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The gradient of a curve is dy/dx + (x^2)/2 = 3x. Find the equation of the curve if it passes through (1, 1/3). Oy=(3/2)x^2 - (x^3)/6 - 1 O y=(3/5)x^2 +(x^4)/12 + 9 O y=(5/2)x^3+(x^2)/9-17 v-(1/21x^2 (x^31/8 5

The gradient of a curve is dy/dx + (x^2)/2 = 3x. Find the equation of the curve if it passes through (1, 1/3). Oy=(3/2)x^2 - (x^3)/6 - 1 O y=(3/5)x^2 +(x^4)/12 + 9 O y=(5/2)x^3+(x^2)/9-17 v-(1/21x^2 (x^31/8 5

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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Question 1 The gradient of a curve is dy/dx + (x^2)/2 = 3x. Find the equation of the curve if it passes through (1, 1/3). Oy=(3/2)x^2 - (x^3)/6 - 1 y=(3/5)x^2 +(x^4)/12 + 9 y=(5/2)x^3+(x^2)/9-17 y=(1/2)x^2-(x^3)/8 - 5
Transcribed Image Text:Question 1 The gradient of a curve is dy/dx + (x^2)/2 = 3x. Find the equation of the curve if it passes through (1, 1/3). Oy=(3/2)x^2 - (x^3)/6 - 1 y=(3/5)x^2 +(x^4)/12 + 9 y=(5/2)x^3+(x^2)/9-17 y=(1/2)x^2-(x^3)/8 - 5
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