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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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