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

Transcript

Example 40 Find absolute maximum and minimum values of a function f given by f (๐‘ฅ) = ใ€–12 ๐‘ฅใ€—^(4/3) โ€“ ใ€– 6๐‘ฅใ€—^(1/3) , ๐‘ฅ โˆˆ [ โ€“ 1, 1] f (๐‘ฅ) = ใ€–12 ๐‘ฅใ€—^(4/3) โ€“ ใ€– 6๐‘ฅใ€—^(1/3) Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=๐‘‘(12๐‘ฅ^(4/3) โˆ’ 6๐‘ฅ^(1/3) )/๐‘‘๐‘ฅ = 12 ร— 4/3 ๐‘ฅ^(4/3 โˆ’1)โˆ’6 ร— 1/3 ๐‘ฅ^(1/3 โˆ’1) = 4 ร— 4 ๐‘ฅ^((4 โˆ’ 3)/3) โˆ’2๐‘ฅ^((1 โˆ’ 3)/3) = 16 ๐‘ฅ^(1/3) โˆ’2๐‘ฅ^((โˆ’2)/3) = 16 ๐‘ฅ^(1/3) โˆ’ 2/๐‘ฅ^(2/3) = (16๐‘ฅ^(1/3) ร— ๐‘ฅ^(2/3) โˆ’ 2)/๐‘ฅ^(2/3) = (16๐‘ฅ^(1/3 + 2/3) โˆ’ 2)/๐‘ฅ^(2/3) = (16๐‘ฅ^(3/3) โˆ’ 2)/๐‘ฅ^(2/3) = (16๐‘ฅ โˆ’ 2)/๐‘ฅ^(2/3) = 2(8๐‘ฅ โˆ’ 1)" " /๐‘ฅ^(2/3) Hence, fโ€™(๐‘ฅ)=2(8๐‘ฅ โˆ’ 1)/๐‘ฅ^(2/3) Putting fโ€™(๐’™)=๐ŸŽ 2(8๐‘ฅ โˆ’ 1)/๐‘ฅ^(2/3) =0 2(8๐‘ฅโˆ’1)=0 ร—๐‘ฅ^(2/3) 2(8๐‘ฅโˆ’1)=0 8๐‘ฅโˆ’1= 0 8๐‘ฅ=1 ๐‘ฅ=1/8 Note that: Since fโ€™(๐‘ฅ)=2(8๐‘ฅ โˆ’ 1)/๐‘ฅ^(2/3) fโ€™(๐‘ฅ) is not defined at ๐‘ฅ= 0 ๐‘ฅ=1/8 & 0 are critical points We are given interval [โˆ’1 , 1] Hence calculating f(๐‘ฅ) at ๐‘ฅ=0, 1/8, โˆ’1 , 1 ๐‘ฅ = 0 f(0)=12(0)^(4/3)โˆ’6(0)^(1/3) = 0 โ€“ 0 = 0 ๐’™ = ๐Ÿ/๐Ÿ– f(1/8)=12(1/8)^(4/3)โˆ’6(1/8)^(1/3) = 12 ร— (1/2)^(3 ร— 4/3)โˆ’6(1/2)^(3 ร— 4/3) = 12(1/2)^4 โ€“ 6(1/2)^1 = (โˆ’๐Ÿ—)/๐Ÿ’ ๐‘ฅ = 1 f(1)=12(1)^(4/3)โˆ’6(1)^(1/3) = 12 โ€“ 6 = 6 Hence, Absolute maximum value of f(x) is 18 at ๐’™ = โ€“1 & Absolute minimum value of f(x) is (โˆ’9)/4 at ๐’™ = ๐Ÿ/๐Ÿ–

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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.