Tutorials › Calculus III › Classifying Polynomial Surfaces in Two Variables

Home  /  Tutorials  /  Critical Points and Local Extrema

Critical Points and Local Extrema · Tutorial 412 of 999

Classifying Polynomial Surfaces in Two Variables

Work through polynomial surfaces from their stationary points to Hessian classifications, including cases where higher-order terms are needed.

Intermediate 9 min read

What You'll Learn

  • Find stationary points by solving the two gradient equations for a polynomial.
  • Evaluate the Hessian entries and determinant at each stationary point.
  • Use the determinant and second-derivative tests to classify minima, maxima, and saddles.
  • Handle a polynomial with several distinct stationary points.
  • Recognize when a degenerate point requires analysis beyond the Hessian test.

From a Polynomial to Its Surface

A polynomial \(f(x,y)\) defines a surface \(z=f(x,y)\). To understand its local shape, begin by finding where the surface is stationary: points where both first partial derivatives vanish. Then evaluate the Hessian there to determine whether the surface curves upward, curves downward, or rises in some directions and falls in others.

Polynomials are useful examples because their derivatives are straightforward to compute, yet their gradient equations can have several solutions and their stationary points can have different local behavior. The goal is to classify every stationary point, not merely to inspect the surface at one convenient location.

Definition: A stationary point of a differentiable function \(f(x,y)\) is a point \((a,b)\) where \(f_x(a,b)=0\) and \(f_y(a,b)=0\). For a polynomial, these partial derivatives exist everywhere.

The Hessian Test at a Stationary Point

The Hessian matrix records the second partial derivatives. For a polynomial \(f\), it is

$$ H_f(x,y)= \begin{pmatrix} f_{xx}(x,y) & f_{xy}(x,y)\\ f_{xy}(x,y) & f_{yy}(x,y) \end{pmatrix}. $$

At a stationary point \((a,b)\), write \(D=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2\). The determinant test classifies the point when \(D\ne 0\).

Theorem: Let \(f\) have continuous second partial derivatives near a stationary point \((a,b)\). If \(D>0\) and \(f_{xx}(a,b)>0\), then \((a,b)\) is a strict local minimum. If \(D>0\) and \(f_{xx}(a,b)<0\), it is a strict local maximum. If \(D<0\), it is a saddle point. If \(D=0\), this test is inconclusive: further analysis is required.

These conditions concern the behavior near the point. A strict local minimum, for example, means that all sufficiently nearby distinct points have larger function values. The conclusion does not by itself say that the point is an absolute minimum over the whole plane.

Why the Determinant Test Works

Here is the key connection between the Hessian and the surface’s local shape. Near a stationary point \((a,b)\), the second-order Taylor expansion has the form

$$ f(a+h,b+k)-f(a,b) = \frac12\left(Ah^2+2Bhk+Ck^2\right) +\text{terms of higher order}, $$

where \(A=f_{xx}(a,b)\), \(B=f_{xy}(a,b)\), and \(C=f_{yy}(a,b)\). There are no constant changes besides \(f(a,b)\), and the first-order terms vanish because the point is stationary. Thus, when the quadratic expression is positive in every nonzero direction, it predicts a local minimum; when it is negative in every direction, it predicts a local maximum. If it takes both positive and negative values, it predicts a saddle.

To see how \(D=AC-B^2\) controls this expression, suppose first that \(A\ne0\). Completing the square gives

$$ Ah^2+2Bhk+Ck^2 = A\left(h+\frac{B}{A}k\right)^2+\frac{D}{A}k^2. $$

If \(D>0\), then \(A\) and \(D/A\) have the same sign, so both squared terms have the sign of \(A\). If \(D<0\), the coefficients \(A\) and \(D/A\) have opposite signs, giving directions of opposite curvature. This explains the minimum, maximum, and saddle cases. When \(D=0\), the quadratic expression may vanish in a direction, so higher-order terms can decide what the surface does there.

