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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

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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.