Multivariable calculus

Critical Points of x^4 - 2x^2 + y^2

Q: Why are there two minima with the same value?

Every x in the formula appears as an even power, so f(-x,y) = f(x,y) and the surface is symmetric across the y axis. That reflection carries (1,0) to (-1,0), so both must have the same height, -1, and the same discriminant, 16.

Q: Is the origin a minimum along the y axis?

Yes, and that is exactly the trap. Restricted to the y axis, f = y^{2}, which has a minimum at y = 0. But restricted to the x axis, f = x^{4} - 2x^{2}, which has a maximum at x = 0. A point that is a minimum in one direction and a maximum in another is a saddle.

f(x,y) = x^4 - 2x^2 + y^2 has three critical points. The origin is a saddle with discriminant D = -8, since the surface falls along the x axis and rises along the y axis. The points (1,0) and (-1,0) are local minima with D = 16, both at height -1, which is the global minimum.

f(x,y) = x^{4} - 2x^{2} + y^{2}

$\left(0, 0\right)$ saddle pointdiscriminant D = -8
$\left(1, 0\right)$ local minimumdiscriminant D = 16
$\left(-1, 0\right)$ local minimumdiscriminant D = 16

A system that comes apart cleanly

\frac{\partial f}{\partial x} = 4x^{3} - 4x = 4x\left(x^{2} - 1\right), \qquad \frac{\partial f}{\partial y} = 2y

The $y$ equation forces $y = 0$ on its own. The $x$ equation factors into $4x(x-1)(x+1) = 0$ , giving $x = 0$ , $x = 1$ and $x = -1$ . So the critical points are $(0,0)$ , $(1,0)$ and $(-1,0)$ , all on the $x$ axis.

The discriminant depends on x alone

f_{xx} = 12x^{2} - 4, \qquad f_{yy} = 2, \qquad f_{xy} = 0

D(x,y) = 2\left(12x^{2} - 4\right) = 24x^{2} - 8

Because $f_{xy} = 0$ , the discriminant is just the product of the two pure second partials, and $f_{yy} = 2$ is a positive constant. So the sign of $D$ is the sign of $f_{xx}$ .

At the origin, $D = -8 < 0$ , a saddle. At $(\pm 1, 0)$ , $D = 24 - 8 = 16 > 0$ with $f_{xx} = 8 > 0$ , so both are local minima at height $f = -1$ .

The mistake students make

The origin gets misread as a minimum, because along the $y$ axis $f = y^{2}$ climbs away from it and the function looks like a sum of squares. Test the other direction: along the $x$ axis, $f = x^{4} - 2x^{2}$ , and $f(0.5,0) = -0.4375$ sits below $f(0,0) = 0$ . Rising one way and falling another is a saddle.

Checking one direction is never enough in two variables. That is the whole reason the second derivative test exists, and here it delivers $D = -8$ without any sampling.

Rewriting the function makes the shape obvious.

f(x,y) = \left(x^{2} - 1\right)^{2} + y^{2} - 1

Both squares are zero exactly when $x = \pm 1$ and $y = 0$ , so $f \ge -1$ everywhere and the two minima are global.

Frequently asked questions

Why are there two minima with the same value?

Every $x$ in the formula appears as an even power, so $f(-x,y) = f(x,y)$ and the surface is symmetric across the $y$ axis. That reflection carries $(1,0)$ to $(-1,0)$ , so both must have the same height, $-1$ , and the same discriminant, $16$ .

Is the origin a minimum along the y axis?

Yes, and that is exactly the trap. Restricted to the $y$ axis, $f = y^{2}$ , which has a minimum at $y = 0$ . But restricted to the $x$ axis, $f = x^{4} - 2x^{2}$ , which has a maximum at $x = 0$ . A point that is a minimum in one direction and a maximum in another is a saddle.