Increasing And Decreasing Functions

let x₁,x₂ (0,)

We have, x₁<x₂

⇒ log_e x₁ < log_e x₂

⇒ f(x₁) < f(x₂)

So, f(x) is increasing in (0,)

case I

When a > 1

let x₁,x₂ (0,)

We have, x₁<x₂

⇒ log_e x₁ < log_e x₂

⇒ f(x₁) < f(x₂)

So, f(x) is increasing in (0,)

case II

When 0 < a < 1

f(x) = log_a x

when a<1 log a < 0

let x₁<x₂

⇒ log x₁ < log x₂

⇒ []

⇒ f(x₁) > f(x₂)

So, f(x) is decreasing in (0,)

we have,

f(x) = ax + b, a > 0

let x₁,x₂ R and x₁ > x₂

⇒ ax₁ > ax₂ for some a > 0

⇒ ax₁ + b> ax₂ + b for some b

⇒ f(x₁) > f(x₂)

Hence, x₁ > x₂⇒ f(x₁) > f(x₂)

So, f(x) is increasing function of R

we have,

f(x) = ax + b, a < 0

let x₁,x₂ R and x₁ > x₂

⇒ ax₁ < ax₂ for some a > 0

⇒ ax₁ + b< ax₂ + b for some b

⇒ f(x₁) < f(x₂)

Hence, x₁ > x₂⇒ f(x₁) < f(x₂)

So, f(x) is decreasing function of R

we have

let x₁,x₂ (0,) We have, x₁ > x₂

⇒

⇒ f(x₁) < f(x₂)

Hence, x₁ > x₂⇒ f(x₁) < f(x₂)

So, f(x) is decreasing function

We have,

Case 1

When x [0,)

Let , (0,] and

⇒

⇒

⇒

⇒ f(x₁)< f(x₂)

f(x) is decreasing on[0,∞).

Case 2

When x (-,0]

Let >

⇒

⇒

⇒

⇒

f(x) is increasing on(-∞,0].

Thus, f(x) is neither increasing nor decreasing on R.

We have,

Case 1

When x [0,)

Let >

⇒

⇒

⇒

f(x₁) < f(x₂)

⇒ f(x) is decreasing on[0,∞).

Case 2

When x (-,0]

Let >

⇒

⇒

⇒

⇒

f(x) is increasing on(-∞,0].

Thus, f(x) is neither increasing nor decreasing on R.

We have,

f(x) = |x| =

(a)Let , (0,) and

⇒

So, f(x) is increasing in (0,)

(b) Let , (-∞, 0)and

⇒

⇒

f(x) is strictly decreasing on(-∞, 0).

Given,

f(x) = 7x – 3

Lets , R and

⇒ 7> 7

⇒ 7> 7

⇒

f(x) is strictly increasing on R.

Given:- Function f(x) = 10 – 6x – 2x²

Theorem:- Let f be a differentiable real function defined on an open interval (a, b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.

Here we have,

f(x) = 10 – 6x – 2x²

⇒

⇒ f’(x) = –6 – 4x

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ –6 –4x > 0

⇒ –4x > 6

⇒

⇒

⇒

Thus f(x) is increasing on the interval

Again, For f(x) to be increasing, we must have

f’(x) < 0

⇒ –6 –4x < 0

⇒ –4x < 6

⇒

⇒

⇒

Thus f(x) is decreasing on interval

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒ f’(x) = 6x³ – 12x² – 90x

⇒ f’(x) = 6x(x² – 2x – 15)

⇒ f’(x) = 6x(x² – 5x + 3x – 15)

⇒ f’(x) = 6x(x – 5)(x + 3)

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 6x(x – 5)(x + 3)> 0

⇒ x(x – 5)(x + 3) > 0

⇒ –3 < x < 0 or 5 < x < ∞

⇒ x ∈ (–3,0)∪(5, ∞)

Thus f(x) is increasing on interval (–3,0)∪(5, ∞)

Again, For f(x) to be decreasing, we must have

f’(x) < 0

⇒ 6x(x – 5)(x + 3)> 0

⇒ x(x – 5)(x + 3) > 0

⇒ –∞ < x < –3 or 0 < x < 5

⇒ x ∈ (–∞, –3)∪(0, 5)

Thus f(x) is decreasing on interval (–∞, –3)∪(0, 5)

Given:- Function f(x) = x² + 2x – 5

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.

