Answer
$0.01$
Work Step by Step
$f_x=2x+y \\ \implies f_x(1,1,2)=2(1)+1=3$
Next, $f_y =x+z \\ \implies f_{y}(1,1,2) =1+2=3$ and $f_z=y+\dfrac{z}{2} \implies f_z(1,1,2)=1+\dfrac{2}{2}=2$
$f_{xx}=2; f_{yy}=0$ and $f_{zz} =\dfrac{1}{2}$ and $f_{xy}=1$ and $f_{xx}= 0$ and $f_{yz}=1$
The error can be found as:
$|E(x,y,z)| \leq \dfrac{1}{2} \times (2) [ |x-1| +|y-1|+|z-2|)^2$
$\implies E \leq \dfrac{1}{2} \times (0.01+0.01+0.08)^2 =0.01$
