Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 9

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Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 10

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Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 11

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

Transcript

Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iv) f (๐‘ฅ) = (๐‘ฅ โˆ’1)2 + 3, ๐‘ฅ โˆˆ [โˆ’3,1] f (๐‘ฅ) = (๐‘ฅ โˆ’1)2 + 3 Finding fโ€™(๐’™) fโ€™(x) = ๐‘‘((๐‘ฅ โˆ’ 1)^2+3)/๐‘‘๐‘ฅ = 2(๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ 2(๐‘ฅโˆ’1)=0 ๐‘ฅโˆ’1=0 ๐‘ฅ=1 Since given interval ๐‘ฅ โˆˆ [โˆ’3 , 1] Hence , calculating f(๐‘ฅ) at ๐‘ฅ = โ€“ 3 , 1 Absolute Minimum value of f(x) is 3 at ๐’™ = 1 & Absolute Maximum value of f(x) is 19 at ๐’™ = โ€“ 3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.