College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.7 - Page 309: 15

Answer

a) Find the inverse equation $f^{-1}(x) = \frac{x-3}{2}$ b) Verify that the inverse equation is correct $f(f^{-1}(x)) = x$ $f(\frac{x-3}{2}) = 2(\frac{x-3}{2}) + 3$ $= \frac{2(x-3)}{2} + 3$ $= x - 3 + 3$ $= x$ $f^{-1}(f(x)) = x$ $f^{-1}(2x + 3) = \frac{(2x+3) -3}{2}$ $= \frac{2x + 3 - 3}{2}$ $= \frac{2x}{2}$ $= x$ Therefore, the inverse equation is correct.

Work Step by Step

a) Find the inverse function Let $f(x) = y$ $y = 2x + 3$ $x = 2y + 3$ $x - 3 = 2y$ $y = \frac{x-3}{2}$ $f^{-1}(x) = \frac{x-3}{2}$ b) Verify that the inverse equation is correct $f(f^{-1}(x)) = x$ $f(\frac{x-3}{2}) = 2(\frac{x-3}{2}) + 3$ $= \frac{2(x-3)}{2} + 3$ $= x - 3 + 3$ $= x$ $f^{-1}(f(x)) = x$ $f^{-1}(2x + 3) = \frac{(2x+3) -3}{2}$ $= \frac{2x + 3 - 3}{2}$ $= \frac{2x}{2}$ $= x$ Therefore, the inverse equation is correct.
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