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Using the Mean Value Theorem

Samuel Baker

Samuel Baker

6 min read

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Study Guide Overview

This study guide covers the Mean Value Theorem (MVT). It explains the theorem, its conditions (continuity and differentiability), and its formula. It includes a walkthrough example and practice problems applying the MVT to find a point c where the instantaneous rate of change equals the average rate of change over a given interval. The guide also reviews the concepts of continuity and differentiability.

5.1 Using the Mean Value Theorem

In the previous unit, we learned all about applying derivatives to different real-world contexts. What else are derivatives useful for? Turns out, we can also use derivatives to determine and analyze the behaviors of functions! πŸ‘€

πŸ“ˆΒ  Mean Value Theorem

The Mean Value Theorem states that if a function f is continuous over the interval [a,b][a, b] and differentiable over the interval (a,b)(a, b), then there exists a point cc within that open interval (a,b)(a,b) where the instantaneous rate of change of the function at cc equals the average rate of change of the function over the interval (a,b)(a, b).

In other words, if a function f is continuous over the interval [a,b][a, b] and differentiable over the interval (a,b)(a, b), there exists some cc on (a,b)(a,b) such that fβ€²(c)=f(b)βˆ’f(a)bβˆ’af'(c)=\frac{f(b)-f(a)}{b-a}.

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Visual representation of mean value theorem

Image Courtesy of Sumi Vora and Ethan Bilderbeek

Yet another way to phrase this theorem is that if the stated conditions of continuity and differentiability are satisfied, there is a point where the slope of the tangent line is equivalent to the slope of the secant line between aa and bb.

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Question 1 of 10

The Mean Value Theorem states that if a function ff is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then there exists a cc in (a,b)(a,b) such that which of the following is true?

fβ€²(c)=f(b)βˆ’f(a)f'(c) = f(b) - f(a)

fβ€²(c)=f(a)βˆ’f(b)bβˆ’af'(c) = \frac{f(a) - f(b)}{b - a}

fβ€²(c)=f(b)βˆ’f(a)bβˆ’af'(c) = \frac{f(b) - f(a)}{b - a}

fβ€²(c)=f(a)+f(b)bβˆ’af'(c) = \frac{f(a) + f(b)}{b - a}