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Critical Points and Local Extrema · Tutorial 401 of 999

Critical Points of Functions of Two Variables

Distinguish zero-gradient stationary points from points where differentiability fails, and learn why both types require investigation when locating local extrema.

Intermediate 10 min read

What You'll Learn

  • Define local maxima, local minima, stationary points, and critical points for functions of two variables.
  • Distinguish a zero gradient from failure of differentiability.
  • Explain why existing partial derivatives do not always guarantee differentiability.
  • Identify stationary and nondifferentiable points that may be local extrema.
  • Recognize that a critical point is a candidate, not a classification.

Why Critical Points Matter

A surface can rise, fall, flatten, or change direction as you move across its domain. Local extrema are points where the function’s value is at least as high, or at least as low, as nearby values. To locate them, it is useful to identify points where ordinary tests based on the gradient do not settle the question immediately.

There are two important possibilities. The function may be differentiable and have zero gradient, or it may fail to be differentiable. The first kind of point is called stationary. Both kinds are included in a broad definition of a critical point. Neither label, by itself, tells us whether the point is a maximum, a minimum, or neither.

Key idea: A critical point is a place to investigate, not a verdict. A zero gradient does not guarantee a local extremum, and a local extremum can occur where the function is not differentiable.

Local Maxima and Minima

Let \(f:D\to\mathbb{R}\), where \(D\) is a subset of the plane, and let \((a,b)\) be an interior point of \(D\). Being interior means that some open disk centered at \((a,b)\) lies entirely within \(D\). This ensures that nearby points in every direction are available for comparison.

Definition: The point \((a,b)\) is a local maximum of \(f\) if there is a radius \(r>0\) such that \(f(x,y)\leq f(a,b)\) for every \((x,y)\in D\) with \(\sqrt{(x-a)^2+(y-b)^2}<r\). It is a local minimum if the inequality is reversed. The extremum is strict if the corresponding inequality is strict whenever \((x,y)\ne(a,b)\).

The comparison is only with nearby points, not with every point in the domain. A local maximum need not be the largest value the function ever takes, and a local minimum need not be the smallest. The interior assumption matters too: points on the boundary may be extrema even though there is no full disk of domain points around them. Here we focus on interior points.

Stationary Points and Critical Points

When \(f\) is differentiable at \((a,b)\), its gradient is the vector of first partial derivatives. A stationary point is a differentiable point where that vector is zero.

$$ \nabla f(a,b) = \left\langle f_x(a,b),f_y(a,b)\right\rangle. $$
Definition: An interior point \((a,b)\) is a stationary point of \(f\) if \(f\) is differentiable there and \(\nabla f(a,b)=\mathbf{0}\). An interior point is a critical point if either \(f\) is differentiable there with \(\nabla f(a,b)=\mathbf{0}\), or \(f\) is not differentiable there.

This definition makes an important distinction. A stationary point has a zero gradient, and the gradient exists. A nondifferentiable critical point does not have a gradient in the usual differentiable sense. In particular, a partial derivative may fail to exist at a critical point. Also, as we will see, the partial derivatives can both exist even when the function is not differentiable.

Some introductory treatments describe critical points using the condition that the first partial derivatives are zero or fail to exist. The definition above also includes points where the partial derivatives exist but the function is not differentiable. This broader wording avoids mistaking the existence of two partial derivatives for differentiability.

Worked Example: A Stationary Point That Is a Minimum

Worked Example: A Stationary Point That Is a Minimum

Find the stationary point of \(f(x,y)=x^2+y^2\), and determine whether it is a local extremum.

First compute both partial derivatives:

$$ f_x(x,y)=2x, \qquad f_y(x,y)=2y. $$

Thus \(\nabla f(x,y)=\langle 2x,2y\rangle\). Setting both components equal to zero gives \(2x=0\) and \(2y=0\), so the only stationary point is \((0,0)\).

