Stitz-Zeager_College_Algebra_e-book

The graph of f has a sharp turn or cusp while g does

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Unformatted text preview: 2. (a) We begin with the assumption that g (c) = g (d) and try to show c = d. g (c) 2c 1−c 2c(1 − d) 2c − 2cd 2c c We have shown that g is one-to-one. = g (d) 2d = 1−d = 2d(1 − c) = 2d − 2dc = 2d =d 298 Further Topics in Functions (b) We can graph g using the six step procedure outlined in Section 4.2. We get the sole intercept at (0, 0), a vertical asymptote x = 1 and a horizontal asymptote (which the graph never crosses) y = −2. We see from that the graph of g passes the Horizontal Line Test. y y 4 3 2 1 2 1 x −2 −1 1 −2 2 −1 −1 x 1 2 −2 −3 −4 −5 −6 −1 −2 y = f (x) y = g (x) 3. (a) We begin with h(c) = h(d). As we work our way through the problem, we encounter a nonlinear equation. We move the non-zero terms to the left, leave a 0 on the right and factor accordingly. h(c) c2 − 2c + 4 c2 − 2c 2 − d2 − 2c + 2d c (c + d)(c − d) − 2(c − d) (c − d)((c + d) − 2) c−d=0 c=d = = = = = = or or h(d) d2 − 2d + 4 d2 − 2d 0 0 0 factor by grouping c+d−2=0 c=2−d We get c = d as one p...
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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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