1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. Miscellaneous
Check sibling questions

Misc 10 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

Transcript

Misc 10 Find the points at which the function f given by f (๐‘ฅ) = (๐‘ฅโˆ’2)^4 (๐‘ฅ+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexionf(๐‘ฅ)= (๐‘ฅโˆ’2)^4 (๐‘ฅ+1)3 Finding fโ€™(๐’™) fโ€™(๐‘ฅ) = (๐‘‘ ((๐‘ฅ โˆ’ 2)^4 (๐‘ฅ + 1)^3 ))/๐‘‘๐‘ฅ = ใ€–((๐‘ฅโˆ’2)^4 )^โ€ฒ (๐‘ฅ+1)ใ€—^3+((๐‘ฅ+1)^3 )^โ€ฒ (๐‘ฅโˆ’2)^4 = 4(๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^3+3(๐‘ฅ+1)^2 (๐‘ฅโˆ’2)^4 = (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 [4(๐‘ฅ+1)+3(๐‘ฅโˆ’2)] = (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 [4๐‘ฅ+4+3๐‘ฅโˆ’6] = (๐’™โˆ’๐Ÿ)^๐Ÿ‘ (๐’™+๐Ÿ)^๐Ÿ [๐Ÿ•๐’™โˆ’๐Ÿ] Putting fโ€™(๐’™)=๐ŸŽ (๐‘ฅโˆ’2)^3 (๐‘ฅ+1)^2 (7๐‘ฅโˆ’2)=0 Hence, ๐‘ฅ=2 & ๐‘ฅ=โˆ’1 & ๐‘ฅ=2/7 = 0.28 (๐‘ฅโˆ’2)^3 = 0 ๐‘ฅ โ€“ 2 = 0 ๐’™=๐Ÿ (๐‘ฅ+1)^2=0 (๐‘ฅ+1)=0 ๐’™ = โ€“1 7๐‘ฅ โ€“ 2 = 0 7๐‘ฅ = 2 ๐’™ = ๐Ÿ/๐Ÿ• Thus, ๐‘ฅ=โˆ’๐Ÿ is a point of Inflexion ๐‘ฅ=๐Ÿ/๐Ÿ• is point of maxima ๐‘ฅ=๐Ÿ is point of minima
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. Miscellaneous

    Case Based Questions (MCQ) →
Case Based Questions (MCQ) →
Chapter 6 Class 12 Application of Derivatives
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