Intervals of Decrease and Increase Calculator
Understanding intervals of decrease and increase is fundamental to calculus and function analysis. This calculator helps you determine where a function is increasing or decreasing by analyzing its derivative. The results are presented in interval notation, which clearly shows the function's behavior across its domain.
What are Intervals of Decrease and Increase?
In calculus, the intervals of increase and decrease describe the behavior of a function over its domain. A function is increasing on an interval if, as the input increases, the output also increases. Conversely, a function is decreasing on an interval if an increase in input leads to a decrease in output.
To determine these intervals, we examine the first derivative of the function. The critical points (where the derivative is zero or undefined) divide the domain into intervals. By testing a point from each interval in the derivative, we can determine whether the function is increasing or decreasing there.
Key Concept: The first derivative test is a fundamental tool for analyzing function behavior. It helps identify local maxima, minima, and the intervals where a function is increasing or decreasing.
How to Calculate Intervals of Decrease and Increase
Calculating intervals of increase and decrease involves these steps:
- Find the derivative of the function.
- Determine the critical points by solving f'(x) = 0 or where f'(x) is undefined.
- Sort the critical points to create intervals.
- Test a point from each interval in the derivative to determine its sign.
- Conclude whether the function is increasing (f'(x) > 0) or decreasing (f'(x) < 0) on each interval.
This process is essential for understanding the shape and behavior of functions in calculus.
Worked Example
Let's find the intervals of increase and decrease for the function f(x) = x³ - 3x².
- Compute the derivative: f'(x) = 3x² - 6x
- Find critical points: 3x² - 6x = 0 → 3x(x - 2) = 0 → x = 0 or x = 2
- Create intervals: (-∞, 0), (0, 2), (2, ∞)
- Test points:
- In (-∞, 0): x = -1 → f'(-1) = 3(-1)² - 6(-1) = 3 + 6 = 9 > 0 → Increasing
- In (0, 2): x = 1 → f'(1) = 3(1)² - 6(1) = 3 - 6 = -3 < 0 → Decreasing
- In (2, ∞): x = 3 → f'(3) = 3(3)² - 6(3) = 27 - 18 = 9 > 0 → Increasing
Therefore, the function f(x) = x³ - 3x² is increasing on (-∞, 0) and (2, ∞), and decreasing on (0, 2).