How To Find The Interval Of Increase

How to Find the Interval of Increase: A Step-by-Step Guide

Understanding where a function rises as you move from left to right—its interval of increase—is a fundamental skill in calculus. This concept reveals the behavior of a function, providing insights into its graph’s shape, trends in real-world data, and solutions to optimization problems. Whether you're analyzing profit margins, population growth, or the trajectory of a projectile, identifying these intervals is crucial. This guide will demystify the process, breaking it down into a clear, repeatable method grounded in the power of the derivative.

The Core Concept: What Does "Increasing" Mean?

Before applying any procedure, we must define our goal precisely. A function f(x) is said to be increasing on an interval if, for any two points x₁ and x₂ within that interval where x₁ < x₂, the function's value at x₂ is greater than at x₁ (f(x₂) > f(x₁)). Visually, the graph of the function moves upward as you travel from left to right across that interval.

The revolutionary tool that allows us to determine this without testing every possible pair of points is the derivative. The derivative f'(x) represents the instantaneous rate of change, or the slope, of the function f(x) at any given point x. This leads us to the first derivative test for increasing intervals:

A function f(x) is increasing on any interval where its derivative f'(x) > 0.

Therefore, our entire task reduces to a two-part problem:

1. Find the algebraic expression for the derivative, f'(x).
2. Solve the inequality f'(x) > 0 to find all x-values that satisfy it. The solution set to this inequality is the interval of increase.

The Step-by-Step Procedure: Your Action Plan

Follow this systematic approach for any differentiable function.

Step 1: Find the First Derivative, f'(x)

Apply the standard rules of differentiation (power rule, product rule, quotient rule, chain rule) to your function f(x). This derivative is your new function to analyze. For example:

• If f(x) = x³ - 6x² + 9x + 1, then f'(x) = 3x² - 12x + 9.
• If f(x) = (x² + 1)/(x - 2), you must use the quotient rule to find f'(x).

Step 2: Identify Critical Numbers

Critical numbers are the x-values where f'(x) is either zero or undefined. These points are the potential boundaries for your intervals of increase and decrease. To find them:

1. Solve f'(x) = 0.
2. Find where f'(x) is undefined (this often occurs where the original function f(x) itself is undefined, such as at vertical asymptotes or points of discontinuity). List all these critical numbers in ascending order. They will partition the real number line (or the domain of f) into separate intervals.

Step 3: Create a Sign Chart for f'(x)

This is the most critical analytical step. On a number line, mark all your critical numbers from Step 2. These points divide the line into distinct intervals. For each interval, you must determine the sign (positive + or negative -) of the derivative f'(x).

How to determine the sign: Choose any convenient test point within each interval (not at the boundaries) and plug it into your derivative f'(x).

• If f'(test point) > 0, then f'(x) > 0 for the entire interval, so f(x) is increasing there.
• If f'(test point) < 0, then f'(x) < 0 for the entire interval, so f(x) is decreasing there.

Mark these signs (+ or -) above each interval on your chart.

Step 4: Write the Interval(s) of Increase

Look at your completed sign chart. The intervals where you marked a + sign are precisely where the function is increasing. Express these intervals using interval notation (e.g., (-∞, -1), (2, 5)). Remember:

• Use parentheses ( ) for open intervals, which is standard when the endpoint is a critical point where the derivative is zero or undefined. The function's increasing behavior is strictly between these points.
• If the interval extends to infinity, use (-∞, a) or (b, ∞).
• Combine disjoint intervals using the union symbol ∪.

Worked Examples: From Simple to Complex

Example 1: A Simple Polynomial

Find the intervals of increase for f(x) = x³ - 3x² - 9x + 5.

