I was wondering if anyone could help me with this question of mine.
I was told to find a relative extrema and saddle point of $f(x,y)=x^2y + 2xy - 3y$
In order to find the critical point, I set $\frac{∂f}{∂y} = 0$ and $\frac{∂f}{∂x} =0$
$\frac{∂f}{∂y}$ = $x^2+2x-3 = 0$
$\frac{∂f}{∂x}$ = $2xy+2y = 0$
Afterwards, I made it into a simultaneous equation to solve it.
where $x = -3$ or $x = 1$ for $x^2+2x-3 = 0$ and $y = 0$ or $x = -1$ for $2xy+2y = 0$
However, I was told that $x = -1$ was rejected, thus the critical point are $(-3, 0, 0)$ and $(1,0,0)$
So my question is, why $x = -1$ was rejected?
For $f$ to have a local extremum both partial derivatives have to be $0$ at that point. $x=-1$ makes the partial derivative w.r.t $x$ zero but it does not make the partial derivative w.r.t $y$ zero.