$x_{1} < x_{2} \Rightarrow f(x_{1}) < f(x_{2})$ 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).

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.

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].

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.

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,$\infty$ )

Home

GMAT

GRE

1st Grade

2nd Grade

3rd Grade

4th Grade

5th Grade

6th Grade

7th grade math

8th grade math

9th grade math

10th grade math

11th grade math

12th grade math

Precalculus

Worksheets

Chapter wise Test

MCQ's

Math Dictionary

Graph Dictionary

Multiplicative tables

Math Teasers

NTSE

Chinese Numbers

CBSE Sample Papers