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5.1 Using the Mean Value Theorem

Syllabus
2020
Topic
5.1
Level
—

Learning objectives

—FUN-1.B—Justify conclusions about functions by applying the Mean Value Theorem over an interval• FUN-1.B Justify conclusions about functions by applying the Mean Value Theorem over an interval.• FUN-1.B.1 If a function f is continuous over the interval [a, b] and differentiable over the interval (a, b), then the Mean Value Theorem guarantees a point within that open interval where the instantaneous rate of change equals the average rate of change over the interval.• Enduring understanding FUN-1: Existence theorems allow us to draw conclusions about a function’s behavior on an interval without precisely locating that behavior.

Match a Tangent Slope to the Interval’s Average Slope

The Mean Value Theorem connects the average rate across an interval to an instantaneous rate inside it. If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), then at least one cc in (a,b)(a,b) has tangent slope equal to the secant slope from aa to bb.

f'(c)=\frac{f(b)-f(a)}{b-a}\qquad\text{for some }c\in(a,b)

  1. Verify continuity on the closed interval [a,b][a,b].\n2. Verify differentiability on the open interval (a,b)(a,b).\n3. Compute the average rate (f(b)−f(a))/(b−a)(f(b)-f(a))/(b-a).\n4. State that MVT guarantees at least one suitable cc; if requested, solve f′(c)f'(c) equal to that average and keep only solutions in (a,b)(a,b).

For f(x)=x2f(x)=x^2 on [1,3][1,3], the polynomial is continuous on [1,3][1,3] and differentiable on (1,3)(1,3). The average rate is f(3)−f(1)3−1=9−12=4.\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4. Since f′(x)=2xf'(x)=2x, solve 2c=42c=4 to get c=2c=2, which lies in (1,3)(1,3).

MVT guarantees at least one point, not exactly one. If continuity or differentiability fails, the theorem gives no guarantee; that does not by itself prove that no matching point exists.