Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.3 Inverse, Exponential, and Logarithmic Functions - 1.3 Exercises: 25

Answer

a. $f^{-1}(x) = x/3 - 5/3$ b. $f^{-1}(f(x)) = (3x + 5)/3 -5/3$ $f^{-1}(f(x)) = 3x/3 + 5/3 -5/3$ $f^{-1}(f(x)) = 3x/3$ $f^{-1}(f(x)) = x$ $f(f^{-1}(x)) = 3(x/3 - 5/3) + 5$ $f(f^{-1}(x)) = x - 5 + 5$ $f(f^{-1}(x)) = x$

Work Step by Step

a. $f(x) = 3x + 5$ $x = 3f^{-1}(x) + 5$ $ x - 5 = 3f^{-1}(x)$ $x/3 - 5/3 = f^{-1}(x)$ b. $f^{-1}(f(x)) = (3x + 5)/3 -5/3$ $f^{-1}(f(x)) = 3x/3 + 5/3 -5/3$ $f^{-1}(f(x)) = 3x/3$ $f^{-1}(f(x)) = x$ $f(f^{-1}(x)) = 3(x/3 - 5/3) + 5$ $f(f^{-1}(x)) = x - 5 + 5$ $f(f^{-1}(x)) = x$
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