Mathematics Increasing and Decreasing Function

### Increasing and Decreasing Function

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.

### Increasing and Decreasing Functions

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

### Definition

\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)}
Q 3165145065

Show that the function given by f (x) = 7x – 3 is strictly increasing on R.
Class 12 Chapter 6 Example 7
Solution:

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.

### First derivative test for increasing and decreasing functions.

\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.
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
Solution:

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
Solution:

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
Solution:

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
Solution:

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
Solution:

We have
f (x) = sin 3x
or f ′(x) = 3cos 3x

Therefore, f ′(x) = 0 gives cos 3x = 0 which in turn gives3x = 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
Solution:

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)