Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Section 2.1 - Introduction to Functions - Exercise Set - Page 112: 28

Answer

$$|f(1) - f(0)| - [g(1)]^{2} + g(1) \div f(-1) \cdot g(2)=0$$

Work Step by Step

$$|f(1) - f(0)| - [g(1)]^{2} + g(1) \div f(-1) \cdot g(2)$$ The values of $f(1), f(0), g(1), f(-1), g(2)$ can be obtained from the given table. The values in the parentheses indicate the value of $f$ or a given value of $x$. Based on the table: $f(1) = -4$ $f(0) = -1$ $g(1) = -3$ $f(-1) = 3$ $g(2) = -6$ Plugging these values to the original equation: $$=|-4 - (-1)| - [-3]^{2} + (-3) \div 3 \cdot -6$$ $$=|-3| - 9 + (-3) \div 3 \cdot -6$$ $$=3 - 9 + (-1) \cdot -6$$ $$=3 - 9 + 6$$ $$=0$$
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