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The value of c that satisfies the Mean Value Theorem on the interval [0, 5] for the function f(x) = x³ - 6x is O 01 0 3

The value of c that satisfies the Mean Value Theorem on the interval [0, 5] for the function f(x) = x³ - 6x is O 01 0 3

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The image contains a question about the Mean Value Theorem in calculus. It asks for the value of \( c \) that satisfies the theorem on the interval \([0, 5]\) for the function \( f(x) = x^3 - 6x \). Options for the value of \( c \) are provided: - \(\frac{5}{\sqrt{3}}\) (this option is selected) - 1 - \(\frac{5}{3}\) - 0 The question requires applying the Mean Value Theorem, which states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Transcribed Image Text:The image contains a question about the Mean Value Theorem in calculus. It asks for the value of \( c \) that satisfies the theorem on the interval \([0, 5]\) for the function \( f(x) = x^3 - 6x \). Options for the value of \( c \) are provided: - \(\frac{5}{\sqrt{3}}\) (this option is selected) - 1 - \(\frac{5}{3}\) - 0 The question requires applying the Mean Value Theorem, which states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
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