Stitz-Zeager_College_Algebra_e-book

If we graph f x e 2 and g x 5 we see that the graphs

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Unformatted text preview: d (−1, 2). If you are experiencing d´j` vu, there is a good ea reason for it but we leave it to the reader to determine the source of this uncanny familiarity. We obtain the graph below. The domain of f −1 is (−3, ∞), which matches the range of f , and the range of f −1 is (−∞, ∞), which matches the domain of f . y y 4 4 3 3 2 2 1 1 −3 −2 −1 −1 1 2 3 4 5 6 7 8 x −2 −1 −1 −2 −3 1 2 3 4 5 6 7 8 x −2 −3 y = j (x) = log2 (x) −− − − − −→ −−−−−− y = f −1 (x) = log2 (x + 3) + 1 4. We now verify that f (x) = 2x−1 − 3 and f −1 (x) = log2 (x + 3) + 1 satisfy the composition requirement for inverses. For all real numbers x, f −1 ◦ f (x) = f −1 (f (x)) = f −1 2x−1 − 3 = log2 2x−1 − 3 + 3 + 1 = log2 2x−1 + 1 = (x − 1) + 1 Since log2 (2u ) = u for all real numbers u =x For all real numbers x > −3, we have9 f ◦ f −1 (x) = f f −1 (x) = f (log2 (x + 3) + 1) = 2(log2 (x+3)+1)−1 − 3 = 2log2 (x+3)...
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