Stitz-Zeager_College_Algebra_e-book

# 1 graph f using transformations and state the domain

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Unformatted text preview: 3 the usual properties of exponents and exercise special care when reducing the 3 power 3 to 1. r(x) = 3(2 − x)1/3 − x(2 − x)−2/3 1 2 = (2 − x)−2/3 3(2 − x) 3 −(− 3 ) − x = (2 − x)−2/3 3(2 − x)3/3 − x = (2 − x)−2/3 3(2 − x)1 − x since √ 3 √3 u3 = ( 3 u) = u = (2 − x)−2/3 (6 − 4x) = (2 − x)−2/3 (6 − 4x) To solve r(x) = 0, we set (2 − x)−2/3 (6 − 4x) = 0, or or x = 3 2. 6−4x (2−x)2/3 = 0. We have 6 − 4x = 0 5.3 Other Algebraic Functions 319 • Common Denominator Approach. We rewrite r(x) = 3(2 − x)1/3 − x(2 − x)−2/3 x = 3(2 − x)1/3 − (2 − x)2/3 x 3(2 − x)1/3 (2 − x)2/3 − = 2/3 (2 − x) (2 − x)2/3 1 = = = = = common denominator 2 3(2 − x) 3 + 3 x − 2/3 (2 − x) (2 − x)2/3 x 3(2 − x)3/3 − 2/3 (2 − x) (2 − x)2/3 3(2 − x)1 x − (2 − x)2/3 (2 − x)2/3 3(2 − x) − x (2 − x)2/3 6 − 4x (2 − x)2/3 since √ 3 √3 u3 = ( 3 u) = u As before, when we set r(x) = 0 we obtain x = 3 . 2 3 We now create our sign diagram an...
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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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