Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.2 Combining Functions - 3.2 Exercises - Page 247: 72

Answer

$(a)$ $f(x)+g(x)= x^2+x+5$ $(b)$ $f(x)-g(x)=-x^2+7x-19$ $(c)$ $f(x)g(x)= 4x^3-19x^2+69x-84$ $(d)$ $\frac{f(x)}{g(x)}=\frac{4x-7}{x^2-3x+12}$

Work Step by Step

$f(x)=4x-7$ $g(x)=x^2-3x+12$ $(a)$ $f(x)+g(x)=4x-7+x^2-3x+12$ $f(x)+g(x)= x^2+4x-3x-7+12$ $f(x)+g(x)= x^2+x+5$ $(b)$ $f(x)-g(x)=4x-7-(x^2-3x+12)=4x-7-x^2+3x-12$ $f(x)-g(x)=-x^2+4x+3x-7-12$ $f(x)-g(x)=-x^2+7x-19$ $(c)$ $f(x)g(x)= (4x-7)(x^2-3x+12)$ $f(x)g(x)= 4x(x^2-3x+12) -7(x^2-3x+12) $ $f(x)g(x)= 4x^3-12x^2+48x-7x^2+21x-84 $ $f(x)g(x)= 4x^3-12x^2-7x^2+48x+21x-84$ $f(x)g(x)= 4x^3-19x^2+69x-84$ $(d)$ $\frac{f(x)}{g(x)}=\frac{4x-7}{x^2-3x+12}$
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