Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Misc 13 Find the points at which the function f given by f (๐‘ฅ) = (๐‘ฅโˆ’2)^4 (๐‘ฅ+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexion f(๐‘ฅ)= (๐‘ฅโˆ’2)^4 (๐‘ฅ+1)3 Finding fโ€™(๐’™) fโ€™(๐‘ฅ) = (๐‘‘ ((๐‘ฅ โˆ’ 2)^4 (๐‘ฅ + 1)^3 ))/๐‘‘๐‘ฅ = ใ€–((๐‘ฅโˆ’2)^4 )^โ€ฒ (๐‘ฅ+1)ใ€—^3+((๐‘ฅ+1)^3 )^โ€ฒ (๐‘ฅโˆ’2)^4 Using product rule as (๐‘ข๐‘ฃ)^โ€ฒ=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข = 4(๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^3+3(๐‘ฅ+1)^2 (๐‘ฅโˆ’2)^4 = (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 [4(๐‘ฅ+1)+3(๐‘ฅโˆ’2)] = (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 [4๐‘ฅ+4+3๐‘ฅโˆ’6] = (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 [7๐‘ฅโˆ’2] Putting fโ€™(๐’™)=๐ŸŽ (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 (7๐‘ฅโˆ’2)=0 Hence, ๐‘ฅ=2 & ๐‘ฅ=โˆ’1 & ๐‘ฅ=2/7 = 0.28 (๐‘ฅโˆ’2)^3 = 0 ๐‘ฅ โ€“ 2 = 0 ๐‘ฅ=2 (๐‘ฅ+1)^2=0 (๐‘ฅ+1)=0 ๐‘ฅ = โ€“1 7๐‘ฅ โ€“ 2 = 0 7๐‘ฅ = 2 ๐‘ฅ = 2/7 Thus ๐‘ฅ=โˆ’1 is a point of inflexion ๐‘ฅ=2/7 is point of maxima & ๐‘ฅ=2 is point of minima

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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.