Question

Find the linearization of \( f(x, y, z)=x^{2}-x y+3 z \) at the point \( (2,1,0) \). Maximum file size: 250MB, maximum number of

Answer 1

The linearization of \( f(x, y, z)=x^{2}-x y+3 z \) at the point \( (2,1,0) \) is given by:

\begin{align*}

L(x,y,z)&=f(2,1,0)+\frac{\partial f}{\partial x}(2,1,0)(x-2)+\frac{\partial f}{\partial y}(2,1,0)(y-1)+\frac{\partial f}{\partial z}(2,1,0)(z-0)\\

&=2^2-2(2)(1)+3(0)+\left(\frac{\partial}{\partial x}(x^{2}-x y+3 z)\bigg|_{(2,1,0)}\right)(x-2)+\left(\frac{\partial}{\partial y}(x^{2}-x y+3 z)\bigg|_{(2,1,0)}\right)(y-1)+\left(\frac{\partial}{\partial z}(x^{2}-x y+3 z)\bigg|_{(2,1,0)}\right)(z-0)\\

&=1-2(x-2)-1(y-1)+3(z-0)\\

&=-2x-y+3z+5.

\end{align*}

Therefore, the linearization of \( f(x, y, z)=x^{2}-x y+3 z \) at the point \( (2,1,0) \) is \( L(x,y,z)=-2x-y+3z+5 \).