Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: esents y as a function of x breaks down if we cannot solve the equation for y in terms of x. However, that does not prevent us from proving that an equation which fails to represent y as a function of x actually fails to do so. What we really need is two points with the same x-coordinate and different y -coordinates which both satisfy the equation so that the graph of the relation would fail the Vertical Line Test 1.1. Discuss with your classmates how you might find such points for the relations given below. (a) x3 + y 3 − 3xy = 0 (c) y 2 = x3 + 3x2 (b) x4 = x2 + y 2 (d) (x2 + y 2 )2 = x3 + y 3 1.4 Introduction to Functions 1.4.2 43 Answers 1. (a) Function domain = {−3, −2, −1, 0, 1, 2 ,3} range = {0, 1, 4, 9} 2. (a) Function domain = {−4, −3, −2, −1, 0, 1} range = {−1, 0, 1, 2, 3, 4} (b) Not a function (b) Not a function (c) Function domain = {−7, −3, 3, 4, 5, 6} range = {0, 4, 5, 6, 9} (c) Function domain = (−∞, ∞) range = [1, ∞) (d) Function domain = {1, 4, 9, 16, 25, 36, . . .} = {x : x is a perfect square} range = {2, 4, 6, 8, 10,...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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