# Misc 13 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 27, 2018 by Teachoo

Last updated at Dec. 27, 2018 by Teachoo

Transcript

Misc 13 Find the points at which the function f given by f (๐ฅ) = (๐ฅโ2)^4 (๐ฅ+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexionf(๐ฅ)= (๐ฅโ2)^4 (๐ฅ+1)3 Step 1: Finding fโ(๐ฅ) fโ(๐ฅ) = (๐ ((๐ฅโ2)^4 (๐ฅ+1)^3 ))/๐๐ฅ using product rule as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข = ใ((๐ฅโ2)^4 )^โฒ (๐ฅ+1)ใ^3+((๐ฅ+1)^3 )^โฒ (๐ฅโ2)^4 = 4(๐ฅโ2)^3 (๐ฅ+1)^3+3(๐ฅ+1)^2 (๐ฅโ2)^4 = (๐ฅโ2)^3 (๐ฅ+1)^2 [4(๐ฅ+1)+3(๐ฅโ2)] = (๐ฅโ2)^3 (๐ฅ+1)^2 [4๐ฅ+4๐ฅ+3๐ฅโ6] = (๐ฅโ2)^3 (๐ฅ+1)^2 [7๐ฅโ2] Step 2: Putting fโ(๐ฅ)=0 (๐ฅโ2)^3 (๐ฅ+1)^2 (7๐ฅโ2)=0 (๐ฅโ2)^3 = 0 ๐ฅ โ 2 = 0 ๐ฅ=2 (๐ฅ+1)^2=0 (๐ฅ+1)=0 ๐ฅ = โ1 7๐ฅ โ 2 = 0 7๐ฅ = 2 ๐ฅ = 2/7 Hence, ๐ฅ=2 & ๐ฅ=โ1 & ๐ฅ=2/7 = 0.28 Thus ๐ฅ=โ1 is a point of inflexion ๐ฅ=2/7 is point of maxima & ๐ฅ=2 is point of minima

Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.