Increasing and Decreasing Functions

Increasing and decreasing functions refer to the behavior of a function's graph as you move from left to right along the x-axis.

A function is considered increasing if for any two values x₁ and x₂​ such that x₁ < x₂ ​, the function value at x₁​ is less than the function value at x₂​ (i.e., f( x₁) < f( x₂)). On the other hand, a function is decreasing if f(x₁) > f( x₂) for x₁ < x₂​.

Some real-life examples of increasing/decreasing functions are given below:

Increasing Function Definition

In simple words, an increasing function is a type of function where, with increasing input (or the independent variable), output also increases (or the value of the function). Now, let's define an increasing function formally.

Now, let us consider I to be an interval that is present in the domain of a real-valued function f, then the function f is increasing on I.

if x₁ < x₂ ⇒ f(x₁) ≤ f(x₂) ∀ x₁ and x₂ ∈ I

Some common examples of increasing functions include linear functions with positive slope (such as y = mx + b), exponential functions (such as y = a^x, where a is a positive constant), and power functions (such as y = xⁿ, where n is a positive integer).

Solved Example: Consider the function: f(x) = 2x + 3. If x₁ = 2 and x₂ = 5. Determine whether the function is increasing or decreasing.

Solution:

A function is considered increasing if, for any two values x₁ and x₂​ such that we have f( x₁) < f( x₂).
Given that x₁ = 2, x₂ = 5.
f( x₁ ​) = f( 2 ) = 2(2) + 3 = 4 + 3 = 7
f( x₂ ​) = f( 5 ) = 2(5) + 3 = 10 + 3 = 13
Since 7 ≤ 13, the function satisfies the definition of an increasing function. Therefore, f(x) = 2x + 3 is increasing.

Strictly Increasing Function

For a function to be strictly increasing, the function should be increasing, but it can't be equal for any two unequal values, i.e.,

if x₁ < x₂ ⇒ f(x₁) < f(x₂) ∀ x₁ and x₂ ∈ I

Decreasing Function Definition

In simple words, a decreasing function is a type of function where, with increasing input (or the independent variable), the output value decreases (or the value of the function). To define a decreasing function formally, let us consider I to be an interval that is present in the domain of a real-valued function f, then the function f is decreasing on I.

if x₁ < x₂ ⇒ f(x₁) ≥ f(x₂) ∀ x₁ and x₂ ∈ I

Some common examples of decreasing functions include exponential decay functions (such as y = a^(-x), where a is a positive constant), and negative power functions (such as y = x^(-n), where n is a positive integer).

Solved Example: Determine whether the function f(x) = -2x + 5 is decreasing, if x₁ = 1 and x₂ = 3.

Solution:

A function is decreasing if f(x₁) > f( x₂) for x₁ < x₂​.
Since x₁ < x₂​, we now calculate f(x₁) and f( x₂):
f(x₁) = f( 1 ) = −2( 1 ) + 5 = −2 + 5 = 3
f(x₂) = f( 3 ) = -2( 3 ) + 5 = -6 + 5 = -1
Clearly, f( 1 ) > f( 3 ) So as x increases, f(x) decreases. This satisfies the condition for a strictly decreasing function.

Strictly Decreasing Function

For a function to be strictly decreasing, the function should be decreasing, but it can't be equal for any two unequal values, i.e.,

if x₁ < x₂ ⇒ f(x₁) > f(x₂) ∀ x₁ and x₂ ∈ I

Constant Function Definition

In simple words, a constant function is a type of function where, regardless of the input or independent variable, the output always remains the same, i.e., for all the inputs output remains constant. To define a constant function more formally, a function f is said to be a constant function if and only if

f(x) = k
Where k is the real number.

➣ Read all about functions in Maths - [Read Here!]

Rules to Check Increasing and Decreasing Functions

In calculus, an increasing function can be defined in terms of the slope of any curve, as an increasing function always has a positive slope, i.e., dy/dx > 0. To define an increasing function more formally, let us consider f to be a function that is continuous on the interval [p, q] and differentiable on the open interval (p, q), then

Function f is increasing in [p, q] if f′(x) > 0 for each x ∈ (p, q).

As a decreasing function always has a negative slope, a decreasing function can be defined in terms of the slope of any curve, i.e., dy/dx < 0. For a more formal definition of the decreasing function, let us consider f to be a function that is continuous on the interval [p, q] and differentiable on the open interval (p, q), then

Function f is decreasing in [p, q] if f′(x) < 0 for each x ∈ (p, q).

