Stitz-Zeager_College_Algebra_e-book

The horizontal asymptotes y 0 on the graphs of the

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Unformatted text preview: +1 +1 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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