Critical Points Notes

Published on: September 29, 2026

This Critical Points Calculator finds where a surface levels off and decides what each of those places is, and shows the working line by line. Both partial derivatives are set to zero first, since a level point has no slope in either direction. The second derivative test then separates a peak from a dip from a saddle. Following the same pattern every time keeps finding the points and classifying them as separate moves.

Step-by-step method

  1. Set up the problem.
  2. Find the first partial derivatives.
  3. Solve f_x = 0 and f_y = 0 for the critical points.
  4. Form the second-derivative test D = f_xx f_yy - (f_xy)^2.
  5. Classify each critical point.

Formula:

\(D = f_{xx}f_{yy} - \left(f_{xy}\right)^{2}\)
See Example 1 Hide Example 1

Problem

\(f = x^{3} - 3 x y + y^{3}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Find the first partial derivatives.
  3. Solve f_x = 0 and f_y = 0 for the critical points.
  4. Form the second-derivative test D = f_xx f_yy - (f_xy)^2.
  5. Classify each critical point.

Step 1

Step 1 - Set up the problem.

In this problem:

\(f = x^{3} - 3 x y + y^{3}\)

Step 2

Step 2 - Find the first partial derivatives.

In this problem:

\(f_x = 3 x^{2} - 3 y,\quad f_y = - 3 x + 3 y^{2}\)

Step 3

Step 3 - Solve f_x = 0 and f_y = 0 for the critical points.

In this problem:

\(\text{Critical points: }(0, 0),\; (1, 1)\)

Step 4

Step 4 - Form the second-derivative test D = f_xx f_yy - (f_xy)^2.

In this problem:

\(f_{xx} = 6 x,\quad f_{yy} = 6 y,\quad f_{xy} = -3\)

Step 5

Step 5 - Classify each critical point.

In this problem:

\(\begin{gathered} (0, 0): \text{a saddle point} \\ (1, 1): \text{a local minimum} \end{gathered}\)

Final Answer

\(\begin{gathered} (0, 0): \text{a saddle point} \\ (1, 1): \text{a local minimum} \end{gathered}\)
See Example 2 Hide Example 2

Problem

\(f = x^{2} + y^{2}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Find the first partial derivatives.
  3. Solve f_x = 0 and f_y = 0 for the critical points.
  4. Form the second-derivative test D = f_xx f_yy - (f_xy)^2.
  5. Classify each critical point.

Step 1

Step 1 - Set up the problem.

In this problem:

\(f = x^{2} + y^{2}\)

Step 2

Step 2 - Find the first partial derivatives.

In this problem:

\(f_x = 2 x,\quad f_y = 2 y\)

Step 3

Step 3 - Solve f_x = 0 and f_y = 0 for the critical points.

In this problem:

\(\text{Critical points: }(0, 0)\)

Step 4

Step 4 - Form the second-derivative test D = f_xx f_yy - (f_xy)^2.

In this problem:

\(f_{xx} = 2,\quad f_{yy} = 2,\quad f_{xy} = 0\)

Step 5

Step 5 - Classify each critical point.

In this problem:

\(\begin{gathered} (0, 0): \text{a local minimum} \end{gathered}\)

Final Answer

\(\begin{gathered} (0, 0): \text{a local minimum} \end{gathered}\)
See Example 3 Hide Example 3

Problem

\(f = x^{2} - 4 x + y^{2} + 2 y\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Find the first partial derivatives.
  3. Solve f_x = 0 and f_y = 0 for the critical points.
  4. Form the second-derivative test D = f_xx f_yy - (f_xy)^2.
  5. Classify each critical point.

Step 1

Step 1 - Set up the problem.

In this problem:

\(f = x^{2} - 4 x + y^{2} + 2 y\)

Step 2

Step 2 - Find the first partial derivatives.

In this problem:

\(f_x = 2 x - 4,\quad f_y = 2 y + 2\)

Step 3

Step 3 - Solve f_x = 0 and f_y = 0 for the critical points.

In this problem:

\(\text{Critical points: }(2, -1)\)

Step 4

Step 4 - Form the second-derivative test D = f_xx f_yy - (f_xy)^2.

In this problem:

\(f_{xx} = 2,\quad f_{yy} = 2,\quad f_{xy} = 0\)

Step 5

Step 5 - Classify each critical point.

In this problem:

\(\begin{gathered} (2, -1): \text{a local minimum} \end{gathered}\)

Final Answer

\(\begin{gathered} (2, -1): \text{a local minimum} \end{gathered}\)
See Example 4 Hide Example 4

Problem

\(f = x^{2} - y^{2}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Find the first partial derivatives.
  3. Solve f_x = 0 and f_y = 0 for the critical points.
  4. Form the second-derivative test D = f_xx f_yy - (f_xy)^2.
  5. Classify each critical point.

Step 1

Step 1 - Set up the problem.

In this problem:

\(f = x^{2} - y^{2}\)

Step 2

Step 2 - Find the first partial derivatives.

In this problem:

\(f_x = 2 x,\quad f_y = - 2 y\)

Step 3

Step 3 - Solve f_x = 0 and f_y = 0 for the critical points.

In this problem:

\(\text{Critical points: }(0, 0)\)

Step 4

Step 4 - Form the second-derivative test D = f_xx f_yy - (f_xy)^2.

In this problem:

\(f_{xx} = 2,\quad f_{yy} = -2,\quad f_{xy} = 0\)

Step 5

Step 5 - Classify each critical point.

In this problem:

\(\begin{gathered} (0, 0): \text{a saddle point} \end{gathered}\)

Final Answer

\(\begin{gathered} (0, 0): \text{a saddle point} \end{gathered}\)