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the resulting sign diagram and corresponding graph. From the graph, it appears as though
x = −1 is a vertical asymptote. Carrying out an analysis as x → −1 as in Section 4.2 conﬁrms
this. (We leave the details to the reader.) Near x = 0, we have a situation similar to x = 2
in the graph of f in number 1 above. Finally, it appears as if the graph of h has a horizontal
asymptote y = 2. Using techniques from Section 4.2, we ﬁnd as x → ±∞, x8x → 8. From
+1
√
this, it is hardly surprising that as x → ±∞, h(x) = 3 x8x ≈ 3 8 = 2.
+1 (+)
−1 (−) 0 (+)
0
y = h(x) 4. To ﬁnd the domain of k , we have both an even root and a denominator to concern ourselves
with. To satisfy the square root, x2 − 1 ≥ 0. Setting r(x) = x2 − 1, we ﬁnd the zeros of r to
be x = ±1, and we ﬁnd the sign diagram of r to be
(+) 0 (−) 0 (+)
−1 1 316 Further Topics in Functions
We ﬁnd x2 − 1 ≥ 0 for (−∞, −1] ∪ [1, ∞). To keep the denominator of k (x) away from zero,
√
we set x2 − 1 = 0. We leave it to the reader to verify the soluti...

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