Calculus Tool

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

Automatic first and second derivative calculation
Identifies local and absolute extrema
Second derivative test classification

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Extrema Calculator
Find local and absolute extrema of functions

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:

x = 0x = 2

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:

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

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

Local Maximum
Highest point in a neighborhood

Test condition

f''(c) < 0

Second derivative negative indicates concave down

Local Minimum
Lowest point in a neighborhood

Test condition

f''(c) > 0

Second derivative positive indicates concave up

Absolute Maximum
Highest value over entire domain

Finding method

Compare all critical points

Evaluate at critical points and endpoints

Absolute Minimum
Lowest value over entire domain

Finding method

Compare all critical points

Find smallest value among candidates

First Derivative Test
Sign change analysis of f'(x)

Method

Check sign changes

+ to - is max, - to + is min

Second Derivative Test
Concavity analysis at critical points

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

Step 1: Find the first derivative f'(x)
Step 2: Set f'(x) = 0 and solve for critical points
Step 3: Find where f'(x) is undefined
Step 4: Apply second derivative test: f''(c) > 0 → min, f''(c) < 0 → max
Step 5: For absolute extrema, check endpoints of closed intervals

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.

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Extrema Finding Example

Step-by-Step Solution
Find extrema of f(x) = x³ - 3x² + 2 using derivative tests

Given Function:

f(x) = x³ - 3x² + 2

Solution Steps:

  1. Step 1: Find the first derivative f'(x)
  2. f'(x) = 3x² - 6x
  3. Step 2: Set f'(x) = 0 to find critical points
  4. 3x² - 6x = 0
  5. 3x(x - 2) = 0
  6. Critical points: x = 0, x = 2
  7. Step 3: Use second derivative test
  8. f''(x) = 6x - 6
  9. At x = 0: f''(0) = -6 < 0 → Local Maximum
  10. 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

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