Critical Point Calculator
Find critical points of single-variable or multivariable functions by solving f′(x)=0 or ∂f/∂x=0 and ∂f/∂y=0. Identify where derivatives are zero or undefined.
Supported: +, -, *, /, ^, sin, cos, tan, sqrt, log, ln, pi, e
How it works:
• If your function only uses x, it finds f'(x) and solves f'(x) = 0
• If your function uses x and y, it finds ∂f/∂x, ∂f/∂y and solves both = 0
Enter function and click Calculate to find critical points.
Enter function and click Calculate to find critical points.
For Single-Variable Functions f(x):
- Find the first derivative f'(x)
- Set f'(x) = 0 and solve for x
- Find points where f'(x) is undefined
- List all critical points
For Multivariable Functions f(x,y):
- Find partial derivatives ∂f/∂x and ∂f/∂y
- Set both partial derivatives equal to zero
- Solve the system of equations simultaneously
- Check for points where partials are undefined
Example: Find critical points of f(x) = 3x² + 4x + 9
Step 1: f(x) = 3x² + 4x + 9
Step 2: f'(x) = 6x + 4
Step 3: Set f'(x) = 0: 6x + 4 = 0
Step 4: Solve: 6x = -4 → x = -4/6 = -2/3
Step 5: Critical Point = -2/3 ≈ -0.6667
The function has one critical point at x = -2/3, where the derivative equals zero.
Example: Find critical points of f(x,y) = 4x² + 6xy + 8y
Step 1: f(x,y) = 4x² + 6xy + 8y
Step 2: ∂f/∂x = 8x + 6y
Step 3: ∂f/∂y = 6x + 8
Step 4: Solve system:
8x + 6y = 0
6x + 8 = 0
Step 5: From equation 2: x = -8/6 = -4/3
Step 6: Substitute: 8(-4/3) + 6y = 0 → y = 16/9
Step 7: Critical Point = (-4/3, 16/9) ≈ (-1.3333, 1.7778)
The function has one critical point at (-4/3, 16/9) where both partial derivatives equal zero.
What is a critical point?
A critical point is a point in the domain of a function where the derivative is either zero or undefined. These points are candidates for local maxima, minima, or inflection points.
How do you find critical points?
For single-variable functions, find f'(x) and solve f'(x) = 0, plus check where f'(x) is undefined. For multivariable functions, find all partial derivatives, set them equal to zero, and solve the system simultaneously.
Why are critical points important?
Critical points help identify local extrema (maxima and minima) of functions, which are essential for optimization problems in calculus, economics, engineering, and many other fields.
What's the difference between critical points and inflection points?
Critical points occur where f'(x) = 0 or is undefined, while inflection points occur where f''(x) = 0 and the concavity changes. A point can be both critical and an inflection point.
Disclaimer: This tool is for educational use only. Verify results for academic or professional work. Complex functions may require advanced techniques not covered by this simplified calculator.