Classifying Nondifferentiable Critical Points

Why Nondifferentiable Critical Points Need a Different Approach

For a differentiable function of two variables, the gradient gives a familiar way to locate possible interior extrema: at a differentiable local extremum, the gradient must be zero. But a function can have a sharp cusp or corner where a first partial derivative does not exist. Such a point can still be a local minimum, a local maximum, or neither.

At these points, there is no gradient equation to solve. Instead, classification comes from the function’s values in a neighborhood. Following selected paths toward the point can quickly rule out an extremum. To prove an extremum, however, we must control the function’s values in every direction nearby, not just along a few paths.

Key idea: A nearby value on the wrong side of the proposed extremum disproves it. Values on both sides, arbitrarily close to the point, establish that the point is neither a local minimum nor a local maximum.

Critical Points and Local Extrema

Let \(f\) be a real-valued function defined on an open subset \(D\) of \(\mathbb{R}^2\). A point \((a,b)\) in \(D\) is a critical point if either both first partial derivatives are zero there, or at least one of the first partial derivatives does not exist there. This tutorial focuses on the second case.

Definition: The point \((a,b)\) is a local minimum of \(f\) if there is a radius \(r>0\) such that \(f(x,y)\geq f(a,b)\) whenever \((x,y)\in D\) and \(0<\sqrt{(x-a)^2+(y-b)^2}<r\). It is a local maximum if the inequality is reversed. It is a saddle point if every neighborhood of \((a,b)\) contains points where \(f(x,y)>f(a,b)\) and points where \(f(x,y)<f(a,b)\).

The inequalities in these definitions concern all nearby points in the domain. A path provides only a selection of nearby points. This makes path tests especially useful for disproving an extremum, but insufficient by themselves to prove one unless the argument also controls all nearby points.

How a Path Tests a Proposed Classification

A path approaching \((a,b)\) can be represented by a pair of functions \((x(t),y(t))\) such that \((x(t),y(t))\to(a,b)\) as \(t\to0\). Substituting this path into \(f\) reduces the comparison to one variable: compare \(f(x(t),y(t))\) with \(f(a,b)\).

To rule out a local minimum.
Find points arbitrarily close to \((a,b)\) where the function value is less than \(f(a,b)\).

To rule out a local maximum.
Find points arbitrarily close to \((a,b)\) where the function value is greater than \(f(a,b)\).

To establish a saddle point.
Find points of both kinds, with values strictly above and strictly below \(f(a,b)\), arbitrarily close to the point.

The reason this works follows directly from the definitions. If \((a,b)\) were a local minimum, every sufficiently close point in the domain would have value at least \(f(a,b)\). One nearby point with a lower value contradicts that requirement. The local-maximum argument is the same with the inequalities reversed. If points of both kinds occur in every neighborhood, neither extremum condition can hold, and the definition of a saddle point is satisfied.

Worked Examples with Cusps and Absolute Values

Worked Example: A Cusp That Is a Local Minimum

Classify the origin for \(f(x,y)=|x|+y^2\).

First calculate the function value at the origin:

$$f(0,0)=|0|+0^2=0.$$

Check the partial derivative with respect to \(x\) using its difference quotient:

\frac{f(h,0)-f(0,0)}{h}=\frac{|h|}{h}=\begin{cases}1,&h>0,\\-1,&h<0.\end{cases}

The one-sided limits differ, so \(f_x(0,0)\) does not exist. For the other partial derivative,

\frac{f(0,h)-f(0,0)}{h}=\frac{h^2}{h}=h\longrightarrow 0,

so \(f_y(0,0)=0\). The origin is a critical point because one first partial derivative fails to exist.

To classify it, compare values at arbitrary nearby points. Since \(|x|\geq0\) and \(y^2\geq0\),

f(x,y)=|x|+y^2\geq0=f(0,0).

Every point other than the origin has either \(x\ne0\), making \(|x|>0\), or \(y\ne0\), making \(y^2>0\). Thus the inequality is strict for every nearby point other than the origin. The origin is a strict local minimum, despite the cusp in the \(x\)-direction.

Check: Along the path \((x,y)=(t,0)\), \(f(t,0)=|t|>0\) for \(t\ne0\); along \((x,y)=(0,t)\), \(f(0,t)=t^2>0\). Both path calculations agree with the full inequality, which proves the minimum.

Worked Example: A Cusp That Is a Local Maximum

Classify the origin for \(g(x,y)=-|x|-y^2\).

At the origin, \(g(0,0)=0\). The partial derivative with respect to \(x\) fails to exist because

\frac{g(h,0)-g(0,0)}{h}=\frac{-|h|}{h}=\begin{cases}-1,&h>0,\\1,&h<0.\end{cases}

The one-sided limits are unequal. For the \(y\)-partial,

\frac{g(0,h)-g(0,0)}{h}=\frac{-h^2}{h}=-h\longrightarrow0,

so \(g_y(0,0)=0\). The origin is therefore a critical point.

For every \((x,y)\), both \(-|x|\leq0\) and \(-y^2\leq0\), so

g(x,y)=-|x|-y^2\leq0=g(0,0).