Worked Example: Two Stationary Points with Different Classifications

Worked Example: Two Stationary Points with Different Classifications

Classify the stationary points of \(f(x,y)=x^3-3x+y^2\).

Step 1: Find the gradient equations.

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

Set both equal to zero:

$$ 3x^2-3=0 \quad\Longrightarrow\quad x^2=1 \quad\Longrightarrow\quad x=1\text{ or }x=-1, \qquad 2y=0 \quad\Longrightarrow\quad y=0. $$

The stationary points are \((1,0)\) and \((-1,0)\). Substitution confirms that the gradient vanishes at each: \(f_x(1,0)=0\), \(f_y(1,0)=0\), \(f_x(-1,0)=0\), and \(f_y(-1,0)=0\).

Step 2: Compute the Hessian data.

$$ f_{xx}=6x,\qquad f_{xy}=0,\qquad f_{yy}=2. $$

At \((1,0)\), \(f_{xx}=6\), \(f_{xy}=0\), and \(f_{yy}=2\), so \(D=(6)(2)-0^2=12>0\). Since \(f_{xx}>0\), \((1,0)\) is a strict local minimum. Its function value is \(f(1,0)=1-3=-2\).

At \((-1,0)\), \(f_{xx}=-6\), \(f_{xy}=0\), and \(f_{yy}=2\), so \(D=(-6)(2)-0^2=-12<0\). Therefore, \((-1,0)\) is a saddle point. Its function value is \(f(-1,0)=-1+3=2\).

Check: The gradient equations gave exactly the two listed points, and direct substitution gives zero for both gradient components at each. The determinant test classifies one as a minimum and the other as a saddle.

Worked Example: Solving a Coupled Gradient System

Worked Example: Solving a Coupled Gradient System

Find and classify all stationary points of \(f(x,y)=x^4+y^4-4xy\).

Step 1: Solve the gradient equations.

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

Setting these derivatives to zero gives \(y=x^3\) and \(x=y^3\). Substituting \(y=x^3\) into \(x=y^3\) yields \(x=(x^3)^3=x^9\). Thus \(x=0\), or, when \(x\ne0\), \(x^8=1\). The real solutions are \(x=0,1,-1\). Using \(y=x^3\), the corresponding points are \((0,0)\), \((1,1)\), and \((-1,-1)\).

Substitution into the gradient equations checks all three: at the origin both components are zero; at \((1,1)\), they are \(4-4=0\) and \(4-4=0\); at \((-1,-1)\), they are \(-4+4=0\) and \(-4+4=0\).

Step 2: Compute and evaluate the Hessian.

$$ f_{xx}=12x^2,\qquad f_{xy}=-4,\qquad f_{yy}=12y^2, \qquad D=(12x^2)(12y^2)-16. $$

At \((0,0)\), \(f_{xx}=0\), \(f_{xy}=-4\), and \(f_{yy}=0\), so \(D=0\cdot0-(-4)^2=-16<0\). The origin is a saddle point, and \(f(0,0)=0\).

At \((1,1)\), \(f_{xx}=12\), \(f_{xy}=-4\), and \(f_{yy}=12\), so \(D=(12)(12)-(-4)^2=144-16=128>0\). Since \(f_{xx}=12>0\), this is a strict local minimum. Its value is \(f(1,1)=1+1-4=-2\).

At \((-1,-1)\), the second derivatives are again \(12,-4,12\), giving \(D=128>0\) and \(f_{xx}=12>0\). This is also a strict local minimum, with \(f(-1,-1)=1+1-4=-2\).

Check: The three points satisfy both gradient equations, and the Hessian test gives a saddle at the origin and strict local minima at the other two points.

Worked Example: When the Hessian Test Is Inconclusive

Worked Example: When the Hessian Test Is Inconclusive

Analyze the stationary point at the origin for \(f(x,y)=x^4+y^4\).

