Critical Points in Multivariable Functions

Core Definition

Let $z = f(x, y)$ be a scalar function of two variables defined on an open set in $\mathbb{R}^2$ . A point $(a, b)$ in the domain of $f$ is a critical point (or stationary point) of $f$ if either:

The gradient vector vanishes simultaneously: $\nabla f(a, b) = \mathbf{0} \iff \frac{\partial f}{\partial x}(a, b) = 0 \quad \text{and} \quad \frac{\partial f}{\partial y}(a, b) = 0$
Or at least one of the first-order partial derivatives $\frac{\partial f}{\partial x}$ or $\frac{\partial f}{\partial y}$ does not exist at $(a, b)$ .

For functions of three variables $w = f(x, y, z)$ , a point $(a, b, c)$ is a critical point if $\nabla f(a, b, c) = \langle 0, 0, 0 \rangle = \mathbf{0}$ or if any first partial derivative fails to exist.

Concept Explanation

Geometric Meaning: The Flat Tangent Plane

In single-variable calculus, critical points occur where the tangent line is horizontal ( $f'(x) = 0$ ) or sharp/undefined. In multivariable calculus, for a surface $z = f(x, y)$ , the tangent plane equation at $(a, b, f(a, b))$ is given by:

z - f(a, b) = f_x(a, b)(x - a) + f_y(a, b)(y - b)

When $(a, b)$ is a critical point where $f_x(a, b) = 0$ and $f_y(a, b) = 0$ , the tangent plane simplifies directly to:

z = f(a, b)

This means the tangent plane is completely horizontal (parallel to the

xy

-plane). At this point, the surface has zero slope along the

x

-axis, zero slope along the

y

-axis, and in fact zero directional derivative in every direction

\vec{u}

D_{\vec{u}} f(a, b) = \nabla f(a, b) \cdot \vec{u} = \mathbf{0} \cdot \vec{u} = 0

               z ^
                 |        Flat Tangent Plane: z = f(a, b)
                 |        +-------------------------+
                 |       /                         /
                 |      /            * (a, b, f)  /
                 |     +-------------------------+
                 |                .-'   '-.
                 |              .'         '.
                 |             /   Surface   \
                 |
                 +----------------------------------> y
                /
               /
            x v

Critical Points vs. Extrema

Locating a critical point is a necessary condition for an interior local extremum of a differentiable function (Fermat's Theorem for Multivariable Functions):

If $f$ has a local maximum or local minimum at an interior point $(a, b)$ , and the first partial derivatives exist, then $\nabla f(a, b) = \mathbf{0}$ .
However, having $\nabla f(a, b) = \mathbf{0}$ is not sufficient to guarantee a maximum or minimum. The point could be a saddle point (where the surface curves upward in one direction and downward in another), or an inflection trough. Classifying the nature of a critical point requires The Second Derivative Test.

Step-by-Step Procedure to Locate Critical Points

Compute First Partial Derivatives: Calculate the symbolic expressions for $f_x(x, y)$ and $f_y(x, y)$ across the domain.
Formulate the Simultaneous System: Set both partial derivatives equal to zero: $\begin{cases} f_x(x, y) = 0 \\ f_y(x, y) = 0 \end{cases}$
Solve the System Algebraically:
- Look for single-variable equations first to isolate numerical values.
- If equations are coupled or non-linear, factor where possible, or substitute expressions (e.g., $y = g(x)$ ) into the other equation.
- Caution: Avoid dividing both sides of an equation by a variable expression without first checking if that variable could equal zero.
Identify Singular Points: Check the original domain of $f$ for points where $f_x$ or $f_y$ ceases to exist (e.g., roots in denominators, cone vertices like $f(x, y) = \sqrt{x^2 + y^2}$ ).

Engineering Applications

Aerospace & Mechanical Engineering: Equilibrium configurations of mechanical mechanisms occur at stationary points of the potential energy surface $V(q_1, q_2)$ . Setting $\nabla V = \mathbf{0}$ yields the equilibrium states of linkages and robotic trusses.
Electrical Engineering: Static bias points in multi-transistor analog circuits correspond to critical points of node-voltage equations, ensuring amplifiers operate in their linear active regions without clipping.
Chemical Engineering: In thermodynamic phase equilibrium, stationary points of the Gibbs free energy function determine the coexistence of vapor and liquid phases in distillation columns.

Example Problems

Example 1: Decoupled Polynomial Critical Points

Find all critical points of the function:

f(x, y) = x^3 - 3x + y^2 - 4y + 7

Solution

Compute first partial derivatives:
$f_x(x, y) = \frac{\partial}{\partial x}(x^3 - 3x + y^2 - 4y + 7) = 3x^2 - 3$ $f_y(x, y) = \frac{\partial}{\partial y}(x^3 - 3x + y^2 - 4y + 7) = 2y - 4$
Set each partial derivative to zero:
$3x^2 - 3 = 0 \implies 3(x^2 - 1) = 0 \implies x = 1 \quad \text{or} \quad x = -1$ $2y - 4 = 0 \implies 2y = 4 \implies y = 2$
Assemble coordinate pairs: Combining the $x$ -solutions with the unique $y$ -solution yields two distinct critical points:
$(1, 2) \quad \text{and} \quad (-1, 2)$

Example 2: Coupled Non-Linear Critical Points