Ex 6.2, 6 (e) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

Ex 6.2,6 - Chapter 6 Class 12 Application of Derivatives - Part 11

Ex 6.2,6 - Chapter 6 Class 12 Application of Derivatives - Part 12
Ex 6.2,6 - Chapter 6 Class 12 Application of Derivatives - Part 13
Ex 6.2,6 - Chapter 6 Class 12 Application of Derivatives - Part 14

Go Ad-free

Transcript

Ex 6.2, 6 Find the intervals in which the following functions are strictly increasing or decreasing: (e) (𝑥 + 1)^3 (𝑥 – 3)^3 f(𝑥) = (𝑥+1)3 (𝑥−3)3 Calculating f’(𝒙) f(𝑥) = (𝑥+1)3 (𝑥−3)3 f’(𝑥)= 〖[(𝑥+1)^3]〗^′ (𝑥−3)^3 +[(𝑥−3)3]^′ (𝑥+1)^3 f’(𝑥)=3(𝑥+1)2(𝑥−3)3 + 3(𝑥−3)2(𝑥+1)3 f’(𝑥)=3(𝑥+1)2(𝑥−3)2 ((𝑥−3)+ (𝑥+1)) f’(𝑥)=3(𝑥+1)2(𝑥−3)2 (2𝑥−2) f’(𝒙)= 6(𝒙+𝟏)𝟐 (𝒙−𝟑)𝟐 (𝒙−𝟏) Putting f’(𝒙)=𝟎 6(𝑥+1)2 (𝑥−3)2 (𝑥−1) = 0 Hence, 𝑥 = –1 , 3 & 1 Plotting values of x on real line. Note that: f’(𝑥) = 6 (𝒙+𝟏)^𝟐 (𝒙−𝟑)^𝟐 (𝑥−1) Hence, f is strictly increasing for 1 < 𝑥 < 3 & 𝑥 > 3 i.e. (1, 3) and (3, ∞) f is strictly decreasing for 𝑥 < –1 & −1<𝑥< 1 i.e. (–∞, –1) and (– 1, 1)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo