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Example 28 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) = ๐Ÿ(๐Ÿ–๐’™ โˆ’ ๐Ÿ)" " /๐’™^(๐Ÿ/๐Ÿ‘) 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’™=๐Ÿ/๐Ÿ– Note that: Since fโ€™(๐‘ฅ)=2(8๐‘ฅ โˆ’ 1)/๐‘ฅ^(2/3) fโ€™(๐‘ฅ) is not defined at ๐’™= 0 ๐’™=๐Ÿ/๐Ÿ– & 0 are critical points Since, we are given interval [โˆ’๐Ÿ , ๐Ÿ] Hence calculating f(๐‘ฅ) at ๐‘ฅ=โˆ’๐Ÿ, 0, 1/8, ๐Ÿ Hence, Absolute maximum value of f(x) is 18 at ๐’™ = โ€“1 & Absolute minimum value of f(x) is (โˆ’๐Ÿ—)/๐Ÿ’ at ๐’™ = ๐Ÿ/๐Ÿ–

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.