Here we have,

f(x) = x² + 2x – 5

⇒

⇒ f’(x) = 2x + 2

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 2x + 2 > 0

⇒ 2x < –2

⇒

⇒ x < –1

⇒ x ∈ (–∞,–1)

Thus f(x) is increasing on interval (–∞,–1)

Again, For f(x) to be increasing, we must have

f’(x) < 0

⇒ 2x + 2 < 0

⇒ 2x > –2

⇒

⇒ x> –1

⇒ x ∈ (–1,∞)

Thus f(x) is decreasing on interval x ∈ (–1, ∞)

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒

⇒

⇒

⇒

⇒

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒

⇒ (x – 2) > 0

⇒ 2 < x < ∞

⇒ x ∈ (2, ∞)

Thus f(x) is increasing on interval (2, ∞)

Again, For f(x) to be decreasing, we must have

f’(x) < 0

⇒

⇒ (x – 2) < 0

⇒ –∞ < x < 2

⇒ x ∈ (–∞, 2)

Thus f(x) is decreasing on interval (–∞, 2)

Given:- Function f(x) = 6 – 9x – x²

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 6 – 9x – x²

⇒

⇒ f’(x) = –9 – 2x

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ –9 – 2x > 0

⇒ –2x > 9

⇒

⇒

⇒

Thus f(x) is increasing on interval

Again, For f(x) to be decreasing, we must have

f’(x) < 0

⇒ –9 – 2x < 0

⇒ –2x < 9

⇒

⇒

⇒

Thus f(x) is decreasing on interval

Given:- Function f(x) = 2x³ – 12x² + 18x + 15

Theorem:- Let f be a differentiable real function defined on an open interval (a, b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ – 12x² + 18x + 15

⇒

⇒ f’(x) = 6x² – 24x + 18

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 24x + 18 = 0

⇒ 6(x² – 4x + 3) = 0

⇒ 6(x² – 3x – x + 3) = 0

⇒ 6(x – 3)(x – 1) = 0

⇒ (x – 3)(x – 1) = 0

⇒ x = 3 , 1

Given:- Function f(x) = 5 + 36x + 3x² – 2x³

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 5 + 36x + 3x² – 2x³

⇒

⇒ f’(x) = 36 + 6x – 6x²

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x² = 0

⇒ 6(–x² + x + 6) = 0

⇒ 6(–x² + 3x – 2x + 6) = 0

⇒ –x² + 3x – 2x + 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ (x – 3)(x + 2) = 0

⇒ x = 3 , – 2

Given:- Function f(x) = 8 + 36x + 3x² – 2x³

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 8 + 36x + 3x² – 2x³

⇒

⇒ f’(x) = 36 + 6x – 6x²

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x² = 0

⇒ 6(–x² + x + 6) = 0

⇒ 6(–x² + 3x – 2x + 6) = 0

⇒ –x² + 3x – 2x + 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ (x – 3)(x + 2) = 0