Graph of Increasing, Decreasing, and Constant Functions

The graphical representation of an increasing function, a decreasing function, and a constant function is,

Example: In this example, we will investigate the graph of f(x) = x². 

Solution:

Function table:

x	-4	-3	-2	-1	0	1	2	3	4
f(x)	16	9	4	1	0	1	4	9	16

As we can see that when x < 0, the value of f(x) is decreasing as the graph moves to the right.

In other words, the "height" of the graph is getting smaller. This is also confirmed by looking at the table of values. When x < 0, as x increases, f(x) decreases. Therefore, f(x) is decreasing on the interval from negative infinity to 0. 

When x > 0, the opposite is happening. When x > 0, the value of f(x) is increasing as the graph moves to the right. In other words, the "height" of the graph is getting bigger. This is also confirmed by looking at the table of values. When x > 0, as x increases, f(x) increases. Therefore, f(x) is increasing on the interval from 0 to infinity. 

Properties of Increasing & Decreasing Functions

Some helpful algebraic properties of increasing & Decreasing Functions are as follows:

• Additive property. If the functions f and g are increasing/decreasing on the interval (a, b), then the sum of the functions f + g is also increasing/decreasing on this interval.
• Opposite property. If the function f is increasing/decreasing on the interval (a, b), then the opposite function, -f, is decreasing/increasing.
• Inverse property. If the function f is increasing/decreasing on the interval (a, b), then the inverse function, 1/f, is decreasing/increasing on this interval.
• Multiplicative property. If the functions f and g are increasing/decreasing and not negative on the interval (a, b), then the product of the functions is also increasing/decreasing.

How to Find Increasing and Decreasing Intervals

Given a function, f(x), we can determine the intervals where it is increasing and decreasing by using differentiation and algebra. 

Step 1: Find the derivative, f'(x), of the function. 

Step 2: Find the zeros of f'(x). Remember, zeros are the values of x for which f'(x) = 0. Set f'(x) = 0 and solve for x. 

Step 3: Determine the intervals. The intervals are between the endpoints of the interval of f(x) and the zeros of f'(x). If the interval of f(x) is not given, assume f(x) is on the interval (-∞, ∞). 

Step 4: Determine whether the function is increasing or decreasing on each interval. Given the interval (a, c), choose a value b, a < b < c. Solve for f'(b). If f'(b) is positive, f(x) is increasing on (a, c). If f'(b) is negative, f(x) is decreasing on (a, c).

Example 1: If g(x) = (x - 5)², find the intervals where g(x) is increasing and decreasing. 

Solution:

Step 1: Find the derivative of the function. 

Using the chain rule, 

g'(x) = 2(5 - x)

Step 2: Find the zeros of the derivative function.  

In other words, find the values of for which g(x) equals zero. You can do this by setting g(x) = 0 and using algebra to solve for x. From the definitions above, we know the function is constant at points where the derivative is zero. 

g'(x) = 0 = 2(5 - x) 
⇒ 0 = 5 - x
⇒ x = 5

Step 3: Use the zeros to determine intervals.

Since x = 5 is the only zero for g'(x), there are just 2 intervals: from negative infinity to 5, and from 5 to negative infinity. 

These can be denoted in inequality notation: 

-∞ < x < 5
⇒ 5 < x < ∞

Or in interval notation: (-∞, 5), (5, ∞)

Remember, the endpoints are NOT inclusive because g(x) is neither increasing nor decreasing at the endpoints. 

Step 4: Determine if the function is increasing or decreasing in each interval.

For the first interval, ((-∞, 5), we'll choose b = 0. -∞ < x < 5
g'(b) = g'(0) = 2(5-0) = 10 
10 > 0 POSITIVE

For the second interval, (5, ∞), we'll choose b = 6. 5 < 6 < ∞
g'(b) = g'(6) = 2(5-6) = -2 
-2 < 0 NEGATIVE

Therefore, g(x) is increasing on (-∞, 5) and decreasing on (5, ∞). We can verify our results visually. In the graph below, you can clearly see that f(x) = (x - 5)² is increasing on the interval (5, ∞) and decreasing on the interval (-∞, 5).

We can visually verify our result by investigating the graph of g(x). 

Looking at the graph, g(x) is indeed increasing in the interval from negative infinity to 5 and decreasing in the interval from 5 to infinity. 

Example 2: Find the intervals in -20 < x < 20 where g(x) is increasing and decreasing, given g'(x) = x² - 100.

