Answer
\[
5.04=f(3.01,3.98)
\]
Work Step by Step
Given that $f(x, y)$ is differentiable at (3,4), partial derivatives exist, and we have to estimate the value of $f(3.01,3.98)$; thus, we use Local Linear Approximation theorem
\[
\begin{aligned}
L(x, y) &=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(-x_{0}+x\right)+f_{y}\left(x_{0}, y_{0}\right)\left(-y_{0}+y\right) \\
L(3.01,3.98) &=f(3,4)+f_{x}(3,4)(-3+3.01)+f_{y}(3,4)(-4+3.98) \\
(0.01)(2)+(-0.02)(-1)+5=L(3.01,3.98) & \\
5.04=L(3.01,3.98) &
\end{aligned}
\]
Thus, the local linear approximation gives us the estimate value of the function at that point:
\[
f(3.01,3.98)=5.04
\]
