Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 16

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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 17

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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 18 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 19 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 20 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 21

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

Transcript

Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) ๐‘“ (๐‘ฅ) = โ€“(๐‘ฅ โ€“ 1)^2+10 f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Finding fโ€™ (x) Diff w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(โˆ’(๐‘ฅโˆ’1)^2+10)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(๐‘‘(๐‘ฅโˆ’1)/๐‘‘๐‘ฅ)+0 fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(1โˆ’0) + 0 fโ€™(๐‘ฅ)=โˆ’2(๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ โ€“2(๐‘ฅโˆ’1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Hence, ๐‘ฅ=1 is point of Maxima & No point of Minima Thus, f(๐‘ฅ) has maximum value at ๐‘ฅ=1 Putting x=1 in f(x) f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 f(1)=โˆ’(1โˆ’1)^2+10 = 0 + 10 = 10 Maximum value of f(๐‘ฅ) is 10 There is no minimum value of f(๐’™) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) ๐‘“ (๐‘ฅ) = โ€“ (๐‘ฅ โ€“ 1)^2+10 f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Finding fโ€™(x) Diff w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(โˆ’(๐‘ฅโˆ’1)^2+10)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(๐‘‘(๐‘ฅโˆ’1)/๐‘‘๐‘ฅ)+0 fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(1โˆ’0) + 0 fโ€™(๐‘ฅ)=โˆ’2(๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ โ€“2(๐‘ฅโˆ’1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=โˆ’2(๐‘ฅโˆ’1) Again diff w.r.t ๐‘ฅ fโ€™โ€™(๐‘ฅ)=๐‘‘(โˆ’2(๐‘ฅ โˆ’ 1))/๐‘‘๐‘ฅ =โˆ’2 ๐‘‘(๐‘ฅ โˆ’ 1)/๐‘‘๐‘ฅ =โˆ’2(1โˆ’0) =โˆ’2 Since fโ€™โ€™(๐‘ฅ) < 0 for ๐‘ฅ=1 Hence, f(๐‘ฅ) has Maximum value at ๐‘ฅ=1 Finding maximum value of f(๐’™) f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Putting ๐‘ฅ=1 f(๐‘ฅ) =โˆ’(1โˆ’1)^2+10 = 0 + 10 = 10 Maximum value of f(๐’™) is 10 There is no minimum value of f(๐’™)

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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.