Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 9

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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 10

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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 11 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 12 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 13 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 14 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 15

  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 (ii) f (๐‘ฅ) = 9๐‘ฅ2+12๐‘ฅ+2Finding fโ€™(x) f (๐‘ฅ)=9๐‘ฅ2+12๐‘ฅ+2 Diff. w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=18๐‘ฅ+12 fโ€™(๐‘ฅ)=6(3๐‘ฅ+2) Putting fโ€™(๐’™)=๐ŸŽ 6(3๐‘ฅ+2)=0 3๐‘ฅ+2=0 3๐‘ฅ=โˆ’2 ๐‘ฅ=(โˆ’2)/( 3) Hence ๐‘ฅ=(โˆ’2)/3 is point of minima of f(๐‘ฅ) Finding minimum value of f(๐‘ฅ) at ๐‘ฅ=(โˆ’2)/3 f(๐‘ฅ)=9๐‘ฅ^2+12๐‘ฅ+2 Putting ๐‘ฅ=(โˆ’2)/3 f(๐‘ฅ)=9((โˆ’2)/3)^2+12((โˆ’2)/3)+2=9(4/3)โˆ’12(2/3)+2=โˆ’2 Thus, Minimum value of f(๐’™)=โˆ’๐Ÿ There is no maximum value Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (๐‘ฅ) = 9๐‘ฅ2+12๐‘ฅ+2Finding fโ€™(๐’™) f (๐‘ฅ)=9๐‘ฅ2+12๐‘ฅ+2 Diff. w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(9๐‘ฅ^2 + 12๐‘ฅ + 2)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ)=18๐‘ฅ+12 fโ€™(๐‘ฅ)=6(3๐‘ฅ+2) Putting fโ€™(๐’™)=๐ŸŽ 6(3๐‘ฅ+2)=0 3๐‘ฅ+2=0 3๐‘ฅ=โˆ’2 ๐‘ฅ=(โˆ’2)/3 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)= 6(3๐‘ฅ+2) Again diff w.r.t ๐‘ฅ fโ€™โ€™(๐‘ฅ)=๐‘‘(6(3๐‘ฅ+2))/๐‘‘๐‘ฅ fโ€™โ€™(๐‘ฅ)=6 ๐‘‘(3๐‘ฅ + 2)/๐‘‘๐‘ฅ fโ€™โ€™(๐‘ฅ)=6(3+0) fโ€™โ€™(๐‘ฅ)=6(3) fโ€™โ€™(๐‘ฅ)=18 So, fโ€™โ€™((โˆ’2)/3)=18 Since fโ€™โ€™(๐‘ฅ)>0 is for ๐‘ฅ=(โˆ’2)/3 ๐‘ฅ=(โˆ’2)/3 is point of local minima Finding minimum value Putting ๐‘ฅ=(โˆ’2)/3 in f(๐‘ฅ) f (๐‘ฅ)=9๐‘ฅ2+12๐‘ฅ+2 f ((โˆ’2)/3)=9((โˆ’2)/3)^2+12((โˆ’2)/3)+2 =9(4/9)+12((โˆ’2)/3)+2 =4โˆ’8+2 =โˆ’2 Hence, minimum value = โ€“2 There is no maximum value

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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.