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 - Got It? - Page 399: 1

Answer

$f(x) + g(x) = 2x^2 + x + 5$ The domain, meaning the possible $x$ values, is the set of all real numbers. $f(x) - g(x) = 2x^2 - x + 11$ The domain, meaning the possible $x$ values, is the set of all real numbers.

Work Step by Step

Add the two functions $g(x)$ and $f(x)$: $f(x) + g(x) = (2x^2 + 8) + (x - 3)$ Use distribute property to get rid of the parentheses, paying attention to the signs: $f(x) + g(x) = 2x^2 + 8 + x - 3$ Combine like terms: $f(x) + g(x) = 2x^2 + x + 5$ The domain, meaning the possible $x$ values, is the set of all real numbers. The second part of the exercise is asking to subtract $g(x)$ from $f(x)$: $f(x) - g(x) = (2x^2 + 8) - (x - 3)$ Distribute to get rid of the parentheses, paying attention to the signs: $f(x) - g(x) = 2x^2 + 8 - x + 3$ Combine like terms: $f(x) - g(x) = 2x^2 - x + 11$ The domain, meaning the possible $x$ values, is the set of all real numbers.
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