Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise - Page 911: 26


$h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}

Work Step by Step

We are given: $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$ The function is continuous everywhere and the $ln$ term is only defined for positive values. Thus, $1-xy\gt 0$ or, $1\gt xy$ Hence, $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}
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