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Unformatted text preview: on, we have g (x) = 2 − x2x−7 6 . For g (x) = 2, we would need x2x−7 6 = 0.
This gives x − 7 = 0, or x = 7. Note that x − 7 is the remainder when 2x2 − 3x − 5 is divided
by x2 − x − 6, and so it makes sense that for g (x) to equal the quotient 2, the remainder from
the division must be 0. Sure enough, we ﬁnd g (7) = 2. Moreover, it stands to reason that g
must attain a relative minimum at some point past x = 7. Calculus veriﬁes that at x = 13,
we have such a minimum at exactly (13, 1.96). The reader is challenged to ﬁnd calculator
windows which show the graph crossing its horizontal asymptote on one window, and the
relative minimum in the other.
In the denominator, we would have (1billion)2 − 1billion − 6. It’s easy to see why the 6 is insigniﬁcant, but to
ignore the 1 billion seems criminal. However, compared to (1 billion)2 , it’s on the insigniﬁcant side; it’s 1018 versus
109 . We are once again using the fact that for polynomial...
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