). c = 1 is an endpoint, so it is not in the open interval. c = 3 is within (1, 4).
6. Conclusion: The Mean Value Theorem guarantees that at c = 3, the instantaneous rate of change f'(3) = 0 equals the average rate of change over [1, 4].

Example 2: A Trigonometric Function

Function: f(x) = sin(x) + cos(x) on the interval [0, π/2].

Solution:

1. Conditions: sin(x) and cos(x) are continuous and differentiable everywhere. Their sum is also continuous on [0, π/2] and differentiable on (0, π/2). Conditions satisfied.
2. Average Rate of Change:
   • f(0) = sin(0) + cos(0) = 0 + 1 = 1
   • f(π/2) = sin(π/2) + cos(π/2) = 1 + 0 = 1
   • Average slope = (1 - 1) / (π/2 - 0) = 0.
3. Derivative: `f'(x) = cos(x) -