Stitz-Zeager_College_Algebra_e-book

Hence g hx 2 simplify g hx 2 3 get common

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Unformatted text preview: for extraneous solutions. Sure enough, we see that x = 1 does not satisfy the original equation and must be discarded. Our solutions are x = − 1 and x = 0. 2 2. To solve the inequality, it may be tempting to begin as we did with the equation − namely by multiplying both sides by the quantity (x − 1). The problem is that, depending on x, (x − 1) may be positive (which doesn’t affect the inequality) or (x − 1) could be negative (which would reverse the inequality). Instead of working by cases, we collect all of the terms on one side of the inequality with 0 on the other and make a sign diagram using the technique given on page 247 in Section 4.2. 268 Rational Functions x3 − 2x + 1 x−1 3 − 2x + 1 x 1 − x+1 x−1 2 3 − 2x + 1 − x(x − 1) + 1(2(x − 1)) 2x 2(x − 1) 2x3 − x2 − x 2x − 2 ≥ 1 x−1 2 ≥0 ≥0 get a common denominator ≥0 expand Viewing the left hand side as a rational function r(x) we make a sign diagram. The only value excluded from the domain of r is x = 1 which is the sol...
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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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