⇒ x = 3 , – 2

Given:– Function f(x) = 5x³ – 15x² – 120x + 3

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 5x³ – 15x² – 120x + 3

⇒

⇒ f’(x) = 15x² – 30x – 120

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 15x² – 30x – 120 = 0

⇒ 15(x² – 2x – 8) = 0

⇒ 15(x² – 4x + 2x – 8) = 0

⇒ x² – 4x + 2x – 8 = 0

⇒ (x – 4)(x + 2) = 0

⇒ x = 4 , – 2

Given:- Function f(x) = x³ – 6x² – 36x + 2

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x³ – 6x² – 36x + 2

⇒

⇒ f’(x) = 3x² – 12x – 36

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 3x² – 12x – 36 = 0

⇒ 3(x² – 4x – 12) = 0

⇒ 3(x² – 6x + 2x – 12) = 0

⇒ x² – 6x + 2x – 12 = 0

⇒ (x – 6)(x + 2) = 0

⇒ x = 6, – 2

Given:- Function f(x) = 2x³ – 15x² + 36x + 1

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ – 15x² + 36x + 1

⇒

⇒ f’(x) = 6x² – 30x + 36

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 30x + 36 = 0

⇒ 6(x² – 5x + 6) = 0

⇒ 3(x² – 3x – 2x + 6) = 0

⇒ x² – 3x – 2x + 6 = 0

⇒ (x – 3)(x – 2) = 0

⇒ x = 3, 2

Given:- Function f(x) = 2x³ + 9x² + 12x + 20

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ + 9x² + 12x + 20

⇒

⇒ f’(x) = 6x² + 18x + 12

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² + 18x + 12 = 0

⇒ 6(x² + 3x + 2) = 0

⇒ 6(x² + 2x + x + 2) = 0

⇒ x² + 2x + x + 2 = 0

⇒ (x + 2)(x + 1) = 0

⇒ x = –1, –2

Given:- Function f(x) = 2x³ – 9x² + 12x – 5

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ – 9x² + 12x – 5

⇒

⇒ f’(x) = 6x² – 18x + 12

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 18x + 12 = 0

⇒ 6(x² – 3x + 2) = 0

⇒ 6(x² – 2x – x + 2) = 0

⇒ x² – 2x – x + 2 = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1, 2

Given:- Function f(x) = -2x³ + 3x² + 12x + 6

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = –2x³ + 3x² + 12x + 6

⇒

⇒ f’(x) = –6x² + 6x + 12

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ –6x² + 6x + 12 = 0

⇒ 6(–x² + x + 2) = 0

⇒ 6(–x² + 2x – x + 2) = 0

⇒ x² – 2x + x – 2 = 0

⇒ (x – 2)(x + 1) = 0

⇒ x = –1, 2

Given:- Function f(x) = 2x³ – 24x + 107

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ – 24x + 107

⇒

⇒ f’(x) = 6x² – 24

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 24 = 0

⇒ 6(x² – 4) = 0

⇒ (x – 2)(x + 2) = 0

⇒ x = –2, 2

clearly, f’(x) > 0 if x < –2 and x > 2

and f’(x) < 0 if –2 < x < 2

Thus, f(x) increases on (–∞, –2) ∪ (2, ∞)

and f(x) is decreasing on interval x ∈ (–2,2)

Given:- Function f(x) = –2x³ – 9x² – 12x + 1

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = –2x³ – 9x² – 12x + 1

⇒

⇒ f’(x) = – 6x² – 18x – 12

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ –6x² – 18x – 12 = 0

⇒ 6x² + 18x + 12 = 0

⇒ 6(x² + 3x + 2) = 0

⇒ 6(x² + 2x + x + 2) = 0

⇒ x² + 2x + x + 2 = 0

⇒ (x + 2)(x + 1) = 0

⇒ x = –1, –2

Given:- Function f(x) = (x – 1) (x – 2)²

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = (x – 1) (x – 2)²

⇒

⇒ f’(x) =(x – 2)² +2(x – 2)(x – 1)

⇒ f’(x) = (x – 2)(x – 2 + 2x – 2)

⇒ f’(x) = (x – 2 )(3x – 4)

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ (x – 2 )(3x – 4) = 0

⇒ x = 2,

Given:- Function f(x) = x³ – 12x² + 36x + 17

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x³ – 12x² + 36x + 17

⇒

⇒ f’(x) = 3x² – 24x + 36

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 3x² – 24x + 36 = 0

⇒ 3(x² – 8x + 12) = 0

⇒ 3(x² – 6x – 2x + 12) = 0

⇒ x² – 6x – 2x + 12 = 0

⇒ (x – 6)(x – 2) = 0

⇒ x = 2, 6

Given:- Function f(x) = 2x³ – 24x + 7

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ – 24x + 7

⇒

⇒ f’(x) = 6x² – 24

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 6x² – 24 > 0

⇒

⇒ x² < 4

⇒ x < –2, +2

⇒ x ∈ (–∞,–2) and x ∈ (2,∞)

Thus f(x) is increasing on interval (–∞, –2) ∪ (2, ∞)

Again, For f(x) to be increasing, we must have

f’(x) < 0

⇒ 6x² – 24< 0

⇒

⇒ x² < 4

⇒ x> –1

⇒ x ∈ (–1,∞)

Thus f(x) is decreasing on interval x ∈ (–1, ∞)

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒

⇒

⇒ x =1, –2, 3

Given:- Function f(x) = x⁴ – 4x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x⁴ – 4x

⇒

⇒ f’(x) = 4x³ – 4

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 4x³ – 4 = 0

⇒ 4(x³ – 1) = 0

⇒ x = 1

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒ f’(x) = x³ + 2x² – 5x – 6

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ x³ + 2x² – 5x – 6 = 0

⇒ (x+1)(x – 2)(x + 3) = 0

⇒ x = –1, 2 , –3

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.

