How to Find Intervals of Increase and Decrease: A Step-by-Step Guide
Calculus is not just about finding derivatives. Equally important is understanding what those derivatives tell you about the behavior of a function. An interval of increase is a range of x-values over which a function’s output increases as the input increases. Likewise, an interval of decrease is a range where the function’s output decreases as the input increases. By analyzing the sign of the derivative, you can determine exactly where these intervals occur and learn how to find intervals of increase and decrease.
In this guide, you’ll learn a step-by-step process for finding intervals of increase and decrease for any differentiable function. We’ll begin by defining increasing and decreasing intervals, then explain how the derivative describes function behavior. Next, you’ll learn how to use critical points and sign charts to identify these intervals, apply the First Derivative Test to classify local maxima and minima, and work through a couple of fully solved examples. By the end of this article, you’ll have a reliable method for analyzing the behavior of functions.
What Are Intervals of Increase and Decrease?
One of the primary goals of differential calculus is to understand how a function behaves as its input changes. In particular, we often want to determine where a function is increasing, decreasing, or changing direction. These regions are called the intervals of increase and decrease of a function.
A function \( f(x) \) is said to be increasing on an interval if its output values increase as the input values increase. In other words, whenever \( x_1 < x_2 \) within the interval, we have
$$f(x_1) < f(x_2).$$
Similarly, a function is decreasing on an interval if its output values decrease as the input values increase. That is, whenever \( x_1 < x_2 \),
$$f(x_1) > f(x_2).$$
In the next section, we’ll explore why the derivative provides information about the behavior of a function and how it allows us to find intervals of increase and decrease.
The Relationship Between the Derivative and Function Behavior
In the article Secant and Tangent Lines in Calculus: A Complete Beginner’s Guide, we learned that the derivative of a function measures its instantaneous rate of change. Geometrically, the derivative gives the slope of the tangent line to the graph of the function at a particular point. Similarly, in the article How to Differentiate a Function Step by Step: A Beginner’s Guide, we show how to compute derivatives. While computing derivatives is an important skill, the real power of derivatives lies in what they tell us about the behavior of functions.
Recall that a positive slope indicates that a graph is rising as we move from left to right, while a negative slope indicates that the graph is falling. In other words
- If \( f'(x) > 0 \), then the function \( f(x) \) is increasing.
- If \( f'(x) < 0 \), then the function \( f(x) \) is decreasing.
- If \( f'(x) = 0 \), then the function may change its behavior and should be investigated further.
Notice that points where the derivative equals zero play a special role. These points are called critical points. A point \( c \) is called a critical point of a function \( f \) if either \( f'(c) = 0 \) or \( f'(c) \) is undefined.
Critical points divide the domain of a function into intervals on which the derivative cannot change sign without first passing through zero or becoming undefined. As a result, to determine where a function is increasing or decreasing, we only need to: find the derivative, determine the critical points, and determine the sign of the derivative on each interval.
In the next section, we’ll develop a procedure to find intervals of increase and decrease for any differentiable function.
How to Find Intervals of Increase and Decrease Step by Step
Now that we understand how the derivative describes the behavior of a function, we can develop a procedure for finding intervals of increase and decrease.
Step 1: Compute the Derivative
The derivative tells us how the function changes at every point in its domain.
Step 2: Find the Critical Points
Next, find all values of \( x \) where the derivative equals zero or is undefined. These values are called critical points.
Step 3: Divide the Number Line into Intervals
The critical points divide the real line into separate intervals: Within each interval, the derivative cannot change sign unless it passes through another critical point.
Step 4: Choose a Test Point from Each Interval
Select one value from each interval and substitute it into the derivative.
Step 5: Determine the Sign of the Derivative
Evaluate the derivative at each test point to determine the sign.
Step 6: State the Intervals of Increase and Decrease
The function is increasing on intervals where the derivative is positive and decreasing on intervals where the derivative is negative
Although this process may seem mechanical at first, it becomes intuitive with practice. In fact, the critical points reveal something even more important: the locations of local maxima and minima. In the next section, we’ll formalize this idea using the First Derivative Test.
Using the First Derivative Test
Finding intervals of increase and decrease allows us to determine more than just where a function is increasing or decreasing. It also helps us identify the locations of local maxima and local minima. The tool that connects these ideas is called the First Derivative Test.
The First Derivative Test examines how the sign of the derivative changes as we pass through a critical point. Since the sign of the derivative determines whether a function is increasing or decreasing, a change in sign indicates a change in the behavior of the function.
Suppose that \( c \) is a critical point of the function \( f \). Then
- If the derivative changes from positive to negative as the function passes through \( c \), then there is a local maximum at \( c \).
- If the derivative changes from negative to positive as the function passes through \( c \), then there is a local minimum at \( c \).
- If the derivative does not change sign, then there is neither a local maximum nor a local minimum at \( c \).
The First Derivative Test connects derivatives with the geometric behavior of functions. In the next section, we’ll work through a couple of examples.
Worked Out Examples
Now let’s put everything together by working through two examples. The first example involves a polynomial function, while the second involves a rational function.
