Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-6 Function Operations - Practice and Problem-Solving Exercises - Page 403: 75

Answer

$f(g(x)) = x - 3$ $g(f(x)) = x - 6$

Work Step by Step

For these types of problems, we are asked to evaluate composite functions. We will use the inner function and substitute it where we see $x$ in the outer function. To evaluate $f(g(x))$, we begin by plugging in the function $g(x)$ where we see $x$ in the outer function, $f(x)$: $f(g(x)) = \frac{(2x - 3) - 3}{2}$ Combine like terms in the numerator: $f(g(x)) = \frac{2x - 6}{2}$ Divide both numerator and denomination by greatest common factor, which is $2$, in this case: $f(g(x)) = x - 3$ To evaluate $g(f(x))$, we begin by plugging in the function $f(x)$ where we see $x$ in the outer function, $g(x)$: $g(f(x)) = 2(\frac{x - 3}{2}) - 3$ Multiply to simplify: $g(f(x)) = \frac{2x - 6}{2} - 3$ Reduce the fraction by dividing numerator and denominator by $2$: $g(f(x)) = (x - 3) - 3$ Combine like terms: $g(f(x)) = x - 6$
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