# What are the values and types of the critical points, if any, of f(x,y)=4y(1 - x^2)?

Jan 3, 2016

There are two critical points, with coordinates $\left(x , y\right) = \left(\pm 1 , 0\right)$, which are both saddle points.

#### Explanation:

Let $z = f \left(x , y\right) = 4 y \left(1 - {x}^{2}\right) = 4 y - 4 {x}^{2} y$, then $\frac{\partial z}{\partial x} = - 8 x y$ and $\frac{\partial z}{\partial y} = 4 - 4 {x}^{2}$.

After setting both of these partial derivatives equal to zero, we get that $4 = 4 {x}^{2}$ so that ${x}^{2} = 1$ and $x = \pm 1$ and we also have $- 8 x y = 0$, so that $y = 0$. The two critical points are therefore $\left(x , y\right) = \left(\pm 1 , 0\right)$.

The second-order partial derivatives are $\frac{{\partial}^{2} z}{\partial {x}^{2}} = - 8 y$, $\frac{{\partial}^{2} z}{\partial {y}^{2}} = 0$, and $\frac{{\partial}^{2} z}{\partial x \partial y} = - 8 x$, giving a discriminant of

$\frac{{\partial}^{2} z}{\partial {x}^{2}} \cdot \frac{{\partial}^{2} z}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} z}{\partial x \partial y}\right)}^{2} = 0 - {\left(- 8 x\right)}^{2} = - 64 {x}^{2}$

This discriminant is negative at both critical points, making them both saddle points.