Here we have,

⇒

⇒ f’(x) = 4x³ – 12x² + 8x

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 4x³ – 12x² + 8x= 0

⇒ 4(x³ – 3x² + 2x) = 0

⇒ x(x² – 3x + 2) = 0

⇒ x(x² – 2x – x + 2) = 0

⇒ x(x – 2)(x – 1)

⇒ x = 0, 1 , 2

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.

Here we have,

⇒

⇒

⇒

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒

⇒

⇒ x = 0, 1

Since x > 0, therefore only check the range on the positive side of the number line.

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒

⇒

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒

⇒

⇒

Since is a complex number, therefore only check range on 0 sides of number line.

Given:- Function f(x) = x³ – 6x² + 9x + 15

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x³ – 6x² + 9x + 15

⇒

⇒ f’(x) = 3x² – 12x + 9

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 3x² – 12x + 9 = 0

⇒ 3(x² – 4x + 3) = 0

⇒ 3(x² – 3x – x + 3) = 0

⇒ x² – 3x – x + 3 = 0

⇒ (x – 3)(x – 1) = 0

⇒ x = 1, 3

Given:- Function f(x)= {x (x – 2)}²

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x)= {x (x – 2)}²

⇒ f(x)= {[x²–2x]}²

⇒

⇒ f’(x)= 2(x²–2x)(2x–2)

⇒ f’(x)= 4x(x–2)(x–1)

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 4x(x–2)(x–1)= 0

⇒ x(x–2)(x–1)= 0

⇒ x =0, 1, 2

Given:- Function f(x) = 3x⁴ – 4x³ – 12x² + 5

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 3x⁴ – 4x³ – 12x² + 5

⇒

⇒ f’(x) = 12x³ – 12x² – 24x

⇒ f’(x) = 12x(x² – x – 2)

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 12x(x² – x – 2)> 0

⇒ x(x² – 2x + x – 2) > 0

⇒ x(x – 2)(x + 1) > 0

⇒ –1 < x < 0 and x > 2

⇒ x ∈ (–1,0)∪(2, ∞)

Thus f(x) is increasing on interval (–1,0)∪(2, ∞)

Again, For f(x) to be decreasing, we must have

f’(x) < 0

⇒ 12x(x² – x – 2)< 0

⇒ x(x² – 2x + x – 2) < 0

⇒ x(x – 2)(x + 1) < 0

⇒ –∞ < x < –1 and 0 < x < 2

⇒ x ∈ (–∞, –1) ∪ (0, 2)

Thus f(x) is decreasing on interval (–∞, –1) ∪ (0, 2)

Given:- Function f(x) = x² – 6x + 9 and a line parallel to y = x + 5

Theorem:- Let f be a differentiable real function defined on an open interval (a, b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x² – 6x + 9

⇒

⇒ f’(x) = 2x – 6

⇒ f’(x) = 2(x – 3)

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 2(x – 3) = 0

⇒ (x – 3) = 0

⇒ x = 3

Given:- Function f(x) = sin x – cos x, 0 < x < 2π

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = sin x – cos x

⇒

⇒ f’(x) = cos x + sin x

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ cos x + sin x = 0

⇒ tan(x) = –1

⇒

Here these points divide the angle range from 0 to 2∏ since we have x as angle

Given:- Function f(x) = e^2x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = e^2x

⇒

⇒ f’(x) = 2e^2x

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 2e^2x > 0

⇒ e^2x > 0

since, the value of e lies between 2 and 3

so, whatever be the power of e (i.e x in domain R) will be greater than zero.