1. Derivative: f'(x) = 3x² - 6x - 9.
2. Critical Numbers: Solve 3x² - 6x - 9 = 0. Divide by 3: x² - 2x - 3 = 0. Factor: (x - 3)(x + 1) = 0. So, x = -1 and x = 3. f'(x) is a polynomial, defined everywhere. Critical numbers: -1, 3.
3. Sign Chart: The number line is split into: (-∞, -1), (-1, 3), (3, ∞).
   • Test x = -2 in (-∞, -1): f'(-2) = 3(4) - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0 → +.
   • Test x = 0 in (-1, 3): f'(0) = 0 - 0 - 9 = -9 < 0 → -.
   • Test x = 4 in (3, ∞): `f'(4) = 3

(16) - 6(4) - 9 = 48 - 24 - 9 = 15 > 0 → +. 4. Conclusion: The function is increasing on the intervals (-∞, -1) and (3, ∞).

Example 2: A Rational Function

Find the intervals of increase for f(x) = (x² - 4)/(x - 1).

1. Derivative: Using the quotient rule, f'(x) = [(2x)(x - 1) - (x² - 4)(1)] / (x - 1)² = (2x² - 2x - x² + 4) / (x - 1)² = (x² - 2x + 4) / (x - 1)².
2. Critical Numbers:
   • Numerator = 0: Solve x² - 2x + 4 = 0. The discriminant is (-2)² - 4(1)(4) = 4 - 16 = -12 < 0. No real solutions, so the numerator is never zero.
   • Denominator = 0: f'(x) is undefined when (x - 1)² = 0, so x = 1. This is also where f(x) is undefined (a vertical asymptote). Critical number: 1.
3. Sign Chart: The number line is split into: (-∞, 1), (1, ∞).
   • Test x = 0 in (-∞, 1): f'(0) = (0 - 0 + 4) / (0 - 1)² = 4/1 = 4 > 0 → +.
   • Test x = 2 in (1, ∞): f'(2) = (4 - 4 + 4) / (2 - 1)² = 4/1 = 4 > 0 → +.
4. Conclusion: The function is increasing on the intervals (-∞, 1) and (1, ∞). Note that the function is not defined at x = 1, so we cannot include it in any interval.

Example 3: A Function with Multiple Critical Points

Find the intervals of increase for f(x) = x⁴ - 4x³.

1. Derivative: f'(x) = 4x³ - 12x² = 4x²(x - 3).
2. Critical Numbers: Solve 4x²(x - 3) = 0. So, x = 0 or x = 3. f'(x) is a polynomial, defined everywhere. Critical numbers: 0, 3.
3. Sign Chart: The number line is split into: (-∞, 0), (0, 3), (3, ∞).
   • Test x = -1 in (-∞, 0): f'(-1) = 4(1)(-4) = -16 < 0 → -.
   • Test x = 1 in (0, 3): f'(1) = 4(1)(-2) = -8 < 0 → -.
   • Test x = 4 in (3, ∞): f'(4) = 4(16)(1) = 64 > 0 → +.
4. Conclusion: The function is increasing on the interval (3, ∞). It is decreasing on (-∞, 0) and (0, 3). Note that at x = 0, the derivative is zero, but the function does not change from increasing to decreasing; it is decreasing on both sides.

Common Pitfalls and How to Avoid Them

• Ignoring Points of Non-Differentiability: Always check where the derivative is undefined. These points are just as important as where it equals zero.
• Incorrect Sign Analysis: Be meticulous when testing points in the derivative. A simple arithmetic error can lead to the wrong conclusion.
• Misinterpreting Critical Points: A critical point (where f'(x) = 0 or is undefined) is a potential location for a change in direction, but it is not guaranteed. Always use the sign chart to confirm.
• Confusing "Increasing at a Point" with "Increasing on an Interval": A function is increasing on an interval, not at a single point. The derivative being positive at a point only tells you about the local behavior, not an interval.

Conclusion: Mastering the Analysis

Finding the intervals of increase is a cornerstone of calculus that provides deep insight into a function's behavior. By systematically applying the derivative test—finding the derivative, locating critical points, and analyzing the sign of the derivative across intervals—you can accurately map out where a function climbs. This process is not just an academic exercise; it is a powerful tool for understanding optimization, modeling real-world phenomena, and solving complex problems in science and engineering. With practice, the steps become intuitive, allowing you to analyze any function with confidence and precision.