Multivariable calculus

Critical Points of x^4 + y^4

Q: Does D = 0 mean there is no maximum or minimum?

No. It means the second derivative test cannot tell. For x^{4} + y^{4} the origin is a global minimum; for x^{4} - y^{4} the origin is a saddle; for -x^{4} - y^{4} it is a global maximum. All three have D = 0 at the origin, so D alone cannot separate them.

Q: How does this compare with x^2 + y^2?

Both have a strict minimum at the origin, but x^{2} + y^{2} has D = 4 > 0 there and is classified in one line. The quartic is so flat near the origin that its second partials all vanish, which is why it needs the direct argument instead.

The only critical point of f(x,y) = x^4 + y^4 is the origin, and there the discriminant is D = 0, so the second derivative test gives no information. Direct reasoning settles it: fourth powers are never negative, so f is at least 0 everywhere and equals 0 only at the origin, a global minimum.

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

$\left(0, 0\right)$ degenerate, the test is inconclusivediscriminant D = 0

One critical point, and the test says nothing

\frac{\partial f}{\partial x} = 4x^{3}, \qquad \frac{\partial f}{\partial y} = 4y^{3}

Both vanish only when $x = 0$ and $y = 0$ , so the origin is the sole critical point. Now take second partials.

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

D(0,0) = (0)(0) - 0^{2} = 0

Every second partial is zero at the origin, so $D = 0$ and the test is inconclusive. This is the degenerate case, and it means the quadratic approximation to $f$ at the origin is flat: the surface is too flat there for second order information to see the shape.

Settling it without the test

When $D = 0$ you go back to the function itself. Here the answer is immediate, because an even power of a real number is never negative.

f(x,y) = x^{4} + y^{4} \ge 0, \quad \text{with equality only when } x = y = 0

So $f(0,0) = 0$ is smaller than the value at every other point in the plane. The origin is a strict global minimum, and therefore a local one, even though the second derivative test refused to say so.

The mistake students make

The big one is reading $D = 0$ as a verdict. It is not a verdict at all. $D < 0$ means saddle and $D > 0$ means extremum, but $D = 0$ means the test has failed and you must argue some other way. Writing that the origin is a saddle because $D$ is not positive is simply wrong here: it is a global minimum.

A useful contrast is $x^{4} - y^{4}$ , which also has $D = 0$ at the origin and is a saddle there, since it rises along the $x$ axis and falls along the $y$ axis. Same value of $D$ , opposite behaviour. That is precisely why $D = 0$ carries no information.

Frequently asked questions

Does D = 0 mean there is no maximum or minimum?

No. It means the second derivative test cannot tell. For $x^{4} + y^{4}$ the origin is a global minimum; for $x^{4} - y^{4}$ the origin is a saddle; for $-x^{4} - y^{4}$ it is a global maximum. All three have $D = 0$ at the origin, so $D$ alone cannot separate them.

How does this compare with x^2 + y^2?

Both have a strict minimum at the origin, but $x^{2} + y^{2}$ has $D = 4 > 0$ there and is classified in one line. The quartic is so flat near the origin that its second partials all vanish, which is why it needs the direct argument instead.