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Solution 1 y = (x +3)(2x − 5) y = 2x² + 6x - 5x - 15 y = 2x2 + x – 15 y = 2(2x)2-1 +1-0 y' = 4x+1 Solution 2 y = (x +3)(2x – 5) y = (x + 3)(2) + (2x − 5)(1) = 2x+6+2x − 5 y' = 4x + 1 y y = (x + 3)(2x - 5) y = 2x2 + x – 15 y = 2(2x)2-1 +1-0 y' = 4x + 1

Solution 1 y = (x +3)(2x − 5) y = 2x² + 6x - 5x - 15 y = 2x2 + x – 15 y = 2(2x)2-1 +1-0 y' = 4x+1 Solution 2 y = (x +3)(2x – 5) y = (x + 3)(2) + (2x − 5)(1) = 2x+6+2x − 5 y' = 4x + 1 y y = (x + 3)(2x - 5) y = 2x2 + x – 15 y = 2(2x)2-1 +1-0 y' = 4x + 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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interpret what the three solutions represent
Solution 1 y = (x +3)(2x − 5) y = 2x² + 6x - 5x - 15 y = 2x2 + x – 15 = 2(2x)2−1 +1-0 y' = 4x + 1 y Solution 2 y = (x +3)(2x − 5) y = (x + 3)(2) + (2x − 5)(1) = 2x+6+2x - 5 y' = 4x + 1 y Solution 3 y y = (x +3)(2x − 5) y = 2x2 + x – 15 = 2(2x)2−1 +1-0 y' = 4x + 1
Transcribed Image Text:Solution 1 y = (x +3)(2x − 5) y = 2x² + 6x - 5x - 15 y = 2x2 + x – 15 = 2(2x)2−1 +1-0 y' = 4x + 1 y Solution 2 y = (x +3)(2x − 5) y = (x + 3)(2) + (2x − 5)(1) = 2x+6+2x - 5 y' = 4x + 1 y Solution 3 y y = (x +3)(2x − 5) y = 2x2 + x – 15 = 2(2x)2−1 +1-0 y' = 4x + 1
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