When the Two-Variable Test Is Inconclusive

When the Determinant Test Stops Short

A stationary point is a point where the first partial derivatives vanish. The two-variable second derivative test uses the Hessian matrix at such a point to help decide whether it is a local minimum, a local maximum, or a saddle point. When the Hessian determinant is zero, however, the test gives no classification. It does not say that the point is an extremum, and it does not say that the point is a saddle.

The right question remains concrete: how does the function compare with its value at the stationary point when we move a small distance away? The signs of those nearby differences determine whether the point is a strict local minimum, a strict local maximum, or a saddle. They can also reveal a non-strict local extremum.

Key takeaway: A zero Hessian determinant means that the two-variable second derivative test is inconclusive. To classify the point, examine the actual nearby function differences; their signs can vary from one function to another even when the Hessians agree.

Definitions in Terms of Nearby Differences

Let \(f\) be defined on an open region containing \((a,b)\). For a displacement \((h,k)\), define the function difference from \((a,b)\) by

\Delta f(h,k)=f(a+h,b+k)-f(a,b).

The point \((a,b)\) is a local minimum if there is a radius \(r>0\) such that \(f(a+h,b+k)\geq f(a,b)\) whenever \(0<\sqrt{h^2+k^2}<r\). It is a strict local minimum if the inequality is strict for every such displacement. For a local maximum, replace \(\geq\) by \(\leq\); for a strict local maximum, replace it by \(<\).

Definition: A stationary point \((a,b)\) is a saddle point if every neighborhood of \((a,b)\) contains points where \(f\) is greater than \(f(a,b)\) and points where \(f\) is less than \(f(a,b)\).

These definitions give useful sign tests. A positive difference for every sufficiently small nonzero \((h,k)\) establishes a strict local minimum. A negative difference for every sufficiently small nonzero \((h,k)\) establishes a strict local maximum. If arbitrarily small displacements give both positive and negative differences, the point is a saddle.

Non-strict extrema require care: if all sufficiently nearby differences are nonnegative, the point is a local minimum, but it need not be strict. Some nearby, distinct points may have the same function value. Similarly, nonpositive differences establish a local maximum, not necessarily a strict one.

What the Hessian Test Tells Us

For a function with continuous second partial derivatives, its Hessian at \((a,b)\) is

\mathbf{H}_f(a,b)= \begin{pmatrix} f_{xx}(a,b) & f_{xy}(a,b)\\ f_{yx}(a,b) & f_{yy}(a,b) \end{pmatrix}, \qquad D=f_{xx}(a,b)f_{yy}(a,b)-\bigl(f_{xy}(a,b)\bigr)^2.

At a stationary point, the two-variable test classifies the point when \(D>0\): a positive \(f_{xx}(a,b)\) gives a local minimum, and a negative \(f_{xx}(a,b)\) gives a local maximum. When \(D<0\), the Hessian is indefinite and the point is a saddle. If \(D=0\), none of these conclusions follows from the determinant test.

The determinant measures a property of the Hessian, not the full behavior of the function in a neighborhood. In particular, a zero determinant means the Hessian is singular; it does not tell us whether nearby function values lie above, below, or on both sides of the stationary value. The next examples check those differences directly.

Worked Examples with Zero Determinant

Worked Example: A Strict Local Minimum Despite a Zero Determinant

Consider \(f(x,y)=x^2+y^4\) at \((0,0)\). First compute the first partial derivatives:

f_x=2x,\qquad f_y=4y^3.

Thus \(f_x(0,0)=0\) and \(f_y(0,0)=0\), so the origin is stationary. The second partial derivatives and Hessian are

f_{xx}=2,\qquad f_{xy}=0,\qquad f_{yy}=12y^2, \qquad \mathbf{H}_f(0,0)= \begin{pmatrix} 2&0\\ 0&0 \end{pmatrix}.

Therefore \(D=(2)(0)-0^2=0\), so the determinant test does not classify the point. Instead, compute the difference from the value \(f(0,0)=0\):

\Delta f(h,k)=f(h,k)-f(0,0)=h^2+k^4.

For any nonzero \((h,k)\), at least one of \(h\) and \(k\) is nonzero. Hence at least one of \(h^2\) and \(k^4\) is positive, while neither is negative. Thus \(h^2+k^4>0\) for every nonzero displacement, however small. The origin is a strict local minimum. Direct substitution verifies the comparison: \(f(h,k)=h^2+k^4>0=f(0,0)\).

Worked Example: A Strict Local Maximum Despite a Zero Determinant

Let \(g(x,y)=-x^2-y^4\). Its first partial derivatives are \(g_x=-2x\) and \(g_y=-4y^3\), so the origin is stationary. Its Hessian is

\mathbf{H}_g(0,0)= \begin{pmatrix} -2&0\\ 0&0 \end{pmatrix}, \qquad D=(-2)(0)-0^2=0.

The difference from \(g(0,0)=0\) is

\Delta g(h,k)=g(h,k)-g(0,0)=-h^2-k^4.

For every nonzero \((h,k)\), \(h^2+k^4>0\), so \(-h^2-k^4<0\). Therefore every sufficiently close, distinct point has a smaller function value: the origin is a strict local maximum. Substitution confirms \(g(h,k)=-h^2-k^4<0=g(0,0)\). This example shows that a zero determinant does not rule out a strict extremum.

Worked Example: A Saddle Point with a Zero Determinant

Now consider \(p(x,y)=x^2-y^4\). Its first partial derivatives are \(p_x=2x\) and \(p_y=-4y^3\), so \((0,0)\) is stationary. The Hessian and determinant at the origin are

\mathbf{H}_p(0,0)= \begin{pmatrix} 2&0\\ 0&0 \end{pmatrix}, \qquad D=(2)(0)-0^2=0.