Thus f(x) is increasing on interval R

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒

⇒

As given x ∈ R , x ≠ 0

⇒

Their ratio is also greater than 0

⇒

⇒ ; as by applying –ve sign change in comparision sign

⇒ f’(x) < 0

Hence, condition for f(x) to be decreasing

Thus f(x) is decreasing for all x ≠ 0

Given:- Function f(x) = log_a x , 0 < a < 1

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = log_a x, 0 < a < 1

⇒

⇒

As given 0 < a < 1

⇒ log(a) < 0

and for x > 0

⇒

Therefore f’(x) is

⇒

⇒ f’(x) < 0

Hence, condition for f(x) to be decreasing

Thus f(x) is decreasing for all x > 0

Given:- Function f(x) = sin x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = sin x

⇒

⇒ f’(x) = cosx

Taking different region from 0 to 2π

a) let

⇒ cos(x) > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

b) let

⇒ cos(x) < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

Therefore, from above condition we find that

⇒ f(x) is increasing in and decreasing in

Hence, condition for f(x) neither increasing nor decreasing in (0,π)

Given:- Function f(x) = log sin x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = log sin x

⇒

⇒

⇒ f’(x) = cot(x)

Taking different region from 0 to π

a) let

⇒ cot(x) > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

b) let

⇒ cot(x) < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

Hence proved

Given:- Function f(x) = x – sin x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x – sin x

⇒

⇒ f’(x) = 1 – cos x

Now, as given

x ϵ R

⇒ –1 < cosx < 1

⇒ –1 > cos x > 0

⇒ f’(x) > 0

hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

Given:- Function f(x) = x³ – 15x² + 75x – 50

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x³ – 15x² + 75x – 50

⇒

⇒ f’(x) = 3x² – 30x + 75

⇒ f’(x) = 3(x² – 10x + 25)

⇒ f’(x) = 3(x – 5)²

Now, as given

x ϵ R

⇒ (x – 5)² > 0

⇒ 3(x – 5)² > 0

⇒ f’(x) > 0

hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

Given:- Function f(x) = cos² x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = cos² x

⇒

⇒ f’(x) = 3cosx(–sinx)

⇒ f’(x) = –2sin(x)cos(x)

⇒ f’(x) = –sin2x ; as sin2A = 2sinA cosA

Now, as given

⇒ 2x ∈ (0,π)

⇒ Sin(2x)> 0

⇒ –Sin(2x)< 0

⇒ f’(x) < 0

hence, Condition for f(x) to be decreasing

Thus f(x) is decreasing on interval

Hence proved

Given:- Function f(x) = sin x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = sin x

⇒

⇒ f’(x) = cosx

Now, as given

That is 4^th quadrant, where

⇒ cosx> 0

⇒ f’(x) > 0

hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval

Given:- Function f(x) = cos x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = cos x

⇒

⇒ f’(x) = –sinx

Taking different region from 0 to 2π

a) let

⇒ sin(x) > 0

⇒ –sinx < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

b) let

⇒ sin(x) < 0

⇒ –sinx > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

Therefore, from above condition we find that

⇒ f(x) is decreasing in and increasing in

Hence, condition for f(x) neither increasing nor decreasing in (–π,π)

Given:- Function f(x) = tan x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = tan x

⇒

⇒ f’(x) = sec²x

Now, as given

That is 4^th quadrant, where

⇒ sec²x> 0

⇒ f’(x) > 0

hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval

Given:- Function f(x) = tan^–1 (sin x + cos x)

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = tan^–1 (sin x + cos x)

⇒

Now, as given

⇒ Cosx – sinx< 0 ; as here cosine values are smaller than sine values for same angle

⇒

⇒ f’(x) < 0

hence, Condition for f(x) to be decreasing

Thus f(x) is decreasing on interval

Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

Now, as given

⇒

⇒

⇒ ;

as here lies in 3^rd quadrant

⇒

⇒

⇒ f’(x) < 0

hence, Condition for f(x) to be decreasing

Thus f(x) is decreasing on interval

Given:- Function f(x) = cot^–1 (sin x + cos x)

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = cot^–1 (sin x + cos x)

⇒

Now, as given

⇒ Cosx – sinx< 0 ; as here cosine values are smaller than sine values for same angle

⇒

⇒ f’(x) < 0

hence, Condition for f(x) to be decreasing

Thus f(x) is decreasing on interval

Given:- Function f(x) = (x – 1) e^x + 1

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = (x – 1) e^x + 1

⇒

⇒ f’(x) = e^x + (x – 1) e^x

⇒ f’(x) = e^x(1+ x – 1)

