`color{red}{✍️}` `color{blue} {"Right of Origin"}` i.e for `color{blue}{ x > 0}` As we move from left to right.

(a) Higher the value of `x` | Higher the value of `y`

(b) Higher the input | higher the output.

(c) i.e as `x` increases , height of the graph continuously increases.

`color{red}{✍️}` `color{blue}{"Left of Origin"}` i.e for `color{blue}{ x < 0}` As we move from right to left.

(a) Higher the value of `x` | Lower the value of `y`

(b) Higher the input | Lower the output.

(c) i.e as `x` increases , height of the graph continuously decreases.

(a) Higher the value of `x` | Higher the value of `y`

(b) Higher the input | higher the output.

(c) i.e as `x` increases , height of the graph continuously increases.

`color{red}{✍️}` `color{blue}{"Left of Origin"}` i.e for `color{blue}{ x < 0}` As we move from right to left.

(a) Higher the value of `x` | Lower the value of `y`

(b) Higher the input | Lower the output.

(c) i.e as `x` increases , height of the graph continuously decreases.

Definition 1 : Let `I` be an open interval contained in the domain of a real valued function `f.` Then `f` is said to be

1. Increasing on I if `color{red}{x_1 < x_2}` in `I => color{blue}{f(x_1) <= f(x_2)}` for all `x_1, x_2 in I`

2. Strictly increasing on `I` if `color{red}{x_1 < x_2}` in `I=>color{blue}{f(x_1) < f(x_2)}` for all `x_1, x_2 in I`

3. Decreasing on I if `color{red}{x_1 < x_2}` in `I => color{blue}{f(x_1) >= f(x_2)}` for all `x_1, x_2 in I`

4. Strictly increasing on `I` if `color{red}{x_1 < x_2}` in `I => color{blue}{f(x_1) > f(x_2)}` for all `x_1, x_2 in I`

5. Neither increasing nor Decreasing on `I` if `color{red}{x_1 < x_2}` in `I` doest not inply either `=> color{blue}{f(x_1) < f(x_2)}` or `color{blue}{f(x_1) > f(x_2)}` for all `x_1, x_2 in I`

1. Increasing on I if `color{red}{x_1 < x_2}` in `I => color{blue}{f(x_1) <= f(x_2)}` for all `x_1, x_2 in I`

2. Strictly increasing on `I` if `color{red}{x_1 < x_2}` in `I=>color{blue}{f(x_1) < f(x_2)}` for all `x_1, x_2 in I`

3. Decreasing on I if `color{red}{x_1 < x_2}` in `I => color{blue}{f(x_1) >= f(x_2)}` for all `x_1, x_2 in I`

4. Strictly increasing on `I` if `color{red}{x_1 < x_2}` in `I => color{blue}{f(x_1) > f(x_2)}` for all `x_1, x_2 in I`

5. Neither increasing nor Decreasing on `I` if `color{red}{x_1 < x_2}` in `I` doest not inply either `=> color{blue}{f(x_1) < f(x_2)}` or `color{blue}{f(x_1) > f(x_2)}` for all `x_1, x_2 in I`

`\color{green} ✍️` Let `x_0` be a point in the domain of definition of a real valued function `f.`

● A function f is said to be increasing at `x_0` if there exists an interval

`color{blue}{I = (x_0 – h, x_0 + h)}`, h > 0 such that for `x_1, x_2 ∈ I`

`color{red}{x_1 < x_2\ \ "in" \ \ I ⇒ f (x_1) ≤ f (x_2)}`

● A function f is said to be decreasing at `x_0` if there exists an interval

`color{blue}{I = (x_0 – h, x_0 + h)}`, h > 0 such that for `x_1, x_2 ∈ I`

`color{red}{x_1 < x_2\ \ "in" \ \ I ⇒ f (x_1) => f (x_2)}`

● A function f is said to be increasing at `x_0` if there exists an interval

`color{blue}{I = (x_0 – h, x_0 + h)}`, h > 0 such that for `x_1, x_2 ∈ I`

`color{red}{x_1 < x_2\ \ "in" \ \ I ⇒ f (x_1) ≤ f (x_2)}`

● A function f is said to be decreasing at `x_0` if there exists an interval

`color{blue}{I = (x_0 – h, x_0 + h)}`, h > 0 such that for `x_1, x_2 ∈ I`

`color{red}{x_1 < x_2\ \ "in" \ \ I ⇒ f (x_1) => f (x_2)}`

Q 3165145065

Show that the function given by f (x) = 7x – 3 is strictly increasing on R.

Class 12 Chapter 6 Example 7

Class 12 Chapter 6 Example 7

Let x1 and x2 be any two numbers in R. Then

`x_1 < x_2 ⇒ 7x_1 < 7x_2 ⇒ 7x_1 – 3 < 7x_2 – 3 ⇒ f (x_1) < f (x_2)`

Thus, by Definition 1, it follows that f is strictly increasing on R.

