Answer
a) $2x+2y+2z-3$
b) $y+z$
c) $0$
Work Step by Step
a) $f_x=y+z \\ \implies f_x(1,1,1)=1+1=2$
Next, $f_y(x,y) =x+z \\ \implies f_{y}(1,1,1) =1+1=2$
and $L(x,y,z)=3+2(x-1)+2(y-1)+2(z-1) =2x+2y+2z-3$
b) $f(1,0,0)=0; f_{x}(1,0,0)=0$ and $f_{xy}(x,y) =0$
and $L(x,y,z)=0+0(x-1)+(y-0)+(z-0) =y+z$
c) $f(0,0,0)=0; f_{x}(1,0,0)=0$ and $f_{y}(0,0,0) =0$ and $f_{z}(0,0,0) =0$
and $L(x,y,z)=0$
