# A surface is represented by the following multivariable function, f(x,y)=x^3+y^3-3x-3y+1 Calculate coordinates of stationary points.

A surface is represented by the following multivariable function,
$f\left(x,y\right)={x}^{3}+{y}^{3}-3x-3y+1$
Calculate coordinates of stationary points.
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Viktor Wiley
Obtain ${f}_{x}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{y}$ as follows.
${f}_{x}\left(x,y\right)=\frac{\partial }{\partial x}\left({x}^{3}+{y}^{3}-3x-3y+1\right)$
$=3{x}^{2}+0-3-0+0$
$=3{x}^{2}-3$
${f}_{y}\left(x,y\right)\frac{\partial }{\partial y}\left({x}^{3}+{y}^{3}-3x-3y+1\right)$
$=0+3{y}^{2}-0-3+0$
$=3{y}^{2}-3$
Now obtain ${f}_{x}=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{y}=0$ as follows.
${f}_{x}=0$
$⇒3{x}^{2}-3=0$
$3\left({x}^{2}-1\right)=0$
${x}^{2}-1=0$
${x}^{2}=1$
$x=\sqrt{1}$
$x=±1$
${f}_{y}=0$
$⇒3{y}^{2}-3=0$
$3\left({y}^{2}-1\right)=0$
${y}^{2}-1=0$
${y}^{2}=1$
$y=\sqrt{1}$
$y=±1$
Thus, the coordinates of stationary points are ${P}_{1}\left(1,1\right),{P}_{2}\left(1,-1\right),{P}_{3}\left(-1,1\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{P}_{4}\left(-1,-1\right)$.