Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 22

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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 23

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Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 24 Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 25

  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 (iv) f(๐‘ฅ) = ๐‘ฅ3 + 1f(๐‘ฅ)=๐‘ฅ^3+1 Finding fโ€™(x) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3 + 1)/๐‘‘๐‘ฅ =3๐‘ฅ^2 Putting fโ€™(๐’™)=๐ŸŽ 3๐‘ฅ^2=0 ๐‘ฅ^2=0 ๐‘ฅ=0 Therefore by first derivate test, the point ๐‘ฅ=0 is neither a point of local maxima nor a point of local Minima Hence ๐’™=๐ŸŽ is point of inflexion Hence, there is no minimum or maximum value Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iv) f(๐‘ฅ) = ๐‘ฅ3 + 1 ๐‘ฅ=0 Finding fโ€™โ€™(x) fโ€™(x) = 3x2 fโ€™โ€™(x) = 6x Finding fโ€™โ€™(x) at x = 0 fโ€™โ€™(0) = 6 ร— 0 = 0 Since fโ€™โ€™(x) = 0 at x = 0 โˆด The point ๐‘ฅ=0 is neither a point of local maxima nor a point of local Minima Hence ๐’™=๐ŸŽ is point of inflexion Hence, there is no minimum or 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.