Misc 7 - Find intervals f(x) = x3 + 1/x3 x = 0 is increasing

Misc 7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 7 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 7 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Misc 7 Find the intervals in which the function f given by f (x) = x3 + 1/๐‘ฅ^3 , ๐‘ฅ โ‰  0 is (i) increasing (ii) decreasing. f(๐‘ฅ) = ๐‘ฅ3 + 1/๐‘ฅ3 Finding fโ€™(๐’™) fโ€™(๐‘ฅ) = ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^3+๐‘ฅ^(โˆ’3) )^. = 3๐‘ฅ2 + (โˆ’3)^(โˆ’3 โˆ’ 1) = 3๐‘ฅ2 โ€“ 3๐‘ฅ^(โˆ’4) = 3๐‘ฅ^2โˆ’3/๐‘ฅ^4 = 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) Putting fโ€™(๐’™) = 0 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) = 0 (๐‘ฅ^6 โˆ’ 1)/๐‘ฅ^4 = 0 ๐’™^๐Ÿ”โˆ’๐Ÿ = 0 (๐‘ฅ^3 )^2โˆ’(1)^2=0 (๐’™^๐Ÿ‘โˆ’๐Ÿ)(๐’™^๐Ÿ‘+๐Ÿ)=๐ŸŽ Hence, ๐’™ = 1 & โ€“1 ๐‘ฅ^3+1 = 0 ๐‘ฅ3 = โˆ’1 ๐’™ = โˆ’1 Plotting points on number line So, f(๐‘ฅ) is strictly increasing on (โˆ’โˆž , โˆ’1) & (1 , โˆž) & f(๐‘ฅ) strictly decreasing on (โˆ’1 , 1) But we need to find Increasing & Decreasing fโ€™(๐‘ฅ) = 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) Thus, f(๐‘ฅ) is increasing on (โˆ’โˆž , โˆ’๐Ÿ] & [๐Ÿ , โˆž) & f(๐‘ฅ) is decreasing on [โˆ’๐Ÿ , ๐Ÿ] For x = โˆ’1 fโ€™(โˆ’1) = 0 For x = 1 fโ€™(1) = 0

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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.