For displacements along the \(x\)-axis, take \((h,k)=(t,0)\), where \(t\ne0\). Then

\Delta p(t,0)=p(t,0)-p(0,0)=t^2>0.

For displacements along the \(y\)-axis, take \((h,k)=(0,t)\), again with \(t\ne0\). Then

\Delta p(0,t)=p(0,t)-p(0,0)=-t^4<0.

Both kinds of displacements can be made arbitrarily small by choosing \(t\) arbitrarily close to zero. Thus every neighborhood of the origin contains points with function values above and below \(p(0,0)=0\). The origin is a saddle point. Substituting the two paths into \(p\) gives \(p(t,0)=t^2>0\) and \(p(0,t)=-t^4<0\), confirming both signs.

The first and third examples have the same Hessian at the origin, yet one is a strict local minimum and the other is a saddle. Their common Hessian has determinant zero. This makes the limitation precise: that Hessian information alone cannot distinguish these outcomes.

Why the Sign Conditions Work

The classification by differences follows directly from the definitions. Here is the full argument for strict extrema and saddle behavior.

Strict minimum.
Suppose there is an \(r>0\) such that \(0<\sqrt{h^2+k^2}<r\) implies \(\Delta f(h,k)>0\). By the definition of \(\Delta f\), this means \(f(a+h,b+k)>f(a,b)\) at every distinct point within distance \(r\). That is exactly the definition of a strict local minimum.

Strict maximum.
If there is an \(r>0\) such that \(0<\sqrt{h^2+k^2}<r\) implies \(\Delta f(h,k)<0\), then \(f(a+h,b+k)<f(a,b)\) at every distinct point within distance \(r\). This is exactly the definition of a strict local maximum.

Saddle point.
If every radius \(r>0\) contains a displacement with \(\Delta f(h,k)>0\) and a displacement with \(\Delta f(h,k)<0\), then every neighborhood has values both above and below \(f(a,b)\). By definition, \((a,b)\) is a saddle point.

The same reasoning also handles non-strict extrema. If \(\Delta f(h,k)\geq0\) for all sufficiently small displacements, then \(f(a,b)\) is a local minimum; equality at some nonzero displacements means the argument does not establish strictness. If \(\Delta f(h,k)\leq0\) throughout such a neighborhood, the point is a local maximum, again not necessarily a strict one.

A Common Mistake: One Negative Difference Is Not a Maximum

A frequent error is to find a nearby point with a smaller function value and conclude that the stationary point is a local maximum. A single negative difference cannot establish that conclusion: a local maximum requires the differences to be nonpositive at all sufficiently close points.

For example, with \(p(x,y)=x^2-y^4\), the calculation along the \(y\)-axis gives \(p(0,t)-p(0,0)=-t^4<0\). The incorrect conclusion would be, “The difference is negative, so the origin is a local maximum.” But along the \(x\)-axis, \(p(t,0)-p(0,0)=t^2>0\). The correction is to compare both signs in every neighborhood: these positive and negative differences establish a saddle point, not a maximum.

A Non-Strict Minimum Can Also Have Zero Determinant

Worked Example: Distinguishing a Local Minimum from a Strict One

Consider \(q(x,y)=x^2\). At the origin, \(q_x=2x\) and \(q_y=0\), so the origin is stationary. The Hessian is

\mathbf{H}_q(0,0)= \begin{pmatrix} 2&0\\ 0&0 \end{pmatrix}, \qquad D=0.

The nearby difference is

\Delta q(h,k)=q(h,k)-q(0,0)=h^2\geq0.

Thus no nearby value is below \(q(0,0)=0\), and the origin is a local minimum. But if \(h=0\) and \(k\ne0\), then \(\Delta q(0,k)=0\). The inequality is not strict at these distinct points, so this calculation does not establish a strict local minimum. Indeed, \(q(0,k)=0=q(0,0)\) for every \(k\), confirming that the minimum is non-strict.

A Reliable Approach

When the Hessian determinant at a stationary point is zero, do not assign a classification from the determinant. Instead, compare nearby function values with the stationary value. A useful sequence is:

Confirm that the point is stationary by evaluating both first partial derivatives.
Compute the Hessian determinant and note that \(D=0\) leaves the second derivative test inconclusive.
Form \(\Delta f(h,k)=f(a+h,b+k)-f(a,b)\).
Determine whether the difference is positive, negative, nonnegative, nonpositive, or has both signs for sufficiently small displacements.
Match that conclusion to the definitions, checking whether equality can occur at distinct nearby points.

When proving a saddle, it is often enough to exhibit two paths approaching the point: one on which the difference is positive and another on which it is negative. Both paths must approach the point, so their signs occur in every neighborhood. A single path or a single nearby value cannot establish a strict extremum.

Summary: A zero Hessian determinant is inconclusive, not a classification. Strictly positive differences throughout a punctured neighborhood give a strict local minimum; strictly negative differences throughout give a strict local maximum; both signs arbitrarily close give a saddle. Nonnegative or nonpositive differences may establish non-strict extrema, so check whether equality occurs nearby.

Check Your Understanding

Use the definitions of nearby differences to answer the following questions.

What does a zero Hessian determinant allow you to conclude about a stationary point, and what does it not allow you to conclude?
What sign condition on every sufficiently small nonzero displacement establishes a strict local maximum?
For \(f(x,y)=x^2-y^4\), what are the signs of the differences along \((t,0)\) and \((0,t)\) when \(t\ne0\), and what do they establish?
Why does \(\Delta f(h,k)\geq0\) near a point establish a local minimum but not necessarily a strict local minimum?
Can two functions with the same Hessian at a stationary point have different classifications when the Hessian determinant is zero? Explain using examples from this tutorial.