Answer
\[
-0.05=f(1.98,0.99,-1.97)
\]
Work Step by Step
Given that $f(x, y)$ is differentiable at (2,1,-2) and partial derivatives exist, we use the Local Linear Approximation theorem:
\[
\begin{aligned}
L(x, y, z) &=f\left(x_{0}, y_{0}, z_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(-x_{0}+x\right)+f_{y}\left(x_{0}, y_{0}, z_{0}\right)\left(-y_{0}+y\right)+f_{z}\left(x_{0}, y_{0}, z_{0}\right)\left(-z_{0}+z\right) \\
&=f(2,1,-2)+f_{z}(1,-2)(-2+1.98)+f_{y}(2,1,-2)(-1+0.99)+f_{z}(2,1,-2)(-(-2)-1.97) \\
&=0+(-0.01)(1)+(0.03) (-2)+(-0.02)(-1)\\
&=-0.01-.06 +0.02\\
&=-0.05
\end{aligned}
\]
Thus:
\[
\begin{array}{l}
-0.05=f(1.98,0.99,-1.97) \\
\end{array}
\]
