Ex 6.2, 6 (e) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by Teachoo

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

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)

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Chapter 6 Class 12 Application of Derivatives (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.

Ex 6.2, 6 (e) - Chapter 6 Class 12 Application of Derivatives (Term 1)

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