г. у = эу! 3 X+l 3 y² = (x+₁).0 - 3 [1] (x+1)² ) - X - 3 (x+1)² + (x-1)0-3[1] (x-1)² (Aus) 3 (x-1)²

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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can you explain how they got the highlighted line in simple terms

2.
so
}}
3
Xtl
, hе
y² = (x+₁).0-3[1]
(x+1)²
3
X-1
3
(x+1)²
-
+
(x-1)0-3[1]
(x-1)²
3
(x-1)² (Aus)
Transcribed Image Text:2. so }} 3 Xtl , hе y² = (x+₁).0-3[1] (x+1)² 3 X-1 3 (x+1)² - + (x-1)0-3[1] (x-1)² 3 (x-1)² (Aus)
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