Book: RD Sharma - Mathematics (Volume 1)
Chapter: 17. Increasing and Decreasing Functions
Subject: Maths - Class 12th
Q. No. 1 of Exercise 17.2
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
Find the intervals in which the following functions are increasing or decreasing.
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
Now, lets check values of f(x) between different ranges
Here points x = 1, –2 , 3 divide the number line into disjoint intervals namely, (–∞, –2),(–2, 1), (1, 3) and (3, ∞)
Lets consider interval (–∞, –2)
In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0
Therefore, f’(x) < 0 when –∞ < x < –2
Thus, f(x) is strictly decreasing on interval x ∈ (–∞, –2)
consider interval (–2, 1)
In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
Therefore, f’(x) > 0 when –2 < x < 1
Thus, f(x) is strictly increases on interval x ∈ (–2, 1)
Now, consider interval (1, 3)
In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 < 0
Therefore, f’(x) < 0 when 1 < x < 3
Thus, f(x) is strictly decreases on interval x ∈ (1, 3)
finally, consider interval (3, ∞)
In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 > 0
Therefore, f’(x) > 0 when x > 3
Thus, f(x) is strictly increases on interval x ∈ (3, ∞)
