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Chapter 5: Chapter 5

Problem 121

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(a) State and prove the Mean Value Theorem for the derivative of a real valued function of a single real variable. (b) Give a geometrical interpretation to this result.

Short Answer

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The Mean Value Theorem states that 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 point \(c \in (a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). In the proof, we define a new function \(g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)\), show that it satisfies the conditions of Rolle's theorem, and use it to demonstrate the theorem's claim. The geometrical interpretation is that there exists a point within the given interval where the tangent line to the curve is parallel to the secant line connecting the points \((a, f(a))\) and \((b, f(b))\), meaning it has the same slope as the average rate of change between the interval's endpoints.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Statement of Mean Value Theorem

The Mean Value Theorem states that: 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 point \(c \in (a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Step 2: Proof of Mean Value Theorem

1. Let's define a new function \(g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)\) 2. Then, \(g(a) = f(a) - \frac{f(b) - f(a)}{b - a}(0) = f(a)\) and \(g(b) = f(b) - \frac{f(b) - f(a)}{b - a}(b - a) = f(a)\) 3. Since \(f(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then \(g(x)\) is also continuous on \([a, b]\) and differentiable on \((a, b)\) as it is a combination of continuous and differentiable functions. 4. Now, we apply Rolle's Theorem on \(g(x)\), which states that if a function is continuous on \([a, b]\), differentiable on \((a, b)\), and \(g(a) = g(b)\), then there exists at least one point \(c \in (a, b)\) such that \(g'(c) = 0\). 5. Differentiate \(g(x)\) with respect to \(x\): \(g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}\) 6. Since Rolle's Theorem states that \(g'(c) = 0\), we plug in \(c\) and solve for \(f'(c)\): \(0 = f'(c) - \frac{f(b) - f(a)}{b - a}\) 7. Therefore, \(f'(c) = \frac{f(b) - f(a)}{b - a}\), proving the Mean Value Theorem. (b) Geometrical interpretation.

Step 3: Interpretation

The Mean Value Theorem states that, given a continuous and differentiable function over an interval, there exists a point within that interval where the derivative (the tangent line) is equal to the average rate of change between the endpoints of the interval. In other words, there is a point in the interval where the tangent line to the curve is parallel to the secant line connecting the points \((a, f(a))\) and \((b, f(b))\). This theorem is extremely useful, as it allows us to understand how the function's derivative behaves over an interval and it also has numerous applications in calculus and real analysis.

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Most popular questions from this chapter

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