Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Slide40.JPG

Slide41.JPG
Slide42.JPG Slide43.JPG

  1. Chapter 6 Class 12 Application of Derivatives
  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 ) 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 ) ๐‘ฅ^3โˆ’1 = 0 ๐‘ฅ^3= โ€“1 ๐‘ฅ = 1 ๐‘ฅ^3+1 = 0 ๐‘ฅ3 = โˆ’1 ๐‘ฅ = โˆ’1 Putting fโ€™(๐’™) = 0 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) = 0 ๐‘ฅ^2โˆ’1/๐‘ฅ^4 = 0 (๐‘ฅ^6 โˆ’ 1)/๐‘ฅ^4 = 0 ๐‘ฅ^6โˆ’1 = 0 (๐‘ฅ^3 )^2โˆ’(1)^2=0 (๐‘ฅ^3โˆ’1)(๐‘ฅ^3+1)=0 Hence, ๐‘ฅ = 1 & โ€“1 Plotting value of ๐’™ on real line Thus, ๐‘ฅ = โ€“1 & 1 divide the real line into three disjoint intervals i.e. (โˆ’โˆž,โˆ’1)(โˆ’1, 1)& (1,โˆž) Hence, f(๐‘ฅ) is strictly increasing on (โˆ’โˆž , โˆ’๐Ÿ) & (๐Ÿ , โˆž) & strictly decreasing on (โˆ’๐Ÿ , ๐Ÿ)

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.