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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.5.80b
Chapter 7, Problem 7.5.80b

80. Find all values of c that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.
b. f(x) = x, g(x) = x², (a, b) arbitrary

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Recall the statement of Cauchy's Mean Value Theorem: If functions \(f\) and \(g\) are continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c \in (a, b)\) such that \[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\]
Identify the given functions and compute their derivatives: \[f(x) = x \implies f'(x) = 1,\] \[g(x) = x^2 \implies g'(x) = 2x.\]
Write the equation from Cauchy's Mean Value Theorem using the derivatives and the function values at \(a\) and \(b\): \[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \implies \frac{1}{2c} = \frac{b - a}{b^2 - a^2}.\]
Simplify the right-hand side by factoring the denominator: \[b^2 - a^2 = (b - a)(b + a),\] so the equation becomes \[\frac{1}{2c} = \frac{b - a}{(b - a)(b + a)} = \frac{1}{b + a}.\]
Solve for \(c\) by cross-multiplying: \[\frac{1}{2c} = \frac{1}{b + a} \implies 2c = b + a \implies c = \frac{b + a}{2}.\]

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem generalizes the Mean Value Theorem by relating two functions f and g that are continuous on [a, b] and differentiable on (a, b). It guarantees the existence of a point c in (a, b) where the ratio of their derivatives equals the ratio of their increments: (f(b)-f(a))/(g(b)-g(a)) = f'(c)/g'(c).
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Differentiability and Continuity of Functions

For Cauchy's Mean Value Theorem to apply, both functions must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). This ensures the derivatives exist and the theorem's conclusion about the existence of c is valid.
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Application to Specific Functions f(x) = x and g(x) = x²

Applying the theorem to f(x) = x and g(x) = x² involves computing f'(x) = 1 and g'(x) = 2x, then solving (f(b)-f(a))/(g(b)-g(a)) = f'(c)/g'(c) for c in (a, b). This step requires algebraic manipulation to find the specific value(s) of c that satisfy the theorem.
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Related Practice
Textbook Question

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