Using properties of exponents we get 23x 241x using

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Unformatted text preview: o be r is (−∞, 1) ∪ (1, ∞). Setting r(x) = 0 gives x = 0. (+) 0 (−) 0 8 See page 44 if you’ve forgotten what this term means. (+) 1 338 Exponential and Logarithmic Functions x We find x−1 > 0 on (−∞, 0) ∪ (1, ∞) to get the domain of g . The graph of y = g (x) confirms this. We can tell from the graph of g that it is not the result of Section 1.8 transformations being applied to the graph y = ln(x), so barring a more detailed analysis using Calculus, the calculator graph is the best we can do. One thing worthy of note, however, is the end behavior of g . The graph suggests that as x → ±∞, g (x) → 0. We can verify this analytically. Using x results from Chapter 4 and continuity, we know that as x → ±∞, x−1 ≈ 1. Hence, it makes sense that g (x) = ln x x−1 ≈ ln(1) = 0. y = f (x) = 2 log(3 − x) − 1 y = g (x) = ln x x−1 While logarithms have some interesting applications of their own which you’ll explore in the exercises, their primary use to us will be to undo exponential functions. (Thi...
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