## Thomas' Calculus 13th Edition

$$0.27$$
$f_x(1,1) =x =1$ and $f_y(1,1) =x-6y =1-(6)(1)=-5$ Now $$f_{xx}=0 \\ f_{yy}=-6 \\ f_{xy}=1$$ We notice that the maximum of $|f_{xx}|,|f_{yy}|,|f_{xy}|$ is $6$. So, $M=6$ Error: $|E(x,y,z)| \leq (\dfrac{1}{2}) (6) [ |x-1| +|y-1|)^2$ or, $$E \leq (3) (0.1+0.2)^2 = 0.27$$