College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Review Exercises - Page 279: 107

Answer

$\left(\dfrac{f}{g}\right)(3)=-\dfrac{23}{4}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the expression, $ \left(\dfrac{f}{g}\right)(3) ,$ given that \begin{array}{l}\require{cancel} f(x)=3x^2-4 \text{ and } \\g(x)=x^2-3x-4 ,\end{array} use the definition of the appropriate function operation. Then substitute $x$ with $3.$ $\bf{\text{Solution Details:}}$ Using $\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)},$ then \begin{array}{l}\require{cancel} \left(\dfrac{f}{g}\right)(x)=\dfrac{3x^2-4}{x^2-3x-4} .\end{array} Sustituting $x$ with $3,$ then \begin{array}{l}\require{cancel} \left(\dfrac{f}{g}\right)(3)=\dfrac{3(3)^2-4}{(3)^2-3(3)-4} \\\\ \left(\dfrac{f}{g}\right)(3)=\dfrac{3(9)-4}{(9)-3(3)-4} \\\\ \left(\dfrac{f}{g}\right)(3)=\dfrac{27-4}{9-9-4} \\\\ \left(\dfrac{f}{g}\right)(3)=\dfrac{23}{-4} \\\\ \left(\dfrac{f}{g}\right)(3)=-\dfrac{23}{4} .\end{array}
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