Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises: 19

Answer

$6.3$

Work Step by Step

Given that the function is a differentiable function with $f(2,5)=6,f_{x}(2,5)=1,f_{y}(2,5)=-1$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ Let $L(x,y)$ at $f(2,5)$ is given by $L(x,y)=f(2,5)+f_{x}(2,5)(x-2)+f_{y}(2,5)(y-5)$ $=6+1(x-2)+(-1)(y-5)$ $=x-y+9$ Then, $f(2.2,4.9)=2.2-4.9+9=6.3$
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