Answer
The linearization to $f\left( {x,y,z} \right) = \frac{{xy}}{z}$ at the point $\left( {2,1,2} \right)$:
$L\left( {x,y,z} \right) = \frac{1}{2}x + y - \frac{1}{2}z$
Using linearization: $f\left( {2.05,0.9,2.01} \right) \approx 0.92$
Using a calculator: $f\left( {2.05,0.9,2.01} \right) \approx 0.91791$.
Work Step by Step
We are given $f\left( {x,y,z} \right) = \frac{{xy}}{z}$. The partial derivatives are
${f_x} = \frac{y}{z}$, ${\ \ }$ ${f_y} = \frac{x}{z}$, ${\ \ }$ ${f_z} = - \frac{{xy}}{{{z^2}}}$
The linearization of $f$ in three variables is given by
$L\left( {x,y,z} \right) = f\left( {a,b,c} \right) + {f_x}\left( {a,b,c} \right)\left( {x - a} \right) + {f_y}\left( {a,b,c} \right)\left( {y - b} \right) + {f_z}\left( {a,b,c} \right)\left( {z - c} \right)$
The linearization of $f$ at $\left( {2,1,2} \right)$:
$L\left( {x,y,z} \right) = f\left( {2,1,2} \right) + {f_x}\left( {2,1,2} \right)\left( {x - 2} \right) + {f_y}\left( {2,1,2} \right)\left( {y - 1} \right) + {f_z}\left( {2,1,2} \right)\left( {z - 2} \right)$
$L\left( {x,y,z} \right) = 1 + \frac{1}{2}\left( {x - 2} \right) + 1\left( {y - 1} \right) - \frac{1}{2}\left( {z - 2} \right)$
$L\left( {x,y,z} \right) = \frac{1}{2}x + y - \frac{1}{2}z$
Next, we estimate $f\left( {2.05,0.9,2.01} \right)$:
From previous result we get
$L\left( {x,y,z} \right) = \frac{1}{2}x + y - \frac{1}{2}z$
Thus,
$f\left( {2.05,0.9,2.01} \right) \approx L\left( {2.05,0.9,2.01} \right) = \frac{1}{2}\cdot2.05 + 0.9 - \frac{1}{2}\cdot2.01$
$f\left( {2.05,0.9,2.01} \right) \approx 0.92$
Using a calculator we obtain $f\left( {2.05,0.9,2.01} \right) \approx 0.91791$.
