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

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