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Stitz-Zeager_College_Algebra_e-book

# In this case the parentheses here do not indicate

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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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