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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
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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
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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
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graph and get − 2 , 0 , (−2, 1) an...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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