To check the function values near that point, compare any \((x,y)\) with the value at the origin:

$$ f(x,y)-f(0,0)=x^2+y^2-0=x^2+y^2\geq 0. $$

Equality holds only when \(x=y=0\). Therefore \((0,0)\) is a strict local minimum; in fact, it is also a global minimum. As a check, \(f(0,0)=0\), and every value \(x^2+y^2\) is nonnegative.

Worked Example: A Stationary Point That Is Not an Extremum

Worked Example: A Stationary Point That Is Not an Extremum

Consider \(g(x,y)=x^2-y^2\). Find its stationary point and test whether it is a local maximum or minimum.

The partial derivatives are

$$ g_x(x,y)=2x, \qquad g_y(x,y)=-2y. $$

Both are zero only when \(x=0\) and \(y=0\). Thus the origin is the only stationary point, and \(g(0,0)=0\).

To test the local-extremum definitions, approach the origin along the \(x\)-axis, where \(y=0\):

$$ g(t,0)=t^2>0 \qquad\text{for }t\ne 0. $$

Along the \(y\)-axis, where \(x=0\), the values instead satisfy

$$ g(0,t)=-t^2<0 \qquad\text{for }t\ne 0. $$

Both types of values occur arbitrarily close to the origin. Therefore the origin is neither a local maximum nor a local minimum. Its value is zero, so the examples above and below zero directly rule out both possibilities. This shows why finding a stationary point is not the same as classifying it.

Worked Example: A Nondifferentiable Point That Is a Minimum

Worked Example: A Nondifferentiable Point That Is a Minimum

Consider \(h(x,y)=|x|+y^2\). Examine the origin.

For the \(x\)-partial derivative at the origin, the difference quotient is

$$ \frac{h(t,0)-h(0,0)}{t} = \frac{|t|}{t}. $$

This quotient equals \(1\) for \(t>0\) and \(-1\) for \(t<0\), so it has no limit as \(t\to 0\). Thus \(h_x(0,0)\) does not exist, and \(h\) is not differentiable there. The origin is a critical point, but not a stationary point.

To check for a minimum, compare values directly:

$$ h(x,y)-h(0,0)=|x|+y^2\geq 0. $$

Equality holds only at \((0,0)\), so the origin is a strict local minimum. Substitution confirms the value \(h(0,0)=0\); all other nearby values are positive. A local extremum can therefore occur at a point where differentiability fails.

Why Partial Derivatives Are Not the Whole Story

Differentiability is a stronger condition than the separate existence of the two partial derivatives. Partial derivatives measure changes along the coordinate axes. Differentiability requires a linear approximation that accounts for changes in every direction, with an error that becomes small compared with the distance moved.

The next example shows why this distinction matters when identifying critical points.

Worked Example: Partial Derivatives Exist but Differentiability Fails

Define

$$ q(x,y)= \begin{cases} \dfrac{xy}{\sqrt{x^2+y^2}}, & (x,y)\ne(0,0),\\[6pt] 0, & (x,y)=(0,0). \end{cases} $$

On either coordinate axis, \(q\) is zero. Therefore the partial derivatives at the origin exist and satisfy

$$ q_x(0,0) = \lim_{t\to0}\frac{q(t,0)-q(0,0)}{t} =0, \qquad q_y(0,0) = \lim_{t\to0}\frac{q(0,t)-q(0,0)}{t} =0. $$

Nevertheless, \(q\) is not differentiable at the origin. If it were differentiable, its linear approximation would have to be \(q_x(0,0)x+q_y(0,0)y=0\). Along \((t,t)\), however,

$$ q(t,t)=\frac{t^2}{\sqrt{2}|t|} =\frac{|t|}{\sqrt{2}}, \qquad \frac{|q(t,t)-0|}{\sqrt{t^2+t^2}} = \frac{|t|/\sqrt{2}}{\sqrt{2}|t|} =\frac12. $$

The ratio does not tend to zero, as the error from a valid linear approximation would have to do. Thus differentiability fails even though both partial derivatives exist. The origin is a critical point under our definition, not a stationary point.

