## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 518: 20

#### Answer

$$f\left( { - 1,2} \right) = - \frac{{\sqrt 5 }}{3}{\text{ and }}f\left( {6, - 3} \right) = \frac{{\sqrt {45} }}{9}$$

#### Work Step by Step

\eqalign{ & f\left( {x,y} \right) = \frac{{\sqrt {{x^2} + {y^2}} }}{{x - y}} \cr & {\text{find }}f\left( { - 1,2} \right).{\text{ substitute }} - 1{\text{ for }}x{\text{ and }}2{\text{ for }}y \cr & f\left( { - 1,2} \right) = \frac{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }}{{\left( { - 1} \right) - \left( 2 \right)}} \cr & {\text{simplifying}} \cr & f\left( { - 1,2} \right) = \frac{{\sqrt {1 + 4} }}{{ - 1 - 2}} \cr & f\left( { - 1,2} \right) = - \frac{{\sqrt 5 }}{3} \cr & \cr & {\text{find }}f\left( {6, - 3} \right).{\text{ substitute 6 for }}x{\text{ and }} - 3{\text{ for }}y \cr & f\left( {6, - 3} \right) = \frac{{\sqrt {{{\left( 6 \right)}^2} + {{\left( { - 3} \right)}^2}} }}{{\left( 6 \right) - \left( { - 3} \right)}} \cr & {\text{simplifying}} \cr & f\left( {6, - 3} \right) = \frac{{\sqrt {36 + 9} }}{{6 + 3}} \cr & f\left( {6, - 3} \right) = \frac{{\sqrt {45} }}{9} \cr}

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