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Find the value of a so that for f(x) = x³ + ax² − 10x+6 and the interval 0 ≤ x ≤ 4 the location satisfying the Mean Value Theorem is at x = = 1.

Find the value of a so that for f(x) = x³ + ax² − 10x+6 and the interval 0 ≤ x ≤ 4 the location satisfying the Mean Value Theorem is at x = = 1.

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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**Problem 4**: Find the value of \( a \) so that for \( f(x) = x^3 + ax^2 - 10x + 6 \) and the interval \( 0 \leq x \leq 4 \) the location satisfying the Mean Value Theorem is at \( x = 1 \).
Transcribed Image Text:**Problem 4**: Find the value of \( a \) so that for \( f(x) = x^3 + ax^2 - 10x + 6 \) and the interval \( 0 \leq x \leq 4 \) the location satisfying the Mean Value Theorem is at \( x = 1 \).
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