Critical Points Calculator

Published on: July 21, 2026
Final Answer: Free Full Steps: Pro

This Critical Points Calculator finds the critical points of \(f(x, y)\) by solving \(f_x = 0\) and \(f_y = 0\), then classifies each one as a local maximum, local minimum, or saddle point using the second-derivative test.

Step-by-step method

  1. Set up the function.
  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:

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

Example 1:

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

Step 1 - Set up the function.

In this problem: We analyze \(f = x^{3} - 3 x y + y^{3}\).

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

Step 2 - Find the first partial derivatives.

In this problem: The first partials are \(f_x = 3 x^{2} - 3 y\) and \(f_y = - 3 x + 3 y^{2}\).

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

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

In this problem: Solving the system gives the critical point(s) \((0, 0),\; (1, 1)\).

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

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

In this problem: The second partials are \(f_{xx} = 6 x\), \(f_{yy} = 6 y\), \(f_{xy} = -3\).

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

Step 5 - Classify each critical point.

In this problem: At \((0, 0)\): \(D = -9\), \(f_{xx} = 0\) → a saddle point. At \((1, 1)\): \(D = 27\), \(f_{xx} = 6\) → a local minimum.

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

Final answer:

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

Example 2:

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

Step 1 - Set up the function.

In this problem: We analyze \(f = x^{2} + y^{2}\).

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

Step 2 - Find the first partial derivatives.

In this problem: The first partials are \(f_x = 2 x\) and \(f_y = 2 y\).

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

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

In this problem: Solving the system gives the critical point(s) \((0, 0)\).

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

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

In this problem: The second partials are \(f_{xx} = 2\), \(f_{yy} = 2\), \(f_{xy} = 0\).

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

Step 5 - Classify each critical point.

In this problem: At \((0, 0)\): \(D = 4\), \(f_{xx} = 2\) → a local minimum.

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

Final answer:

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