Example 1: Determine the local maxima and minima of \( f(x) = x^4 – 4x^3 \).
Solution: Taking the derivative, we get
$$f'(x) = (x^4 – 4x^3)’.$$
Using the sum rule, this is
$$f'(x) = (x^4)’ – (4x^3)’.$$
Taking the derivatives, we get
$$f'(x) = 4x^3 – 12x^2.$$
Setting the derivative equal to 0, we obtain
$$0 = 4x^3 – 12x^2.$$
Factoring gives
$$0 = 4x^2(x – 3).$$
Setting each factor equal to 0, we find
$$4x^2 = 0, x – 3 = 0.$$
Dividing both sides of the first equation by 4, we get
$$x^2 = 0, x – 3 = 0.$$
Taking the square root of both sides of the first equation gives
$$x = 0, x – 3 = 0.$$
Adding 3 to both sides of the second equation, we obtain
$$x = 0, x = 3.$$
This breaks the real line into the three intervals
$$(-\infty, 0), (0, 3), (3, \infty).$$
For the first interval, we test
$$-1 \in (-\infty, 0).$$
Substituting this into the derivative, we get
$$f'(-1) = 4(-1)^3 – 12(-1)^2.$$
Exponentiating gives
$$f'(-1) = 4(-1) – 12(1).$$
Multiplying we find
$$f'(-1) = -4 – 12.$$
Subtracting, we arrive at
$$f'(-1) = -16.$$
For the second interval, we test
$$1 \in (0, 3).$$
Substituting this into the derivative, we get
$$f'(1) = 4(1)^3 – 12(1)^2.$$
Exponentiating gives
$$f'(1) = 4(1) – 12(1).$$
Multiplying we find
$$f'(1) = 4 – 12.$$
Subtracting, we arrive at
$$f'(1) = -8.$$
For the third interval, we test
$$4 \in (3, \infty).$$
Substituting this into the derivative, we get
$$f'(4) = 4(4)^3 – 12(4)^2.$$
Exponentiating gives
$$f'(4) = 4(64) – 12(16).$$
Multiplying we find
$$f'(4) = 256 – 192.$$
Subtracting, we arrive at
$$f'(4) = 64.$$
Finally, we conclude that the function is decreasing on the interval \( (-\infty, 3) \) and increasing on the interval \( (3, \infty) \). Furthermore, there is a local minimum at \( x = 3 \).
This example demonstrates that not every critical point produces a local maximum or minimum.
Example 2: Determine the local maxima and minima of \( f(x) = \frac{x + 1}{x – 2} \).
Solution: Taking the derivative, we get
$$f'(x) = (\frac{x + 1}{x – 2})’.$$
Using the quotient rule, this is
$$f'(x) = \frac{(x + 1)'(x – 2) – (x +1)(x – 2)’}{(x – 2)^2}.$$
Taking the derivatives, we get
$$f'(x) = \frac{(x – 2) – (x + 1)}{(x – 2)^2}.$$
Simplifying we find
$$f'(x) = -\frac{3}{(x – 2)^2}.$$
The derivative is undefined when
$$0 = (x – 2)^2.$$
Taking the square root of both sides of the equation gives
$$x – 2 = 0.$$
Adding 2 to both sides of the equation, we obtain
$$x = 2.$$
This breaks the real line into the two intervals
$$(-\infty, 2), (2, \infty).$$
For the first interval, we test
$$0 \in (-\infty, 2).$$
Substituting this into the derivative, we get
$$f'(0) = -\frac{3}{(0 – 2)^2}.$$
Subtracting, we arrive at
$$f'(0) = -\frac{3}{(–2)^2}.$$
Exponentiating gives
$$f'(0) = -\frac{3}{4}.$$
For the second interval, we test
$$3 \in (2, \infty).$$
Substituting this into the derivative, we get
$$f'(3) = -\frac{3}{(3 – 2)^2}.$$
Subtracting, we arrive at
$$f'(3) = -\frac{3}{(1)^2}.$$
Exponentiating gives
$$f'(3) = -\frac{3}{1}.$$
Dividing we find
$$f'(3) = -3.$$
Finally, we conclude that the function is decreasing on the intervals \( (-\infty, 2) \) and \( (2, \infty) \).
This example illustrates that rational functions may have intervals separated by points where the function is undefined. Even when a function is undefined at the critical points, analyzing the sign of the derivative still allows us to determine where the function is increasing or decreasing.
Conclusion
Learning how to find intervals of increase and decrease is important because it allows us to understand how functions behave without relying solely on their graphs. By analyzing the sign of the derivative, you can determine where a function is increasing, where it is decreasing, and where it changes direction.
In this guide, we explored the relationship between derivatives and function behavior, developed a step-by-step process for finding intervals of increase and decrease, applied the First Derivative Test, and used critical points to identify local maxima and minima. We also worked through examples involving both polynomial and rational functions.
Further Reading
Secant and Tangent Lines in Calculus: A Complete Beginner’s Guide – Secant and tangent lines are another geometric property that can be investigated with the derivative.