We shall now give the first derivative test for increasing and decreasing functions.

The proof of this test requires the Mean Value Theorem studied in Chapter 5.

`\color{green} ✍️` Let `f` be continuous on `[a, b]` and differentiable

on the open interval (a,b). Then

`1. color{blue}{ f \ \ "is increasing in" [a,b] "if" \ \ f′(x) > 0 \ \ "for each"\ \ x ∈ (a, b)}`

`2. color{blue}{ f \ \ "is decreasing in" \ \ [a,b] \ \ "if" \ \ f ′(x) < 0 \ \ "for each" \ \ x ∈ (a, b)}`

`3. color{blue}{ f \ \ "is a constant function in" \ \[a,b] "if" \ \ f ′(x) = 0 \ \ "for each" \ \ x ∈ (a, b)}`

`color{red}{"Proof: "}` (a) Let `x_1, x_2 ∈ [a, b]` be such that `x_1 < x_2`.

Then, by Mean Value Theorem , there exists a point c between `x_1` and `x_2` such that

`f (x_2) – f (x_1) = f ′(c) (x_2 – x_1)`

i.e. `f (x_2) – f (x_1) > 0` (as f ′(c) > 0 (given))

i.e. `f (x_2) > f (x_1)`

Thus, we have

`x_1 < x_2 ⇒ f (x_1) < f (x_2 )`, for all `x_1, x_2 ∈ [a,b]`

Hence, f is an increasing function in [a,b].

The proofs of part (b) and (c) are similar manner.

`color{green} {✍️ "Key Points"}`

(i) `f` is strictly increasing in `(a, b)` if `f ′(x) > 0` for each `x ∈ (a, b)`.

(ii) `f` is strictly decreasing in `(a, b)` if `f ′(x) < 0` for each `x ∈ (a, b)`.

(iii) A function will be increasing (decreasing) in `R` if it is so in every interval of `R`.

on the open interval (a,b). Then

`1. color{blue}{ f \ \ "is increasing in" [a,b] "if" \ \ f′(x) > 0 \ \ "for each"\ \ x ∈ (a, b)}`

`2. color{blue}{ f \ \ "is decreasing in" \ \ [a,b] \ \ "if" \ \ f ′(x) < 0 \ \ "for each" \ \ x ∈ (a, b)}`

`3. color{blue}{ f \ \ "is a constant function in" \ \[a,b] "if" \ \ f ′(x) = 0 \ \ "for each" \ \ x ∈ (a, b)}`

`color{red}{"Proof: "}` (a) Let `x_1, x_2 ∈ [a, b]` be such that `x_1 < x_2`.

Then, by Mean Value Theorem , there exists a point c between `x_1` and `x_2` such that

`f (x_2) – f (x_1) = f ′(c) (x_2 – x_1)`

i.e. `f (x_2) – f (x_1) > 0` (as f ′(c) > 0 (given))

i.e. `f (x_2) > f (x_1)`

Thus, we have

`x_1 < x_2 ⇒ f (x_1) < f (x_2 )`, for all `x_1, x_2 ∈ [a,b]`

Hence, f is an increasing function in [a,b].

The proofs of part (b) and (c) are similar manner.

`color{green} {✍️ "Key Points"}`

(i) `f` is strictly increasing in `(a, b)` if `f ′(x) > 0` for each `x ∈ (a, b)`.

(ii) `f` is strictly decreasing in `(a, b)` if `f ′(x) < 0` for each `x ∈ (a, b)`.

(iii) A function will be increasing (decreasing) in `R` if it is so in every interval of `R`.

Q 3175145066

Show that the function f given by

`f (x) = x^3 – 3x^2 + 4x, x ∈ R`

is strictly increasing on R.

Class 12 Chapter 6 Example 8

`f (x) = x^3 – 3x^2 + 4x, x ∈ R`

is strictly increasing on R.

Class 12 Chapter 6 Example 8

Note that

`f ′(x) = 3x^2 – 6x + 4`

`= 3(x^2 – 2x + 1) + 1`

`= 3(x – 1)^2 + 1 > 0`, in every interval of R

Therefore, the function f is strictly increasing on R.

Q 3115245160

Prove that the function given by f (x) = cos x is

(a) strictly decreasing in (0, π)

(b) strictly increasing in (π, 2π), and

(c) neither increasing nor decreasing in (0, 2π).

Class 12 Chapter 6 Example 9

(a) strictly decreasing in (0, π)

(b) strictly increasing in (π, 2π), and

(c) neither increasing nor decreasing in (0, 2π).

Class 12 Chapter 6 Example 9

Note that f ′(x) = – sin x

(a) Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is strictly

decreasing in (0, π).

(b) Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is strictly

increasing in (π, 2π).

(c) Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π).

