Find Critical point and nature

$4x({x}^{2}+{y}^{2}-1)=0$

$4y({x}^{2}+{y}^{2}+1)=0$

$4x({x}^{2}+{y}^{2}-1)=0$

$4y({x}^{2}+{y}^{2}+1)=0$

allucinemsj
2022-08-09
Answered

Find Critical point and nature

$4x({x}^{2}+{y}^{2}-1)=0$

$4y({x}^{2}+{y}^{2}+1)=0$

$4x({x}^{2}+{y}^{2}-1)=0$

$4y({x}^{2}+{y}^{2}+1)=0$

You can still ask an expert for help

Evelin Castillo

Answered 2022-08-10
Author has **12** answers

From the second equation, $y=0$. That's because

${x}^{2}+{y}^{2}+1\ge 1$

for any real $x$ and $y$. So than either $x=0$ or ${x}^{2}-1=0$. So the critical points are $(0,0)$, $(-1,0)$, and $(1,0)$.

${x}^{2}+{y}^{2}+1\ge 1$

for any real $x$ and $y$. So than either $x=0$ or ${x}^{2}-1=0$. So the critical points are $(0,0)$, $(-1,0)$, and $(1,0)$.

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