Answer
\[
4.14=f(1.01,2.02,3.03)
\]
Work Step by Step
Given that $f(x, y, z)$ is differentiable at (1,2,3) and partial derivatives exist, we get to find $f(1.01,2.02,3.03)$. Thus, we use the Local Linear Aproximation theorem:
\[
\begin{aligned}
&f\left(x_{0}, y_{0}=L(x, y, z), z_{0}\right)+f_{x}\left(x_{0}, y_{0}, z_{0}\right)\left(-x_{0}+x\right)+f_{y}\left(x_{0}, y_{0}, z_{0}\right)\left(-y_{0}+y\right)+f_{z}\left(x_{0}, y_{0}, z_{0}\right)\left(-z_{0}+z\right) \\
&=f(1,2,3)+f_{x}(1,2,3)(-1+1.01)+f_{v}(1,2,3)(-2+2.02)+f_{z}(1,2,3)(-3+3.03) \\
&=4+(0.01)(1)+(0.02)(2)+(0.03)(3) \\
&=4.14
\end{aligned}
\]
\[
4.14=f(1.01,2.02,3.03)
\]
