Find the linearization of the function at the given point.
-$$f ( x , y , z ) = \ln ( 6 x - 10 y - 4 z ) \text { at } \left( \frac { 1 } { 6 } , - \frac { 1 } { 10 } , - \frac { 1 } { 4 } \right)$$
A) $L ( x , y , z ) = 3 x - 5 y - 2 z + \ln 3 - 1$
B) $L ( x , y , z ) = 2 x - \frac { 10 } { 3 } y - \frac { 4 } { 3 } z + \ln 3 - 1$
C) $\mathrm { L } ( \mathrm { x } , \mathrm { y } , \mathrm { z } ) = 2 \mathrm { x } - \frac { 10 } { 3 } \mathrm { y } - \frac { 4 } { 3 } \mathrm { z } + \ln 2 - 1$
D) $L ( x , y , z ) = 3 x - 5 y - 2 z + \ln 2 - 1$

Question

Quizplus · Accepted Answer

The linearization of $f(x,y,z)$ at the point $(\frac{1}{6},-\frac{1}{10},-\frac{1}{4})$ is given by:
\begin{align*}
L(x,y,z)&=f\left(\frac{1}{6},-\frac{1}{10},-\frac{1}{4}ight)\
&\quad+\frac{\partial f}{\partial x}\Bigg|_{\substack{x=\frac{1}{6} \ y=-\frac{1}{10} \ z=-\frac{1}{4}}}(x-\frac{1}{6})+\frac{\partial f}{\partial y}\Bigg|_{\substack{x=\frac{1}{6} \ y=-\frac{1}{10} \ z=-\frac{1}{4}}}(y+\frac{1}{10})\
&\quad+\frac{\partial f}{\partial z}\Bigg|_{\substack{x=\frac{1}{6} \ y=-\frac{1}{10} \ z=-\frac{1}{4}}}(z+\frac{1}{4})\
&=\ln 2+\frac{3}{\frac{1}{6}}(x-\frac{1}{6})-\frac{5}{-\frac{1}{10}}(y+\frac{1}{10})-\frac{2}{-\frac{1}{4}}(z+\frac{1}{4})\
&=2x-\frac{10}{3}y-\frac{4}{3}z+\ln 2-1
\end{align*}

Find the Linearization of the Function at the Given Point f(x,y,z)=ln⁡(6x−10y−4z) at (16,−110,−14)f ( x , y , z ) = \ln ( 6 x - 10 y - 4 z ) \text { at } \left( \frac { 1 } { 6 } , - \frac { 1 } { 10 } , - \frac { 1 } { 4 } \right)f(x,y,z)=ln(6x−10y−4z) at (61​,−101​,−41​)

Find the Linearization of the Function at the Given Point $f ( x , y , z ) = \ln ( 6 x - 10 y - 4 z ) \text { at } \left( \frac { 1 } { 6 } , - \frac { 1 } { 10 } , - \frac { 1 } { 4 } \right)$