Step 1: Check stationarity and the Hessian.

$$ f_x=4x^3,\qquad f_y=4y^3, \qquad f_{xx}=12x^2,\qquad f_{xy}=0,\qquad f_{yy}=12y^2. $$

Both first partial derivatives vanish at \((0,0)\), so it is stationary. At that point all three second partial derivatives are zero, and \(D=(0)(0)-0^2=0\). The determinant test does not classify it.

Step 2: Examine the polynomial itself.

$$ f(x,y)=x^4+y^4\ge 0=f(0,0), $$

and equality holds only if \(x=0\) and \(y=0\). Therefore, every other point has a strictly larger function value. The origin is a strict local minimum (in fact, an absolute minimum), even though its Hessian test is inconclusive.

Contrast: The zero Hessian at the origin does not force a minimum. For \(g(x,y)=x^4-y^4\), the gradient also vanishes at the origin and all second partial derivatives there are zero, so \(D=0\). Along the \(x\)-axis, \(g(t,0)=t^4>0\) for \(t\ne0\); along the \(y\)-axis, \(g(0,t)=-t^4<0\). Points arbitrarily close to the origin therefore give values both above and below \(g(0,0)=0\). The origin is a saddle point.

Check: The exact inequalities for \(f\), and the two axis paths for \(g\), settle the classifications without relying on a determinant that equals zero.

A Common Classification Mistake

A common error is to treat \(D=0\) as though it meant “saddle.” It does not. Consider the incorrect calculation for \(f(x,y)=x^4+y^4\): “At the origin, \(D=0\), so the origin is a saddle.” The conclusion is unsupported, because the determinant test gives no classification in this case.

The correction is to examine the function itself or use other information. Here, \(x^4+y^4\ge0\), with equality only at the origin, so the origin is a strict minimum. By contrast, the example \(x^4-y^4\) has a saddle at the origin despite also having \(D=0\). The same inconclusive Hessian outcome can accompany different local behavior.

A Reliable Workflow for Polynomial Surfaces

When classifying a polynomial surface, organize the calculation so that no stationary point is overlooked.

1
Compute the first partial derivatives.
Write down \(f_x\) and \(f_y\) before solving anything.
2
Solve both gradient equations together.
List every real solution and check each one in both equations.
3
Compute the second partial derivatives.
Form the determinant \(D=f_{xx}f_{yy}-[f_{xy}]^2\) and evaluate it at each stationary point.
4
Apply the test point by point.
Use \(D<0\) for a saddle and \(D>0\) together with the sign of \(f_{xx}\) for a strict extremum.
5
Investigate every case with \(D=0\).
Use higher-order terms, inequalities, or carefully chosen paths; do not assign a classification from the determinant alone.

Why Classifying Polynomial Surfaces Matters

For a polynomial, finding stationary points reduces a geometric question to algebraic equations, while the Hessian translates local curvature into a test at each solution. A single surface can contain several types of stationary behavior, as the examples show. Keeping the two tasks separate—first find all candidates, then classify each one—makes the reasoning easier to verify.

The distinction between a decisive and an inconclusive determinant is just as important as the sign rules themselves. When \(D\ne0\), the quadratic part determines the local type. When \(D=0\), the polynomial may still have a clear minimum or saddle, but that conclusion must come from additional analysis.

Check Your Understanding

For each question, state the reasoning that supports your answer.

  1. What equations must a stationary point of a differentiable function of two variables satisfy?
  2. If \(D>0\) at a stationary point and \(f_{xx}<0\), what classification follows?
  3. For \(f(x,y)=x^3-3x+y^2\), what are the stationary points, and which is a saddle?
  4. Why does \(D=0\) fail to determine the classification by itself?
  5. For \(x^4-y^4\), what do the \(x\)-axis and \(y\)-axis values near the origin show?

Previous

Next

Back to Tutorials