Stitz-Zeager_College_Algebra_e-book

Since u 5x we have 5x 2 or 5x 3 since 5x 2 has no

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Unformatted text preview: 1 from a procedural perspective. If we break f (x) = 2x−1 − 3 into a series of steps, we find f takes an input x and applies the steps (a) subtract 1 (b) put as an exponent on 2 (c) subtract 3 Clearly, to undo subtracting 1, we will add 1, and similarly we undo subtracting 3 by adding 3. How do we undo the second step? The answer is we use the logarithm. By definition, log2 (x) undoes exponentiation by 2. Hence, f −1 should (a) add 3 (b) take the logarithm base 2 (c) add 1 In symbols, f −1 (x) = log2 (x + 3) + 1. 3. To graph f −1 (x) = log2 (x + 3) + 1 using transformations, we start with j (x) = log2 (x). We 1 track the points 2 , −1 , (1, 0) and (2, 1) on the graph of j along with the vertical asymptote x = 0 through the transformations using Theorem 1.7. Since f −1 (x) = j (x + 3) + 1, we first 5 subtract 3 from each of the x values (including the vertical asymptote) to obtain − 2 , −1 , 340 Exponential and Logarithmic Functions (−2, 0) and (−1, 1) with a vertical asymptote x = −3. Next, we add 1 to the y values on the 5 graph and get − 2 , 0 , (−2, 1) an...
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