F is one to one continuous and smooth ba c if and only

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Unformatted text preview: ons are x = ±1, both of which must be excluded from the domain. Hence, the domain of k is (−∞, −1) ∪ (1, ∞). To 2x build the sign diagram for k , we need the zeros of k . Setting k (x) = 0 results in √x2 −1 = 0. We get 2x = 0 or x = 0. However, x = 0 isn’t in the domain of k , which means k has no zeros. We construct our sign diagram on the domain of k below alongside the graph of k . It appears that the graph of k has two vertical asymptotes, one at x = −1 and one at x = 1. The gap in the graph between the asymptotes is because of the gap in the domain of k . Concerning end behavior, there appear to be two horizontal asymptotes, y = 2 and y = −2. To see why this is the case, we think of x → ±∞. The radicand of the denominator x2 − 1 ≈ x2 , and as 2x 2 x x such, k (x) = √x2 −1 ≈ √ x2 = |2x| . As x → ∞, we have |x| = x so k (x) ≈ 2x = 2. On the other x 2 hand, as x → −∞, |x| = −x, and as such k (x) ≈ −x = −2. Finally, it appears as thoug...
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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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