Ex 6.2, 15 - Let I be any interval disjoint from [–1, 1]. Prove

Ex 6.2,15 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,15 - Chapter 6 Class 12 Application of Derivatives - Part 3

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

Transcript

Ex 6.2, 15 Let I be any interval disjoint from [โ€“1, 1]. Prove that the function f given by ๐‘“(๐‘ฅ) = ๐‘ฅ + 1/๐‘ฅ is strictly increasing on I.I is any interval disjoint from [โ€“1, 1] Let I = (โˆ’โˆž, โˆ’๐Ÿ)โˆช(๐Ÿ, โˆž) Given f(๐‘ฅ) = ๐‘ฅ + 1/๐‘ฅ We need to show f(๐‘ฅ) is strictly increasing on I i.e. we need to show fโ€™(๐’™) > 0 for ๐‘ฅ โˆˆ (โˆ’โˆž, โˆ’๐Ÿ)โˆช(๐Ÿ, โˆž) Finding fโ€™(๐’™) f(๐‘ฅ) = ๐‘ฅ + 1/๐‘ฅ fโ€™(๐‘ฅ) = 1 โ€“ 1/๐‘ฅ2 fโ€™(๐‘ฅ) = (๐‘ฅ2 โˆ’ 1)/๐‘ฅ2 Putting fโ€™(๐’™) = 0 (๐‘ฅ2 โˆ’ 1)/๐‘ฅ2 = 0 ๐‘ฅ2โˆ’1 = 0 (๐‘ฅโˆ’1)(๐‘ฅ+1)=0 So, ๐’™=๐Ÿ & ๐’™=โˆ’๐Ÿ Plotting points on number line The point ๐‘ฅ = โ€“1 , 1 into three disjoint intervals โˆด f(x) is strictly increasing on (โˆ’โˆž , โˆ’๐Ÿ) & (๐Ÿ , โˆž) Hence proved

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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.