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


Ex 6.5, 7 Find both the maximum value and the minimum value of 3๐‘ฅ4 โ€“ 8๐‘ฅ3 + 12๐‘ฅ2 โ€“ 48๐‘ฅ + 25 on the interval [0, 3]. Let f(x) = 3๐‘ฅ4 โ€“ 8๐‘ฅ3 + 12๐‘ฅ2 โ€“ 48๐‘ฅ + 25, where ๐‘ฅ โˆˆ [0, 3] Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=๐‘‘(3๐‘ฅ^4 โˆ’ 8๐‘ฅ^3 + 12๐‘ฅ^2 โˆ’ 48๐‘ฅ + 25)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ)=3 ร—4๐‘ฅ^3โˆ’8 ร—3๐‘ฅ^2+12 ร—2๐‘ฅโˆ’48+0 fโ€™(๐‘ฅ)=12๐‘ฅ^3โˆ’24๐‘ฅ^2+24๐‘ฅโˆ’48 fโ€™(๐‘ฅ)=12(๐‘ฅ^3โˆ’2๐‘ฅ^2+2๐‘ฅโˆ’4) Putting fโ€™(๐’™)=๐ŸŽ 12(๐‘ฅ^3โˆ’2๐‘ฅ^2+2๐‘ฅโˆ’4)=0 ๐‘ฅ^3โˆ’2๐‘ฅ^2+2๐‘ฅโˆ’4=0 ๐‘ฅ^2 (๐‘ฅโˆ’2)+2(๐‘ฅโˆ’2)=0 (๐‘ฅ^2+2)(๐‘ฅโˆ’2)=0 Since ๐‘ฅ^2=โˆ’2 is not possible Thus ๐‘ฅ=2 is only critical point v๐‘ฅ^2+2=0 ๐‘ฅ^2=โˆ’2 ๐‘ฅโˆ’2=0 ๐‘ฅ=2 Square of number can not be negative Since are given interval ๐‘ฅ โˆˆ [0 , 3] Hence , calculating f(๐‘ฅ) at ๐‘ฅ = 0 , 2 & 3 Hence, Minimum value of f(๐‘ฅ) is โ€“39 at ๐’™ = 2 Maximum value of f(๐‘ฅ) is 25 at ๐’™ = 0

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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.