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239 LearnersLast updated on October 22, 2025

Walking downstairs requires less effort than climbing up, similar to how functions in calculus increase or decrease. In calculus, a function is considered increasing when its value rises with increasing x, and decreasing when its value drops as x increases.
The increasing or decreasing intervals of a function show where its graph rises or falls. We can find these intervals by checking the sign of the first derivative in each interval.
To write intervals of increase and decrease, you can follow these basic mathematical rules:
You can also use the first derivative to identify these intervals:
This method makes it easier to analyze the behavior of functions and represent their increasing or decreasing nature.
Now that we understand what increasing and decreasing intervals mean, let’s learn how to find them for a function. We’ll use an example to understand the process step by step.
Consider the function: \(f(x) = x³ + 3x² – 45x + 9\)
Step 1: Differentiate the function
Find the first derivative, \(f'(x)\):
\(f '(x) = 3x² + 6x – 45\\ f '(x)= 3(x² + 2x – 15)\\ f '(x)= 3(x + 5)(x – 3)\)
Step 2: Find critical points
Set \(f '(x) = 0\):
\(3(x + 5)(x – 3) = 0\\ x+5 = 0\\ x = -5\\and, \\ x-3 = 0\\ x= 3\\ ⇒ x = –5 \ and\ x = 3\)
Here, -5 and 3 are critical points. These are the points where the slope of the function changes.
Step 3: Identify intervals
Using the critical points, divide the number line into three intervals:
\((–∞, –5), (–5, 3), and (3, ∞)\)
|
Interval |
Value of X |
f '(x) |
Increasing/ Decreasing |
|
\((–∞, –5)\) |
\(x = –7\) |
\(f '(–7) = 3(–2)(–10) = 60 > 0\) |
Increasing |
|
\((–5, 3)\) |
\(x = –2\) |
\(f '(–2) = 3(3)(–5) = –45 < 0\) |
Decreasing |
|
\((3, ∞)\) |
\(x = 5\) |
\(f '(5) = 3(10)(2) = 60 > 0\) |
Increasing |
As a result, \(f(x) = x^3 + 3x^2 - 45x + 9\) has increasing intervals (-∞, -5) and (3, ∞) and decreasing intervals (-5, 3).
A critical point of a function occurs when the first derivative of the function is either zero or undefined. These points are important for identifying extrema, which are the local maximum or minimum values of the function.
Critical point: A point \(x = c\) is a critical point if either \(f '(c) = 0\) or \(f '(c)\) is undefined.
Extrema: If the function changes from increasing to decreasing or vice versa around a critical point, then it is called an extrema — either a local maximum or a local minimum.
Let's examine a graph of a curve. The curve will turn at certain points, either from increasing to decreasing or the other way around. These points are called extrema, which may be:
It becomes clear from the above figures that every extremum of a function is a point where its derivative changes sign. In other words, the function either goes from increasing to decreasing or vice versa. When identifying regions where a function is increasing or decreasing, it is important to examine the behavior around the extrema.
To find these intervals for any function \(f(x)\) over a given interval, follow these steps:
For a function to have increasing or decreasing intervals, it must be continuous and differentiable in that interval.
While every differentiable function is continuous, not all continuous functions are differentiable.
In this section, we will learn how to visually represent increasing and decreasing intervals on a graph. This helps you identify these intervals simply by looking at the graph.
Below are two sample graphs of different functions. The first graph shows an increasing function because the curve goes up as we move from left to right along the x-axis. The second graph shows a decreasing function because the curve goes down as we move from left to right.
Here are some of the tips and tricks for students to master increasing and decreasing intervals:
Understand the core idea, that it is an increasing Interval if the function goes up as 𝑥 increases. Decreasing Interval if the function goes down as 𝑥 increases.
Test intervals with simple numbers. Pick a number in each interval instead of complicated calculations. Only the sign of 𝑓′(𝑥) matters, not the exact value.
Use graphing for visualization. Draw a rough sketch of increasing sections slope upward, decreasing slope downward. For cubic or quartic functions, mark critical points and check slope signs. Visualization helps remember the logic rather than just formulas.
Remember that critical points are the key.
Critical points = where derivative = 0 or undefined.
