College Algebra (6th Edition)

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

Chapter 5 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-5) - Page 588: 29

Answer

$$f(g(x)) = 2x^{2} -3x + 2$$ $$and$$ $$g(f(x)) = 2x^{2} - x - 2$$

Work Step by Step

The exercise in question is basically asking us to find $f(g(x))$ and $g(f(x))$. Knowing that $g(x) = 1 - x$, and that $f(x) = 2x^{2} - x -1$, to find $f(g(x))$ we simply substitute $g(x)$ inside the values of $x$ from $f(x)$ as so: $f(g(x)) = 2[x - 1]^{2} - [x - 1] - 1$ $f(g(x)) = 2[x^{2} - 2x +1] - x + 1 - 1$ $f(g(x)) = 2x^{2} - 4x +2 - x$ $f(g(x)) = 2x^{2} -3x + 2$ Following the same principles, we can find the following: $g(f(x)) = [2x^{2} - x - 1] - 1$ $g(f(x)) = 2x^{2} - x - 1 - 1$ $g(f(x)) = 2x^{2} - x - 2$
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