⇒ f’(x) = xe^x

as given

x > 0

⇒ e^x > 0

⇒ xe^x > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x > 0

Given:- Function f(x) = x² – x + 1

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = x² – x + 1

⇒

⇒ f’(x) = 2x – 1

Taking different region from (0, 1)

a) let

⇒ 2x – 1 < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

b) let

⇒ 2x – 1 > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

Therefore, from above condition we find that

⇒ f(x) is decreasing in and increasing in

Hence, condition for f(x) neither increasing nor decreasing in (0, 1)

Given:- Function f(x) = x⁹ + 4x⁷ + 11

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.

Here we have,

f(x) = x⁹ + 4x⁷ + 11

⇒

⇒ f’(x) = 9x⁸ + 28x⁶

⇒ f’(x) = x⁶(9x² + 28)

as given

x ϵ R

⇒ x⁶ > 0 and 9x² + 28 > 0

⇒ x⁶(9x² + 28) > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

Given:- Function f(x) = x³ – 6x² + 12x – 18

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.

Here we have,

f(x) = x³ – 6x² + 12x – 18

⇒

⇒ f’(x) = 3x² – 12x + 12

⇒ f’(x) = 3(x² – 4x + 4)

⇒ f’(x) = 3(x – 2)²

as given

x ϵ R

⇒ (x – 2)²> 0

⇒ 3(x – 2)² > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

Given:- Function f(x) = f(x) = x² – 6x + 3

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = f(x) = x² – 6x + 3

⇒

⇒ f’(x) = 2x – 6

⇒ f’(x) = 2(x – 3)

Here A function is said to be increasing on [a,b] if f(x) > 0

as given

x ϵ [4, 6]

⇒ 4 ≤ x ≤ 6

⇒ 1 ≤ (x–3) ≤ 3

⇒ (x – 3) > 0

⇒ 2(x – 3) > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ [4, 6]

we have,

f(x) = sin x – cos x

f'(x) = cos x + sin x

Now,

⇒ f'(x)>0

Hence, f(x) is an increasing function on (–π /4, π /4)

we have,

f(x) = tan^–1 x – x

Now,

Hence, f(x) is an decreasing function for R

we have,

Now,

Hence, f(x) is an increasing function on (–π /3, π /3)

we have

Critical points

⇒x = 0, –1

Clearly,f'(x) > 0 if x>0

And f'(x) < 0 if –1 < x < 0 or x < –1

Hence, f(x) increases in (0,), decreases in (–, –1) U (–1, 0)

we have,

f(x) = (x + 2)e^–x

f'(x) = e^–x – e^–x (x+2)

= e^–x (1 – x – 2)

= – e^–x (x+1)

Critical points

f'(x) = 0

⇒ – e^–x (x + 1) = 0

⇒x = – 1

Clearly

Hence f(x) increases in (–∞,–1), decreases in (–1, ∞)

we have,

f(x) = 10^x

∴f' (x) = 10^x log10

Now,

⇒

⇒

⇒ f'(x)>0

Hence, f(x) in an increasing function for all x

we have,

f(x) = x – [x]

∴f'(x) = 1 > 0

∴ f(x) is an increasing function on (0,1)

(i) we have

f(x) = 3x⁵ + 40x³ + 240x

Now,

xR

Hence, f(x) is an increasing function for all x

(ii) we have

f(x) = 4x³ – 18x² + 27x – 27

Now,

x R

Hence, f(x) is an increasing fuction for all x

we have,

In Interval (0,), tan x > 0

In interval (), tan x < 0

given

f(x)

Hence f(x) is strickly increasing on R

Given f(x) =cos x

(i) Since for each x (),sin x > 0

⇒

So f is strictly decreasing in (0,)

(ii) Since for each x (),sin x <0

⇒

So f is strictly increasing in (,2)

(iii) Clearly from (1) and (2) above, f is neither increasing nor decreasing in (0,)

We have,

f(x) = – x sinx

f ’(x) = 2x – sin x – x cos x

Now,

x ()

⇒ 0 sin x 1, 0 cos x 1,

⇒ 2x–sin x –x cos x > 0

⇒ f ’(x) ≥ 0

Hence,f(x) is an increasing function on ().

