Answer
\[
2.07=f(-0.99,2.02)
\]
Work Step by Step
Given that $f(x, y)$ is differentiable at (-1,2), partial derivatives exists, and we have to find $\mathrm{f}(-0.99,2.02)$. Thus, we use the Local Linear Aproximation theorem:
\[
\begin{aligned}
L(x, y) &=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(-x_{0}+x\right)+f_{y}\left(x_{0}, y_{0}\right)\left(-y_{0}+y\right) \\
L(-0.99,2.02) &=f(-1,2)+f_{x}(-1,2)(-(-1)-0.99)+f_{y}(-1,2)(-2+2.02) \\
(3)(0.02) +(1)(0.01)+2 =L(-0.99,2.02) & \\
2.07=L(-0.99,2.02) &
\end{aligned}
\]
\[
2.07=f(-0.99,2.02)
\]
