Intervals of Increase and Decrease Calculator Webassign

This calculator helps you determine where a function is increasing or decreasing. It's compatible with WebAssign assignments and provides clear results with visual representation.

What are Intervals of Increase and Decrease?

In calculus, the intervals of increase and decrease of a function describe the regions where the function's value is rising or falling as the input variable changes. These concepts are fundamental to understanding the behavior of functions and are essential in many applications, including optimization problems and curve sketching.

Key Concept: A function is increasing on an interval if, for any two points in that interval, the function value at the right point is greater than at the left point. Conversely, a function is decreasing if the function value decreases as the input increases.

To determine these intervals, we typically analyze the first derivative of the function. The first derivative tells us the slope of the tangent line to the function at any point, which directly indicates whether the function is increasing or decreasing at that point.

How to Find Intervals of Increase and Decrease

The standard method for finding intervals of increase and decrease involves these steps:

Find the first derivative of the function, f'(x).
Determine the critical points by solving f'(x) = 0 or where f'(x) is undefined.
Test the intervals between critical points to determine where f'(x) > 0 (increasing) and where f'(x) < 0 (decreasing).

f'(x) > 0 on interval (a, b) → f(x) is increasing on (a, b) f'(x) < 0 on interval (a, b) → f(x) is decreasing on (a, b)

This method works for continuous functions on closed intervals. For functions with vertical asymptotes or other discontinuities, additional analysis may be required.

Worked Example

Let's find the intervals of increase and decrease for the function f(x) = x³ - 3x².

Step 1: Find the first derivative

f'(x) = d/dx (x³ - 3x²) = 3x² - 6x

Step 2: Find critical points

Set f'(x) = 0: 3x² - 6x = 0 → 3x(x - 2) = 0 → x = 0 or x = 2

Step 3: Test intervals

For x < 0: Test x = -1 → f'(-1) = 3(-1)² - 6(-1) = 3 + 6 = 9 > 0 → Increasing
For 0 < x < 2: Test x = 1 → f'(1) = 3(1)² - 6(1) = 3 - 6 = -3 < 0 → Decreasing
For x > 2: Test x = 3 → f'(3) = 3(3)² - 6(3) = 27 - 18 = 9 > 0 → Increasing

Result

f(x) is increasing on (-∞, 0) and (2, ∞), and decreasing on (0, 2).

FAQ

What if the first derivative is zero over an entire interval?: If f'(x) = 0 for all x in an interval, the function is constant on that interval and neither increasing nor decreasing.
How do I handle functions with vertical asymptotes?: Vertical asymptotes occur where the function is undefined. You should consider the intervals on either side of the asymptote separately when testing for increase/decrease.
What if the function is not continuous?: For discontinuous functions, you should analyze the intervals of continuity separately. The function may have different increasing/decreasing behavior on each continuous piece.
Can I use this calculator for WebAssign assignments?: Yes, this calculator is designed to be compatible with WebAssign assignments. The results it provides match the expected format for such assignments.