`"Note :"` One may note that the function in Example is neither strictly increasing in `[π, 2π]` nor strictly decreasing in `[0, π].` However, since the function is continuous at the end points 0 and π, by Theorem , f is increasing in [π, 2π] and decreasing in [0, π].

Q 3125245161

Find the intervals in which the function f given by `f (x) = x^2 – 4x + 6` is

(a) strictly increasing (b) strictly decreasing

Class 12 Chapter 6 Example 10

(a) strictly increasing (b) strictly decreasing

Class 12 Chapter 6 Example 10

We have

`f (x) = x^2 – 4x + 6`

or f ′(x) = 2x – 4

Therefore, f ′(x) = 0 gives x = 2. Now the point x = 2 divides the real line into two

disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6.3). In the interval (– ∞, 2),

f ′(x) = 2x – 4 < 0.

Therefore, f is strictly decreasing in this

interval. Also, in the interval (2,∞) , f ′(x) > 0

and so the function f is strictly increasing in this

interval.

`"Note:"` that the given function is continuous at 2 which is the point joining

the two intervals. So, by Theorem 1, we conclude that the given function is decreasing

in `(– ∞, 2]` and increasing in `[2, ∞).`

Q 3135245162

Find the intervals in which the function f given by` f (x) = 4x^3 – 6x^2 – 72x + 30`

is (a) strictly increasing (b) strictly decreasing.

Class 12 Chapter 6 Example 11

is (a) strictly increasing (b) strictly decreasing.

Class 12 Chapter 6 Example 11

We have

`f (x) = 4x^3 – 6x^2 – 72x + 30`

or `f ′(x) = 12x^2 – 12x – 72`

`= 12(x^2 – x – 6)`

= 12(x – 3) (x + 2)

Therefore, f ′(x) = 0 gives x = – 2, 3. The

points x = – 2 and x = 3 divides the real line into

three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)

and (3, ∞).

In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),

f ′(x) is negative. Consequently, the function f is strictly increasing in the intervals

(– ∞, – 2) and (3, ∞) while the function is strictly decreasing in the interval (– 2, 3).

However, f is neither increasing nor decreasing in R.

Q 3135345262

Find intervals in which the function given by `f (x) = sin 3x , x ∈ [0, pi/2] ` is

(a) increasing (b) decreasing.

Class 12 Chapter 6 Example 12

(a) increasing (b) decreasing.

Class 12 Chapter 6 Example 12

We have

f (x) = sin 3x

or f ′(x) = 3cos 3x

Therefore, f ′(x) = 0 gives cos 3x = 0 which in turn gives`3x = pi/2, (3 pi)/2` ( as `x ∈ [0,pi/2] `

implies `3x ∈ [ 0, (3 pi)/2] ` ) . So `x= pi/6` and `pi/2` The point ` x= pi/6` divides the interval `[ 0, pi/2]`

into two disjoint intervals ` [ 0, pi/6)` and ` ( pi/6, pi/2 ]` .

Now, f ′(x) > 0 for all `x ∈ [ 0, pi/6)` as `0 le x < pi/6 => 0 le 3x < pi/2` and ` f'(x) < 0` for

all `x ∈ (pi/6, pi/2)` as ` pi/6 < x < pi/2 => pi/2 < 3x < (3 pi )/2`

Therefore, f is strictly increasing in ` [ 0, pi/6)` and strictly decreasing in `(pi/6, pi/2)`

Also, the given function is continuous at x = 0 and `x = pi/6`. Therefore, by Theorem 1,

f is increasing on ` [0, pi/6] ` and decreasing on `[ pi/6, pi/2]`

Q 3145345263

Find the intervals in which the function f given by

f (x) = sin x + cos x, 0 ≤ x ≤ 2π

is strictly increasing or strictly decreasing.

Class 12 Chapter 6 Example 13

f (x) = sin x + cos x, 0 ≤ x ≤ 2π

is strictly increasing or strictly decreasing.

Class 12 Chapter 6 Example 13

We have

f(x) = sin x + cos x,

or f ′(x) = cos x – sin x

Now f ′(x) = 0 gives sin x = cos x which gives that `x = pi/4, (5 pi)/4` as ` 0 le x le 2 pi`

The points `x = pi/4` and `x = (5 pi)/4` divide the interval [0, 2π] into three disjoint intervals, namely,

` [ 0, pi/4) , ( pi/4, (5 pi)/4)` and ` ( (5 pi )/4 ,2 pi ]` .

Note that `f ' (x) > 0` if `x ∈ [ 0, pi/4 ) ∪ ( (5 pi)/4 , 2 pi ]`

or f is strictly increasing in the intervals `[ 0, pi/4) ` and ` ( (5 pi)/4 , 2 pi ]`

Also `f' (x) < 0` if `x ∈ ( pi/4, (5 pi )/4 )`

or f is strictly decreasing in ` (pi/4 , (5 pi )/4)`