These points divide the domain into intervals. Each interval has a consistent increasing or decreasing trend. Label intervals left to right, then test one number in each.
Try to connect the concepts with some real-life examples. For example, the speed of a car going uphill, where speed rises as time passes, is an example of increasing interval. At the same time, Battery percentage draining where battery drops as time passes, is an example of decreasing interval.
Understanding increasing and decreasing intervals is an essential algebra skill. However, students often make mistakes when learning this concept. Here are a few common mistakes and ways to avoid them:
Increasing and decreasing intervals have real-life applications across various fields. Here are a few examples:
Determine the intervals where the function f(x) = x² increases and decreases.
Increasing on: (0, ∞)
Decreasing on: (−∞, 0)
We first find the derivative → \(f′(x) = 2x\)
Set \(f′(x) = 0 → 2x = 0 → x = 0\) (critical point)
Test sign of \(f′(x)\) around \(x = 0\)
\(f′(−1) = −2\) → negative → decreasing
\(f′(1) = 2\) → positive → increasing
So, \(f(x) = x^2\) is:
Increasing on: \((0, ∞)\)
Decreasing on: \((−∞, 0)\)
For the function f(x) = x³, determine the intervals of increase and decrease.
Increasing on: \((−∞, ∞)\)
Decreasing: None.
Let \(f′(x) = 3x²\)
Set \(f′(x) = 0 → 3x² = 0 → x = 0\)
We now test \(f′(x)\) around \(x = 0\)
\(f′(−1) = 3\) → positive
\(f′(1) = 3\) → positive
So, the \(f(x) = x^3\) is:
Increasing on: \((−∞, ∞)\)
Decreasing: None.
Find where the function f(x) = sin(x) is increasing or decreasing in the interval [0, 2π].
Increasing on: \((0, \frac {π}{2}) ∪ (\frac {3π}{2}, 2π)\)
Decreasing on: \((\frac {π}{2}, \frac {3π}{2})\)
Let \(f′(x) = cos(x)\)
We need to set \(f′(x) = 0 → cos(x) = 0 → x = \frac {π}{2}, \frac {3π}{2}\)
Test the sign in intervals:
\((0, \frac {π}{2}): cos(x) > 0\) → increasing
\((\frac {π}{2}, \frac {3π}{2}): cos(x) < 0\) → decreasing
\((\frac {3π}{2}, 2π): cos(x) > 0\) → increasing
So, the \(f(x) = sin(x)\) is:
Increasing on: \((0, \frac {π}{2}) ∪ (\frac {3π}{2}, 2π)\)
Decreasing on: \((\frac {π}{2}, \frac {3π}{2})\)
Find the increasing/decreasing intervals for f(x) = √x on (0, ∞)
Increasing on: \((0, ∞)\)
Decreasing: None
Let \(f(x) = x(\frac {1}{2})\)
We then need to set \(f′(x) = (\frac {1}{2})x(\frac {−1}{2}) = \frac 1 {(2\sqrt x)}\)
Since \(\sqrt x > 0\) for all \(x > 0\), \(f′(x) > 0\) for all \(x > 0\)
So, the \(f(x) = \sqrt x\) is:
Increasing on: \((0, ∞)\)
Decreasing: None
Find intervals of increase and decrease for: f(x)=x^3 −3x^2 −9x+5
Increasing: \((-\infty, -1) \cup (3, \infty)\)
Decreasing: \((-1, 3)\)
\(f(x)=x^3 −3x^2 −9x+5\)
Let us find the derivative of the equation.
\(f'(x) = 3x^2 - 6x -9\)
Solve for f'(x) = 0
\(3x^2 - 6x -9 = 0 \\ x^2 - 2x - 3 = 0\\ (x-3)(x+ 1) = 0\\ x = 3, x = -1\)
Now test their intervals
Interval 1: \(x < -1\), choose
\(x = -2 → f'(-2) = 3(-2)^2 - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0\) → increasing
Interval 2: \(-1 < x < 3\), choose
\(x = 0 → f'(0) = -9 < 0\) → decreasing
Interval 3: \(x > 3\), choose
\(x = 4 → f'(4) = 48 - 24 - 9 = 15 > 0\) → increasing
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






