# Monotonocity of function , Increasing Function , decreasing Function

Let y = f (x) be a given function with ‘ D ‘ as it’s domain. Let D1 ⊆ D then;

Increasing Function: f(x) is said to be increasing in D1 if for every x1 , x2 ∈ D1, x1 > x2 ⇒ f(x1) > f (x2)

It means that there is a certain increase in the value of f (x) with an increase in the value of x. Refer to fig.2

Non-Decreasing Function: f (x) is said to be non-decreasing in D1 if for every x1 , x2 ∈ D1 , x1 > x2

⇒ f(x1) ≥ f (x2 ). It means that the value of f (x) would never decrease with an increase in the value of x. Refer to fig.

Decreasing Function: f(x) is said to be decreasing in D1 if for every x1, x2 ∈ D1, x1 > x2

⇒  f (x1) < f (x2 It means that there is a certain decrease in the value of f (x) with an increase in the value of x. Refer to fig.

Non-Increasing Function: f (x) is said to be non-increasing in D1 if for every x1, x2 ∈ D1, x1 > x2

=> f (x1) ≤ f (x2 ). It means that the value of f (x) would never increase with an increase in the value of x.Refer to fig.

### Increasing function , decreasing function

Let y = f (x) be a given function, continuous in [a, b] and differentiable in (a, b). Then

∎ f (x) is increasing in [a , b] if f'(x) > 0 ∀ x ∈(a, b)

∎ f (x) is non-decreasing in [a , b] if f'(x) ≥ 0 ∀ x ∈(a, b)

∎ f (x) is decreasing in [a , b] if f'(x) < 0 ∀ x ∈(a, b)

∎ f (x) is non-increasing in [a , b] if f'(x) ≤ 0 ∀ x ∈(a, b)

### Remarks:

∎ If f'(x) ≥ 0 ∀ x ∈(a, b) and points which make f'(x) equal to zero (in between (a, b)) don’t form an interval, then f (x) would be increasing in [a, b].

∎ If f'(x) ≤ 0 ∀ x ∈(a, b) and points which make f'(x) equal to zero (in between (a, b)) don’t form an interval, f (x) would be decreasing in [a, b].

∎ If f(0) = 0 and f'(x) ≥ 0 ∀ x ∈R then f'(x) ≤ 0 ∀ x ∈(-∞,0) and f(x) ≥ 0 ∀ x ∈ (0, ∞).

∎ If f (0) = 0 and f'(x) ≤ 0 ∀ x ∈R then f'(x) ≥ 0 ∀ x ∈ (-∞, 0) and f (x) ≤ 0 ∀ x ∈ (0, ∞).

∎ A function is said to be monotonic if it’s either increasing or decreasing.

∎ The points for which f'(x) is equal to zero or doesn’t exist are called critical points. Here it should also be noted that critical points are the interior points of an interval.

∎ The stationary points are the points where f'(x) = 0, in the domain.

The following illustrations would make the above mentioned theorems and remarks clear.

Illustration : Find the critical points and the intervals of increase and decrease for f(x) = 3x4 – 8x3 – 6x2 + 24x + 7

Solution:

f(x) = 3x4 – 8x3 – 6x2 + 24x + 7

f'(x) = 12x3 – 24x2 -12x + 24 = 0

=> 12(x3 – 2x2 – x + 2) = 0

=> 12(x – 1) (x – 2) (x + 1) = 0

Critical points are – 1, 1 and 2

The wavy curve of the derivative is given in figure Hence function increases in the interval [-1, 1] ∪ [2, ∞) and decreases in the interval (-∞ , -1] ∪[1 , 2]

Illustration : Find the number of critical points for the following functions in their respective intervals provided

(i) min(x , cosx) ; x ∈ [-π, π]

(ii) min ({x}, {-x}) ; x ∈ [-3, 3], where {.} denotes the fractional part

Solution: (i) Given function is min(x, cos x). In order to check the points we plot the graph of curve, which is more convenient in these examples. The number of critical points are to be two one between (0,π/2) and one at π.

(ii) Similarly we plot min({x}, {-x}) we get number of critical points are eleven. As 3 and -3 are not critical points. Illustration : Prove that the following functions are increasing for the given intervals,

(i) y = ex + sin x , x ∈ R+

(ii) y = sin x + tan x – 2 x , x ∈ (0 , π/2)

(iii) y = x + sin x , x ∈ R

Solution: (i) f (x) = ex + sin x , x ∈ R+

=> f'(x) = ex + cos x

Clearly f'(x) > 0 ∀ x ∈ R+ (as ex >1 , x ∈ R+ and –1 ≤ cosx ≤ 1, x ∈ R+)

Hence f (x) is increasing.

(ii) f (x) = sin x + tan x – 2 x , x ∈ (0 , π/2)

=> f'(x) = cos x + sec2 x – 2

as cos x > cos2 x , x ∈ (0 , π/2)

=> f'(x) > cos2 x + sec2 x – 2 = (cos x – sec x)2 > 0 , x ∈ (0 , π/2)

Hence f (x) is increasing in (0 , π/2)

(iii) f (x) = x + sin x , x ∈ R

=> f'(x) = 1 + cos x => f'(x) ≥ 0 , as -1 ≤ cos x ≤ 1

Here f'(x)  = 0 => cos x = -1 => x = (2n + 1) π, n ∈ I

=> Zeros of don’t form an interval. Hence f(x) would be increasing for all real values of x.

Illustration : Investigate the behaviour of the following functions for monotonicity in the given intervals,

(i) f(x) = -sin3 x + 3 sin2 x + 5, x ∈ [- π/2, π/2]

(ii) f(x) = sec x – cosec x, x ∈ (0, π/2)

Solution:

(i) f (x) = – sin3 x + 3 sin2 x + 5, x ∈ [- π/2, π/2]

=> f'(x) = – 3 sin2 x. cos x + 6 sin x. cos x

= 3 sin x cos x (2 – sin x)

As cos x > 0 and 2 – sin x > 0 ∀ x ∈ (- π/2 , π/2)

and sin x > 0 ∀ x ∈ (0 , π/2), sin x < 0 ∈ x ∈ (- π/2 , 0)

f'(x) > 0, x ∈ (0, π/2) and < 0, x ∈ (- π/2, 0)

=> f (x) is increasing in (0, π/2) and decreasing in (- π/2 , 0)

(ii) f (x) = sec x – cosec x , x ∈ (0, π/2)

=> f'(x) = sec x tan x + cosec x cot x > 0 ∀ x ∈ (0, π/2)

Thus f (x) is increasing in (0, π/2)

Illustration : Prove that : 2 sin x + tan x ≥ 3x, for x ∈ [0, π/ 2)

Solution: Let f (x) = 2 sin x + tan x – 3x

f'(x) = sec2 x + 2 cos x – 3

Now, f”(x) = 2 sec2 x tan x – 2 sin x = 2 sinx [sec3 x – 1] ≥ 0 ∀ x ∈ [0, π/ 2)

=> f’ (x) is increasing function.

Also f'(0) = 0 => f'(x) ≥ 0 which means f(x) is an increasing function.

Now, since f (0) = 0 => f (x) ≥ 0 => 2 sin x + tan x ≥ 3x