Check sibling questions

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

Last updated at May 29, 2023 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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

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)

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.