Solution:

If the derivative is given, we can skip the first step and go straight to finding the zeroes. 

g'(x)= 0 = x² - 100
⇒ x² = 100
⇒ x = 10, -10

Intervals: (-20, -10), (-10, 10), (10, 20) 

For (-20, -10), we'll choose b = -12. -20 < -12 < -10 
g'(-12) = 44 > 0 

For (-10, 10), we'll choose b = 0. -10 < 0 < 10 
g(0) = -100 < 0 

For (10, 20), we'll choose b = 12. 10 < 12 < 20 
g(12) = 44 > 0

Hence, for -20 < x < 20, g(x) is increasing on (-20, -10) and (10, 20) and decreasing on (-10, 10). 

Read More,

• Types of Functions
• Graphing of Function

Solved Question on Increasing and Decreasing Functions

Question 1: Given the function g(x) = 3x² - 12, find the intervals on -3 < x < 3 where g(x) is increasing and decreasing. 

Solution:

Given function: g(x) = 3x² - 12

Differentiate w.r.t. x, we get 

g'(x) = 6x 

For increasing and decreasing 

Put g'(x) = 0 
⇒ g'(x) = 6x = 0
⇒ x = 0 

Intervals: (-3, 0), (0, 3) 

• At x = -2, g'(-2) = -12 < 0 
• At x = 2, g'(2) = 12 > 0 

So, for -3 < x < 3, g(x) is decreasing on (-3, 0) and increasing on (0, 3). 

Question 2: Given the derivative of f(x), f'(x) = -10x² + 40x, find the intervals where f(x) is increasing and decreasing. 

Solution:

Given: f'(x) = -10x² + 40x

For increasing and decreasing 

Put f'(x) = 0 
⇒ f'(x) =  -10x² + 40x = 0
⇒ x = 4, 0

Intervals: (−∞, 0), (0, 4), (4, ∞) 

So, at x = -1, f'(-1) = -50 < 0 

• at x = 1, f'(1) = 30 > 0 
• at x = 5, f'(5) = -50 < 0 

So, f(x) is increasing on (0, 4) and decreasing on (−∞, 0), (4, ∞)

Question 3: Given the function g(x) = 5x² - 20x + 100, find the intervals where g(x) is increasing and decreasing. 

Solution:

Given: g(x) = 5x² - 20x + 100
⇒ g'(x) = 10x - 20 

For increasing and decreasing 

Put g'(x) = 0 
⇒ g'(x) = 10x - 20 = 0
⇒ x = 2 

Intervals: (−∞, 2), (2, ∞)

• At, x = 1, g'(1) = -10 < 0 
• At x = 3, g'(3) = 10 > 0 

So, g(x) is decreasing on (-∞, 2) and increasing on (2, ∞) 

Question 4: Given the functions s(x) = 6x³ - x², find the intervals on 0 < x < 10 where s(x) is increasing and decreasing. 

Solution:

Given: s(x) = 6x³ - x²
⇒ s'(x) = 18x² - 2x 

For increasing and decreasing 

Put s'(x) = 0 
⇒ s'(x) = 18x² - 2x = 0
⇒ x = 1/9, -1/9

Intervals: (0, 1/9)  

Here, -1/9 is not in the given interval, 0 < x < 10 
For 0 < x < 1/9: Choose x=0.05:
s'(0.05) = 18(0.05)² - 2(0.05) = 0.045 - 0.1 = -0.055
So, s(x)s is decreasing on (0,1/9).
For x > 1/9: Choose x=1:
s'(1) = 18(1)² - 2(1) = 18 - 2 = 16
So, s(x) is increasing on (1/9,10).

Question 5:

Solution:

Given: g'(x) = 7x² - 8

For increasing and decreasing 

Put g'(x) = 0 
⇒ g'(x) =  7x² - 8 = 0  
⇒ x = {√(8/7), -√(8/7)}

Intervals: (-∞,√(8/7)), (-√(8/7), √(8/7)), (√(8/7), ∞)

So, At x = -10, g'(-10) = 692 > 0 

• At x = 0, g'(0) = -8 < 0 
• At x = 10, g'(10) = 692 < 0 

Hence, g(x) is increasing on (-∞,√(8/7))and (√(8/7), ∞), and decreasing on (-√(8/7), √(8/7))

Practice More

• Practice Questions on Functions
• Quiz on Functions

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Article Tags :

• Mathematics
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• Class 12
• Maths-Class-12