# Increasing and Decreasing Functions

Definition of Increasing and decreasing functions

Increasing functions A function f(x) is said to be increasing function on (a,b), if
$x_{1} < x_{2} \Rightarrow f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in$(a,b).
Thus, f(x) is increasing on (a,b) if the values of f(x) increase with increase in the values of x.

Geometrically, f(x) is increasing on (a,b) if the graph y=f(x) moves up as x moves to the right then the graph is strictly increasing function on (a,b). increasing function
Decreasing functions
A function f(x) is said to be a decreasing function on (a,b) if,

$x_{1} < x_{2} \Rightarrow f(x_{1}) > f(x_{2})$ for all $x_{1}, x_{2} \in$(a,b).
Thus, f(x) is decreasing function on (a,b) if the graph moves down as x moves to the right. The graph below shows strictly decreasing function. Decreasing function
Test for Increasing and Decreasing Functions :

Let 'f' be a continuous function in the closed interval [a,b] and differentiable in the open interval (a,b) then
1) If f '(x) > 0 for all x in (a,b) then f is increasing on [a,b].
2) If f '(x) < 0 for all x in (a,b) then f is decreasing on [a,b].
3) If f '(x) = 0 for all x in (a,b) then f is constant on [a,b].

## Examples on Increasing and Decreasing functions

Example 1 : Determine the given function f(x) = $(2x-5)^{3}$ + x increasing or decreasing for all x.
Solution : f(x) = $(2x-5)^{3}$ + x
f '(x) = $3(2x-5)^{2}$(2) + 1
Since, $(2x-5)^{2}$ > 0
f '(x) > 0
So the given function is increasing for x $\in$ R.

Example 2 : For f(x) = $x^{4}$ − $8 x^{2}$ determine all intervals where f is increasing or decreasing.
Solution : f(x) = $x^{4}$ − $8 x^{2}$
First we will find the derivative of the given function to find the critical points
f '(x) = $4 x^{3}$ − 16x
$4 x^{3}$ − 16x = 0
4x ($x^{2}$ − 4) = 0
$x^{2}$ − 4 = 0
4x(x +2)(x - 2) = 0
So critical points are -2,0 and 2
So the interval will be (-$\infty$ , -2), (-2,0) , (0,2) and (2,$\infty$ )
Now we will check whether the function is increasing or decreasing for the above interval.
Let us consider x = -3 from the interval (-$\infty$ , -2)
f '(x) = $4 x^{3}$ − 16x
f '(-3) = $4 (-3)^{3}$ − 16(-3)
= -108 + 48 = -60
f '(-3) < 0 so the function is decreasing for the interval (-$\infty$ , -2)
Let us consider x = -1 from the interval (-2 , 0)
f '(x) = $4 x^{3}$ − 16x
f '(-1) = $4 (-1)^{3}$ − 16(-1)
= -4 + 16 = 12
f '(-1) > 0 so the function is increasing for the interval (-2 , 0)
Let us consider x = 1 from the interval (0 , 2)
f '(x) = $4 x^{3}$ − 16x
f '(1) = $4 (1)^{3}$ − 16(1)
= 4 - 16 = -12
f '(1) < 0 so the function is decreasing for the interval (0 , 2)
Let us consider x = 3 from the interval (2,$\infty$)
f '(x) = $4 x^{3}$ − 16x
f '(3) = $4 (3)^{3}$ − 16(3)
= 108 - 48 = 60
f '(3) > 0 so the function is increasing for the interval (2,$\infty$ )