Multivariable calculus

Critical Points of x^3 + y^3 - 3x - 3y

Q: Why do both saddle points have the same discriminant?

Because D = 36xy and both (1,-1) and (-1,1) have xy = -1. Their heights match too: f = 0 at each. The function's odd symmetry maps one to the other, so they are genuinely mirror images.

Q: Is the local maximum at (-1,-1) a global maximum?

No. Along y = x the function is 2x^{3} - 6x, which grows without bound as x \to \infty, so f has no global maximum. The value f(-1,-1) = 4 is only the largest in a neighbourhood of that point.

f(x,y) = x^3 + y^3 - 3x - 3y has four critical points, at (1,1), (1,-1), (-1,1) and (-1,-1). The discriminant is D = 36xy, so (1,1) is a local minimum with D = 36, (-1,-1) is a local maximum with D = 36, and the two mixed sign points are saddles with D = -36.

f(x,y) = x^{3} + y^{3} - 3x - 3y

$\left(1, 1\right)$ local minimumdiscriminant D = 36
$\left(1, -1\right)$ saddle pointdiscriminant D = -36
$\left(-1, 1\right)$ saddle pointdiscriminant D = -36
$\left(-1, -1\right)$ local maximumdiscriminant D = 36

A separable system gives four points, not two

Neither partial contains the other variable, so the two equations are solved independently.

\frac{\partial f}{\partial x} = 3x^{2} - 3, \qquad \frac{\partial f}{\partial y} = 3y^{2} - 3

So $x^{2} = 1$ and $y^{2} = 1$ , giving $x = \pm 1$ and $y = \pm 1$ . The choices are independent, so every combination is a critical point: $(1,1)$ , $(1,-1)$ , $(-1,1)$ and $(-1,-1)$ .

The discriminant is just the sign of xy

f_{xx} = 6x, \qquad f_{yy} = 6y, \qquad f_{xy} = 0

D(x,y) = (6x)(6y) - 0^{2} = 36xy

At the four points $|D| = 36$ , and only the sign of the product $xy$ varies. When $x$ and $y$ have the same sign, $D = 36 > 0$ and $f_{xx} = 6x$ decides: at $(1,1)$ it is positive, so a local minimum with $f = -4$ ; at $(-1,-1)$ it is negative, so a local maximum with $f = 4$ .

$(1,1)$ : $D = 36$ , $f_{xx} = 6$ , local minimum, $f = -4$
$(-1,-1)$ : $D = 36$ , $f_{xx} = -6$ , local maximum, $f = 4$
$(1,-1)$ and $(-1,1)$ : $D = -36$ , saddles, both with $f = 0$

The mistake students make

Two errors show up here. The first is writing the answer as $(\pm 1, \pm 1)$ and counting two points. The signs are chosen independently, so there are four, and they do not all behave the same way.

The second is assuming that symmetry forces the same classification. The function does satisfy $f(-x,-y) = -f(x,y)$ , but that symmetry turns minima into maxima, not into copies of themselves. Check each point separately, since $f_{xx} = 6x$ changes sign with $x$ .

Frequently asked questions

Why do both saddle points have the same discriminant?

Because $D = 36xy$ and both $(1,-1)$ and $(-1,1)$ have $xy = -1$ . Their heights match too: $f = 0$ at each. The function's odd symmetry maps one to the other, so they are genuinely mirror images.

Is the local maximum at (-1,-1) a global maximum?

No. Along $y = x$ the function is $2x^{3} - 6x$ , which grows without bound as $x \to \infty$ , so $f$ has no global maximum. The value $f(-1,-1) = 4$ is only the largest in a neighbourhood of that point.