Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 17

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 18

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 19

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

Transcript

Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (v) ๐‘“ (๐‘ฅ)=๐‘ฅ3 โ€“6๐‘ฅ2+9๐‘ฅ+15๐‘“ (๐‘ฅ)=๐‘ฅ3 โ€“6๐‘ฅ2+9๐‘ฅ+15 Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ3 โ€“ 6๐‘ฅ2 + 9๐‘ฅ + 15" " )/๐‘‘๐‘ฅ fโ€™(๐‘ฅ)=3๐‘ฅ^2โˆ’12๐‘ฅ+9 fโ€™(๐‘ฅ)=3(๐‘ฅ^2โˆ’4๐‘ฅ+3) Putting fโ€™(๐’™)=๐ŸŽ 3(๐‘ฅ^2โˆ’4๐‘ฅ+3)=0 ๐‘ฅ^2โˆ’4๐‘ฅ+3=0 ๐‘ฅ^2โˆ’3๐‘ฅโˆ’๐‘ฅ+3=0 ๐‘ฅ(๐‘ฅโˆ’3)โˆ’1(๐‘ฅโˆ’3)=0 (๐‘ฅโˆ’1)(๐‘ฅโˆ’3)=0 So, x = 1 & x = 3 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=3(๐‘ฅ^2โˆ’4๐‘ฅ+3) fโ€™โ€™(๐‘ฅ)=๐‘‘(3(๐‘ฅ^2 โˆ’ 4๐‘ฅ+3))/๐‘‘๐‘ฅ = 3(2๐‘ฅโˆ’4+0) = 6๐‘ฅโˆ’12 Putting ๐’™=๐Ÿ in fโ€™โ€™(๐’™) fโ€™โ€™(1)=6(1)โˆ’12 = 6 โ€“ 12 = โ€“ 6 < 0 Since fโ€™โ€™(๐‘ฅ)<0 when ๐‘ฅ=1 โ‡’ ๐‘ฅ=1 is point of local maxima โˆด f(๐‘ฅ) is maximum at ๐’™=๐Ÿ Maximum value of f(๐‘ฅ) at ๐‘ฅ = 1 f(๐‘ฅ)=๐‘ฅ^3โˆ’6๐‘ฅ^2+9๐‘ฅ+15 f(1)=(1)^3โˆ’6(1)^2+9(1)+15 = 1 โ€“ 6 + 9 + 15 = 19 Putting ๐’™=๐Ÿ‘ in fโ€™โ€™(x) fโ€™โ€™(๐‘ฅ)=6๐‘ฅโˆ’12 fโ€™โ€™(3)=6(3)โˆ’12 = 18 โ€“ 12 = 6 > 0 Since fโ€™โ€™(๐‘ฅ)>0 when ๐‘ฅ=3 โ‡’ ๐‘ฅ=3 is point of local minima โˆด f(๐‘ฅ) is minimum at ๐’™=๐Ÿ‘ Minimum value of f(๐‘ฅ) at ๐‘ฅ = 3 f(๐‘ฅ)=๐‘ฅ^3โˆ’6๐‘ฅ^2+9๐‘ฅ+15 f(3)=(3)^3โˆ’6(3)^2+9(3)+15 = 27 โ€“ 54 + 27 + 15 = 15

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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.