Linear Algebra
12.1310 min read

Calculating Critical Points

To find all critical points of ff, solve the system f(x)=0\nabla f(\mathbf{x}) = \mathbf{0}, i.e., all nn partial derivatives equal to zero simultaneously. This is an nn-equation system in nn unknowns.

For polynomial or rational functions, this typically reduces to solving polynomial equations. The system may have zero, finitely many, or infinitely many solutions. Each solution is a critical point — but we must classify it further to determine if it is a local minimum, local maximum, or saddle point.

Example: for f(x,y)=x33x+y22yf(x,y) = x^3 - 3x + y^2 - 2y: f/x=3x23=0x=±1\partial f/\partial x = 3x^2 - 3 = 0 \Rightarrow x = \pm 1 f/y=2y2=0y=1\partial f/\partial y = 2y - 2 = 0 \Rightarrow y = 1 Critical points: (1,1)(1, 1) and (1,1)(-1, 1).

Formal View

Example 12.2 — Finding Critical Points
For f(x,y)=x2+xy+y23xf(x,y) = x^2 + xy + y^2 - 3x: f/x=2x+y3=0\partial f/\partial x = 2x + y - 3 = 0 f/y=x+2y=0x=2y\partial f/\partial y = x + 2y = 0 \Rightarrow x = -2y Substituting: 2(2y)+y3=03y=3y=12(-2y) + y - 3 = 0 \Rightarrow -3y = 3 \Rightarrow y = -1, x=2x = 2 Unique critical point: (2,1)(2, -1).

Why This Matters

Solving f=0\nabla f = \mathbf{0} analytically gives the exact critical points when possible.

  • Finding MLE estimators: set gradient of log-likelihood to zero and solve
  • Deriving least squares solutions: Axb2=0\nabla \|A\mathbf{x}-\mathbf{b}\|^2 = 0 gives normal equations
  • Physics: equilibrium positions solve V=0\nabla V = \mathbf{0} where VV is potential energy

Quiz

Question 1

For f(x,y)=x2+4y22x+8yf(x,y) = x^2 + 4y^2 - 2x + 8y, the critical point is at:

Question 2

A differentiable function of two variables always has at least one critical point.

Common Mistakes

  • Solving each equation independently without checking they are simultaneously satisfied.
  • Stopping after finding one critical point without checking if others exist.
  • Assuming every system f=0\nabla f = \mathbf{0} has a unique solution — it can have zero, one, or many solutions.