It is not a local extremum: along \((t,t)\), \(q(t,t)=|t|/\sqrt{2}>0\), while along \((t,-t)\), \(q(t,-t)=-|t|/\sqrt{2}<0\) for every \(t\ne0\). These values approach zero from both sides. Substitution into the defining formula verifies both signs and the value \(q(0,0)=0\).

Why a Zero Gradient Means No First-Order Change

The gradient summarizes directional rates of change at a differentiable point. If \(f\) is differentiable at \((a,b)\) and \(\mathbf{u}=\langle u_1,u_2\rangle\) is a unit vector, the directional derivative in that direction is

$$ D_{\mathbf{u}}f(a,b) = \nabla f(a,b)\cdot\mathbf{u} = f_x(a,b)u_1+f_y(a,b)u_2. $$

This formula explains the role of a zero gradient. The dot product of the zero vector with every direction is zero, so a stationary point has zero directional derivative in every direction. Conversely, if every directional derivative is zero, choosing the coordinate directions \(\langle1,0\rangle\) and \(\langle0,1\rangle\) shows that \(f_x(a,b)=0\) and \(f_y(a,b)=0\), hence \(\nabla f(a,b)=\mathbf{0}\).

This argument identifies stationary points as points with no first-order directional change. It does not say that nearby function values must be above or below the value at the point: the second example showed changes of both signs. And it only applies when the function is differentiable, so points where differentiability fails must be checked separately.

A Practical Way to Think About Candidates

When searching for interior local extrema, the distinction between the two types of critical points helps organize the investigation:

  • At a differentiable point, check whether both components of the gradient are zero. If so, the point is stationary and needs further analysis.
  • At a point where differentiability fails, do not discard it. It may still be a local extremum, as with \(|x|+y^2\).
  • Do not treat a zero gradient as proof of an extremum. Compare nearby values or use an appropriate classification method.
  • Do not assume that existing partial derivatives guarantee differentiability. The function \(q\) demonstrates why that check matters.

These are candidate-finding principles, not a complete method for classifying every candidate. The examples establish the main limitation: either type of critical point may or may not be a local extremum. Further analysis is needed to decide which case holds for a particular function.

Common Mistake: Setting a Missing Derivative Equal to Zero

A common error is to differentiate \(|x|\) at zero as though its derivative were zero because the function value there is zero. That confuses a function value with a derivative. The incorrect calculation would be \(h_x(0,0)=0\) on the grounds that \(h(0,0)=0\).

The difference quotient shows the problem precisely:

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

The one-sided limits disagree, so the partial derivative does not exist. The correction is to test the defining limit, not infer a derivative from the function’s value. The origin is critical because differentiability fails there; it is not stationary.

What the Labels Do—and Do Not—Tell You

Point type What is true What is not guaranteed
Stationary point The function is differentiable there and its gradient is zero. It need not be a local maximum or minimum.
Nondifferentiable critical point The function fails to be differentiable there. It need not be a local maximum or minimum.
Local extremum Nearby function values lie on one side of the value at the point. The point need not be differentiable.

Critical points matter because they flag places where local behavior deserves attention. The label is a starting point: to determine whether a local extremum actually occurs, inspect the function’s values near the point and keep the differentiability assumptions clear.

Check Your Understanding

Use the definitions and examples above to answer these questions.

  1. What condition makes a stationary point different from a nondifferentiable critical point?
  2. Why is \((0,0)\) a local minimum of \(x^2+y^2\)?
  3. How do the values of \(x^2-y^2\) near the origin show that its stationary point is not a local extremum?
  4. For \(h(x,y)=|x|+y^2\), what does the difference quotient for the \(x\)-partial derivative at the origin show?
  5. Why does the existence of both partial derivatives at a point not, by itself, establish differentiability there?

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