We just plug in the values ( x , y ) = ( − 1.02 , 0.97 ) {\displaystyle \displaystyle (x,y)=(-1.02,0.97)} into the linear approximation found in part b) f ( x , y ) ≈ − 1 − 3 2 ( y − 1 ) {\displaystyle \displaystyle f(x,y)\approx -1-{\frac {3}{2}}(y-1)}
f ( − 1.02 , 0.97 ) ≈ − 1 − 3 2 ( 0.97 − 1 ) = − 0.955 {\displaystyle \displaystyle f(-1.02,0.97)\approx -1-{\frac {3}{2}}(0.97-1)=-0.955}
[As a quick check, recall that f ( − 1 , 1 ) = − 1 {\displaystyle \displaystyle f(-1,1)=-1} , so our answer should be close to this (assuming our approximation is good).]
into
the linear approximation found in part b)
, so our answer should be close to this (assuming our approximation is good).]