We have,

f(x) = – ax

Given that f(x) is on increasing function

for all x R

⇒ for all x R

⇒ a < for all x R

But the last value of 3x² = 0 for x = 0

∴a ≤ 0

We have,

f(x) = sin x – bx +c

Given that f(x) is on decreasing function on R

for all x R

⇒ for all x R

⇒ b < for all x R

But the last value of cos x in 1

We have,

f(x) = x + cos x – a

Now,

x R

⇒ > 0

⇒ > 0

⇒

Hence,f(x) is an increasing function for x R

As f(0) = f(1) and f is differentiable, hence by Rolles theorem:

for some c [0,1]

let us now apply LMVT (as function is twice differentiable) for point c and x [0,1],

hence,

f ”(d)

⇒ f ”(d)

⇒ f ”(d)

A given that | f ”(d)| <=1 for x [0,1]

⇒

⇒

Now both x and c lie in [0,1], hence [0,1]

(i): Consider the given function,

f(x) = x |x|, x ϵ R

⇒

⇒

⇒ f> 0

Therefore, f(x) is an increasing function for all real values.

(ii): Consider the given function,

f(x) = sin x +|sin x|, 0< x 2

⇒

⇒

The function 2cos x will be positive between (0,)

Hence the function f(x) is increasing in the interval (0,)

The function 2cos x will be negative between ()

Hence the function f(x) is decreasing in the interval ()

The value of f= 0, when,

Therefore, the function f(x) is neither increasing nor decreasing in the interval ()

(iii): consider the function,

f(x) = sin x(1 + cos x), 0 < x <

⇒ f ’(x) = cos x + sin x( – sin x ) + cos x ( cos x )

⇒ f ’(x) = cos x – sin² x + cos² x

⇒ f ’(x) = cos x + (cos² x – 1) + cos² x

⇒ f ’(x) = cos x + 2 cos² x – 1

⇒ f ’(x)=(2cos x – 1)(cos x + 1)

for f(x) to be increasing, we must have,

f’(x)> 0

⇒ f )=(2

⇒ 0 < x<

So, f(x) to be decreasing, we must have,

f< 0

⇒ f )=(2cos x – 1)(cos x + 1)

⇒ < x <

⇒ x,

So,f(x) is decreasing in ,

Formula:- The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

Now

f^’(x)>0

1-e

x>0

X<0

Formula:- The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = cos^–1 x + x

Now

f^’(x)>0

xR

x(–∞, ∞)

Formula:- The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = x^x

now for decreasing

f^’(x)<0

x^x(1+logx)<0

(1+logx)<0

logx<-1

x<e^-1

Formula:- The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = 2log(x – 2) – x² + 4x + 1

f^’(x)=

now for increasing

f^’(x)>0

3<0 and x-2>0

and x>2

Formula:- The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

f(x) = 2x² – kx + 5

f^’(x)>0

4x-k>0

K<4x

For x=1

K<4

Formula:- (i) ax²+bx+c>0 for all x a>0 and b²-4ac<0

(ii) ax²+bx+c<0 for all x a<0 and b²-4ac<0

(iii)The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = x³ + ax² + bx + 5 sin²x

For increasing function f’(x)>0

3x²+2ax+b+5sin2x>0

Then

3x²+2ax+b-5<0

And b²-4ac<0

4a²-12(b-5)<0

a²-3b+15<0

a² – 3b + 15 < 0

Formula:- (i)if f(-x)=f(x) then function is even

(ii) if f(-x)=-f(x) then function is odd

(iii) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x)=)

f^’(x)>0

hence function is increasing function

f(-x)=-log()

f(-x)=-f(x) is odd function

Formula:- (i) ax²+bx+c>0 for all x a>0 and b²-4ac<0

(ii) ax²+bx+c<0 for all x a<0 and b²-4ac<0

(iii) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = 2tanx+(2a+1)log_e |sec x|+(a – 2)x

f^’(x)=2sec²x+ (2a+1) tanx + (a-2)

f^’(x)=2(tan²+1) + (2a+1).tanx +(a-2)

f^’(x)=2tan²x+2atanx+tanx+a

For increasing function

f’(x)>0

2tan²x+2atanx+tanx+ a>0

From formula (i)

