Extrema Calculator
Free extrema calculator for finding local and absolute maximum and minimum values. Get step-by-step solutions with derivative tests and critical point analysis. Perfect for calculus students learning optimization and extrema identification.
Last updated: December 15, 2024
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Use x² for squared, x³ for cubed, etc.
Specify interval for absolute extrema
Extrema Analysis
First Derivative:
f'(x) = 3x² - 6x
Second Derivative:
f''(x) = 6x - 6
Critical Points:
Local Extrema:
Local Maximum at x = 0
f(0) = 2
Second derivative: -6 < 0
Local Minimum at x = 2
f(2) = -2
Second derivative: 6 > 0
Absolute Extrema:
Absolute Maximum:
x = 0, f(0) = 2
Absolute Minimum:
x = 2, f(2) = -2
Step-by-Step Solution:
Tips for Finding Extrema:
- • Critical points occur where f'(x) = 0 or f'(x) is undefined
- • Use second derivative test: f''(c) > 0 → minimum, f''(c) < 0 → maximum
- • Check endpoints of closed intervals for absolute extrema
- • Local extrema are highest/lowest points in a neighborhood
- • Absolute extrema are highest/lowest over entire domain
Types of Extrema & Tests
Test condition
f''(c) < 0
Second derivative negative indicates concave down
Test condition
f''(c) > 0
Second derivative positive indicates concave up
Finding method
Compare all critical points
Evaluate at critical points and endpoints
Finding method
Compare all critical points
Find smallest value among candidates
Method
Check sign changes
+ to - is max, - to + is min
Method
Evaluate f''(c)
Faster than first derivative test when applicable
Quick Example Result
For function f(x) = x³ - 3x² + 2:
Local Maximum
x = 0, f(0) = 2
Local Minimum
x = 2, f(2) = -2
How to Find Extrema
Finding extrema is a fundamental calculus skill for optimization problems. The process involves using derivatives to identify critical points where the function reaches maximum or minimum values, then classifying these points using derivative tests.
The Extrema-Finding Process
This systematic approach ensures all extrema are identified and properly classified.
Second Derivative Test
The second derivative test is a quick method for classifying critical points. At a critical point c where f'(c) = 0, evaluate the second derivative f''(c). If positive, the function is concave up, indicating a local minimum. If negative, it's concave down, indicating a local maximum.
- If f''(c) > 0, the function has a local minimum at x = c
- If f''(c) < 0, the function has a local maximum at x = c
- If f''(c) = 0, the test is inconclusive (use first derivative test)
- The test only applies where f'(c) = 0
Sources & References
- Calculus: Early Transcendentals - James Stewart (9th Edition)Comprehensive coverage of extrema and optimization
- Thomas' Calculus - Weir, Hass, Giordano (14th Edition)Detailed explanations of derivative tests and applications
- Khan Academy - Extrema and OptimizationVideo tutorials and practice problems on finding extrema
Need help with other calculus topics? Check out our critical numbers calculator and derivative calculator.
Get Custom Calculator for Your PlatformExtrema Finding Example
Given Function:
Solution Steps:
- Step 1: Find the first derivative f'(x)
- f'(x) = 3x² - 6x
- Step 2: Set f'(x) = 0 to find critical points
- 3x² - 6x = 0
- 3x(x - 2) = 0
- Critical points: x = 0, x = 2
- Step 3: Use second derivative test
- f''(x) = 6x - 6
- At x = 0: f''(0) = -6 < 0 → Local Maximum
- At x = 2: f''(2) = 6 > 0 → Local Minimum
Extrema Found:
Local Maximum:
x = 0, f(0) = 2
Local Minimum:
x = 2, f(2) = -2
Quadratic Example
f(x) = x² - 4x + 5
Minimum at x = 2, f(2) = 1
Absolute Extrema
On [0, 3]: f(x) = x³ - 3x² + 2
Check x = 0, 2, 3
Frequently Asked Questions
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