Chapter 14 - Section 14.4 - Tangent Planes and Linear Approximation - 14.4 Exercise - Page 935: 19

$6.3$

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 the 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, $f(2.2,4.9)=2.2-4.9+9=6.3$