Answer
The estimations:
$f\left( {4.1,4} \right) \simeq 2.9$
$f\left( {3.88,4.03} \right) \simeq 3.09$
Work Step by Step
We have $f\left( {4,4} \right) = 3$ and ${f_x}\left( {4,4} \right) = {f_y}\left( {4,4} \right) = - 1$.
Let $x=a+h$ and $y=b+k$. The linear approximation to $f\left( {x,y} \right)$ (Eq.(3) of Section 15.4) is given by
(1) ${\ \ \ }$ $f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
Write $\left( {a,b} \right) = \left( {4,4} \right)$ and $\left( {h,k} \right) = \left( {0.1,0} \right)$. Using equation (1), we get
$f\left( {4.1,4} \right) \simeq f\left( {4,4} \right) + {f_x}\left( {4,4} \right)\cdot0.1 + {f_y}\left( {4,4} \right)\cdot0$
$f\left( {4.1,4} \right) \simeq 3 + \left( { - 1} \right)\cdot0.1 = 2.9$
Write $\left( {a,b} \right) = \left( {4,4} \right)$ and $\left( {h,k} \right) = \left( { - 0.12,0.03} \right)$. Using equation (1), we get
$f\left( {3.88,4.03} \right) \simeq f\left( {4,4} \right) + {f_x}\left( {4,4} \right)\cdot\left( { - 0.12} \right) + {f_y}\left( {4,4} \right)\cdot0.03$
$f\left( {3.88,4.03} \right) \simeq 3 + \left( { - 1} \right)\cdot\left( { - 0.12} \right) + \left( { - 1} \right)\cdot0.03 = 3.09$
So, the estimations are $f\left( {4.1,4} \right) \simeq 2.9$ and $f\left( {3.88,4.03} \right) \simeq 3.09$.
