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Approximate √9.1 using tangent line approximation First note that √9.1≈ √9. Let f(x)=√x. Then, f'(x) = Let 20 = 9. Then f'(9) = 0 0 f(x), the line tangent to √ at xo = 9 is: f(x)=y= Use this result to approximate √9.1. √9.1

Approximate √9.1 using tangent line approximation First note that √9.1≈ √9. Let f(x)=√x. Then, f'(x) = Let 20 = 9. Then f'(9) = 0 0 f(x), the line tangent to √ at xo = 9 is: f(x)=y= Use this result to approximate √9.1. √9.1

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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Approximate √9.1 using tangent line approximation. First note that √9.1≈ √9. Let ƒ(x) = √x. Then, ƒ'(x) = 0 Let o 9. Then f'(9) 9. Then f'(9) = Q f), the line tangent to √ at xo = 9 is: f(x)=y= Use this result to approximate √9.1. √9.12 0
Transcribed Image Text:Approximate √9.1 using tangent line approximation. First note that √9.1≈ √9. Let ƒ(x) = √x. Then, ƒ'(x) = 0 Let o 9. Then f'(9) 9. Then f'(9) = Q f), the line tangent to √ at xo = 9 is: f(x)=y= Use this result to approximate √9.1. √9.12 0
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