(2a+1)²-8a<0

Formula:-

(i)The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:- f(x) = tan^–1 (g(x))

For increasing function

f’(x)>0

Formula:- (i)The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

(ii)If f(x) is strictly increasing function on interval [a, b], then f^-1 exist and it is also a strictly increasing function

Given:- f(x) = x³ – 6x² + 15x + 3

=3x²-12x+15=f’(x)

Therefore f’(x) will increasing

Also f^-1(x) is possible

Therefore f(x) is invertible function.

f(x) = x² e^-x

=xe^-x(2-x)=f’(x)

for

f’(x)=0

x² e^-x=0

x(2-x)=0

x=2,x=0

f(x) is increasing in (0,2)

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly decreasing on (a,b) is that f’(x)<0 for all x(a,b)

Given:-

f(x) = cosx – 2λ x

=-sinx-2λ =f’(x)

for decreasing function f’(x)<0

-sinx-2λ <0

Sinx+2λ >0

2λ>-sinx

2λ>1

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly decreading on (a,b) is that f’(x)<0 for all x(a,b)

Given:-

f(x)=2(x-1)+3(2-x)

f(x)=-x+4

Therefore f’(x) <0

Hence decreasing function

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

f(x)= x³– 27x +5

=3x²– 27=f’(x)

for increasing function f’(x)>0

3x²– 27>0

(x+3)(x-3)>0

|x|>3

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly decreasing on (a, b) is that f’(x)<0 for all x(a,b)

Given:-

f(x) = 2x³ – 9x² + 12x + 29

f’(x)=6(x-1)(x-2)

for decreasing function f’(x)<0

f’(x)<0

6(x-1)(x-2)<0

1<x<2

Formula:- (i) ax²+bx+c>0 for all x a>0 and b²-4ac<0

(ii) ax²+bx+c<0 for all x a<0 and b²-4ac<0

(iii) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = kx³ – 9x² + 9x + 3

f’(x)=3kx²-18x+9

for increasing function f’(x)>0

f’(x)>0

3kx²-18x+9>0

kx²-6x+3>0

using formula (i)

36-12k<0

k>3

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f’(x)>0 for all x(a,b)

Given:-

For increasing function f’(x)>0

x ϵ R

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

For x<0

f(x)=-1

for 0<x<1

f(x)=2x-1

for x>1

f(x)=1

Hence f(x) will increasing in 0<x<1

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

If f(x) is strictly increasing function on interval [a, b], then f^-1 exist and it is also a strictly increasing function

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly decreasing on (a, b) is that f’(x)<0 for all x(a,b)

Given:-

f(x)=2(x-1)+3(2-x)

f(x)=-x+4

f’(x)=-1

Therefore f’(x) <0

Hence decreasing function

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = cos|x| – 2ax + b

For increasing f’(x)>0

-sinx-2a>0

2a<-sinx

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

For x>0

For x<0

Both are increasing for f’(x)>0

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

For increasing function f’(x)<0

>2

Let x₁<x₂ and both are real number

f(x₁)<f(x₂)

x₁<x₂

only possible on a>1

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

f(x) = log_a x

For increasing f’(x)>0

For log a>1

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

ϕ(x) = f(x) + f(2a – x)

ϕ’(x) = f’(x)-f’(2a – x)

ϕ’’(x) = f’’(x) + f’’(2a – x)

checking the condition

ϕ(x) is decreasing in [0,a]

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = x² – kx + 5

For increasing function f’(x)>o

2x-k>0

K<2x

Putting x=2

K<4

kϵ (–∞, 4)

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

f(x)=

=

checking the value of x

hence increasing

Formula:- (i) ax²+bx+c>0 for all x a>0 and b²-4ac<0

(ii) ax²+bx+c<0 for all x a<0 and b²-4ac<0

(iii) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = x³ – 9k x² + 27x + 30

f’(x)=3x²-18kx+27

for increasing function f’(x)>0

3x²-18kx+27>0

x²-6kx+9>0

Using formula (i)

36k²-36>0

K²>1

Therefore –1 <k < 1

Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on (a, b) is that f’(x)>0 for all x(a,b)

Given:-

f(x) = x⁹ + 3x⁷ + 64

For increasing f’(x)>o

9x⁸+21x⁶>0

xR