If \((x,y)\ne(0,0)\), at least one of \(|x|\) and \(y^2\) is positive, so \(g(x,y)<0\). Hence the origin is a strict local maximum. Along the \(x\)-axis, \(g(t,0)=-|t|<0\) for \(t\ne0\); along the \(y\)-axis, \(g(0,t)=-t^2<0\). The all-points inequality, rather than these two paths alone, establishes the maximum.

Check: Substitution at \((0,0)\) gives \(g(0,0)=0\), and the inequality shows every other point has a smaller value, as required for a strict local maximum.

Worked Example: A Nondifferentiable Saddle Point

Define \(q(x,y)=|x|+|y|-3\min(|x|,|y|)\). Classify the origin.

First, \(q(0,0)=0\). Along the \(x\)-axis,

q(h,0)=|h|+0-3\min(|h|,0)=|h|.

Thus the \(x\)-partial difference quotient is \(|h|/h\), which is \(1\) for \(h>0\) and \(-1\) for \(h<0\). It has no two-sided limit, so \(q_x(0,0)\) does not exist. Similarly, \(q(0,h)=|h|\), and \(q_y(0,0)\) does not exist. The origin is a critical point.

Now compare values along two paths. On the \(x\)-axis, for \(t>0\),

q(t,0)=t+0-3\min(t,0)=t>0=q(0,0).

On the diagonal \(x=y=t\), also for \(t>0\),

q(t,t)=t+t-3\min(t,t)=2t-3t=-t<0=q(0,0).

Both paths approach the origin as \(t\to0^+\). The first supplies arbitrarily nearby values greater than \(q(0,0)\); the second supplies arbitrarily nearby values less than it. The origin is therefore a saddle point.

Check: For every \(t>0\), the axis value is \(t\) and the diagonal value is \(-t\). These have opposite signs for every such \(t\), so the comparison holds arbitrarily close to the origin, not just at one distance.

Why a Few Positive Paths Do Not Prove a Minimum

The saddle-point example also illustrates an important limitation. Along each coordinate axis, the function is positive away from the origin. If we inspected only those paths, we might incorrectly guess that the origin is a local minimum. The diagonal path disproves that guess.

This is a general logical distinction: to disprove a local minimum, one lower nearby value is enough; to prove a local minimum, the inequality must hold throughout some neighborhood. Testing a finite selection of paths cannot rule out a different path, or other nearby points, with lower values.

Common pitfall: Checking that a function increases along the \(x\)- and \(y\)-axes and concluding that the point is a local minimum. For \(q\), the axis calculations give \(q(t,0)=t>0\) and \(q(0,t)=t>0\) when \(t>0\), but the diagonal calculation gives \(q(t,t)=-t<0\). The correction is to treat path evidence as a way to test and often disprove a classification; prove an extremum with an inequality valid for all nearby points.

A Precise Path Argument

The classification logic can be stated directly. Suppose two paths \(\gamma_+(t)\) and \(\gamma_-(t)\) lie in the domain for \(t>0\), both approach \((a,b)\) as \(t\to0^+\), and satisfy

f(\gamma_+(t))>f(a,b) \qquad\text{and}\qquad f(\gamma_-(t))<f(a,b)

for all sufficiently small \(t>0\). Let any neighborhood of \((a,b)\) be given. Because both paths approach \((a,b)\), each path has points inside that neighborhood for sufficiently small positive \(t\). The displayed inequalities give one point in the neighborhood with a higher value and one with a lower value. Since the neighborhood was arbitrary, every neighborhood contains both kinds of points. By definition, \((a,b)\) is a saddle point.

If only the lower-value path is available, it rules out a local minimum but does not by itself rule out a local maximum. The function might still be no greater than \(f(a,b)\) at all other nearby points. Likewise, a higher-value path rules out a local maximum but does not by itself establish a minimum.

A Reliable Classification Procedure

Evaluate the function. Find \(f(a,b)\), the value against which nearby values will be compared.
Check the partial derivatives. Use the definition of a partial derivative when a cusp or absolute value is present. Unequal one-sided difference-quotient limits show that the partial derivative does not exist.
Choose informative paths. Start with simple paths such as \(x=a+t,\ y=b\), \(x=a,\ y=b+t\), and, when useful, a diagonal path. Substitute into the function and simplify fully.
Interpret the inequalities carefully. A lower nearby value rules out a minimum; a higher nearby value rules out a maximum; both signs arbitrarily close establish a saddle.
Prove an extremum for all nearby points. If the evidence suggests a minimum or maximum, seek an inequality that holds throughout a neighborhood, not only on the paths tested.

Nondifferentiability does not determine the classification on its own. The cusps in the first two examples lead to opposite types of extrema, while the third nondifferentiable point is a saddle. What matters is how the function compares with its value at the point throughout the nearby region.

Check Your Understanding

Use the definitions and path comparisons above to answer these questions.

For \(f(x,y)=|x|+y^2\), which first partial derivative fails to exist at the origin, and what inequality proves the classification?
For \(g(x,y)=-|x|-y^2\), compute the \(x\)-partial difference quotient from the right and from the left. What do the two values show?
Why does finding one nearby point with a value below \(f(a,b)\) rule out a local minimum?
What two kinds of nearby values must be found to establish that a point is a saddle point?
Why do positive function values along both coordinate axes fail, by themselves, to prove a local minimum?