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Using y' = 6 - m = f' 1 = 6 Step 3 5 2 1 5 ,x-1/2 -1/2, we can now find the slope of the tangent line at (1, 1) by finding the following. Now since 1-1/2 m = f'(1) = 1 1 -1/2 Submit Skip (you cannot come back) we have the following.

Using y' = 6 - m = f' 1 = 6 Step 3 5 2 1 5 ,x-1/2 -1/2, we can now find the slope of the tangent line at (1, 1) by finding the following. Now since 1-1/2 m = f'(1) = 1 1 -1/2 Submit Skip (you cannot come back) we have the following.

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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Using y' = 6 m = f' 1 = 6 Step 3 5/(0 || 1 5 2-x-1/2₁ Now since 1-1/2 m = f'(1) = 1 we can now find the slope of the tangent line at (1, 1) by finding the following. 1 -1/2 Submit Skip (you cannot come back) we have the following.
Transcribed Image Text:Using y' = 6 m = f' 1 = 6 Step 3 5/(0 || 1 5 2-x-1/2₁ Now since 1-1/2 m = f'(1) = 1 we can now find the slope of the tangent line at (1, 1) by finding the following. 1 -1/2 Submit Skip (you cannot come back) we have the following.
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