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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

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

About the Author

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 and Computer Science at Teachoo