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Stitz-Zeager_College_Algebra_e-book

# What is the signicance of the y intercept of the

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Unformatted text preview: tions y 6 5 y = j (x) y=x 4 3 2 1 1 2 3 4 5 6 x −1 y = j −1 (x) 2. We graph y = k (x) = √ x + 2 − 1 using what we learned in Section 1.8 and see k is one-to-one. y 2 1 −2 −1 1 2 x −1 −2 y = k (x) We now try to ﬁnd k −1 . y y x x+1 (x + 1)2 2 + 2x + 1 x y = = = = = = = k (x) √ x+2−1 √ y + 2 − 1 switch x and y √ y+2 √ 2 y+2 y+2 x2 + 2 x − 1 We have k −1 (x) = x2 + 2x − 1. Based on our experience, we know something isn’t quite right. We determined k −1 is a quadratic function, and we have seen several times in this section that these are not one-to-one unless their domains are suitably restricted. Theorem 5.2 tells us that the domain of k −1 is the range of k . From the graph of k , we see that the range is [−1, ∞), which means we restrict the domain of k −1 to x ≥ −1. We now check that this works in our compositions. 5.2 Inverse Functions k −1 ◦ k (x) = = = = = = 307 k −1 (k (x)) √ x + 2 − 1 , x ≥ −2 k −1 √ √ 2 x+2−1 +2 x+2−1 −1 √ √ √ 2 x+2 −2 x+2+1+2 x+2−2−1 x+2−2 x and k ◦ k −1 (x) = = = = = = = k x2 + 2x − 1 x ≥ −1 2 √ (x + 2x − 1) + 2 − 1 2 + 2x + 1 − 1 x (x + 1)2 − 1 |x + 1| − 1 x+1−1 since x ≥ −1 x Graphically, everything checks out as well, provided that we remember the domain re...
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