Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 20

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 21

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 22 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 23

  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: (vi) g (๐‘ฅ) = ๐‘ฅ/2 + 2/๐‘ฅ , ๐‘ฅ > 0g (๐‘ฅ) = ๐‘ฅ/2 + 2/๐‘ฅ, ๐‘ฅ > 0 Finding gโ€™(๐’™) gโ€™(๐‘ฅ)=๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ/2+2/๐‘ฅ) =๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ/2)+๐‘‘/๐‘‘๐‘ฅ (2๐‘ฅ^(โˆ’1) ) =1/2โˆ’2๐‘ฅ^(โˆ’2) =1/2โˆ’2/๐‘ฅ^2 Putting gโ€™(๐’™)=๐ŸŽ 1/2โˆ’2/๐‘ฅ^2 =0 (๐‘ฅ^2โˆ’ 4)/(2๐‘ฅ^2 )=0 ๐‘ฅ^2โˆ’4=0 ร—2๐‘ฅ^2 (๐‘ฅโˆ’2)(๐‘ฅ+2)=0 So, ๐‘ฅ=2 & ๐‘ฅ=โˆ’2 Since ๐‘ฅ>0 is given, we consider only ๐‘ฅ=2 Finding gโ€™โ€™(๐’™) gโ€™(๐‘ฅ)=1/2โˆ’2/๐‘ฅ^2 gโ€™โ€™(๐‘ฅ)=๐‘‘/๐‘‘๐‘ฅ (1/2โˆ’2/๐‘ฅ^2 ) = 0 โ€“ 2 . (โˆ’2) ๐‘ฅ^(โˆ’2โˆ’1) = 4๐‘ฅ^(โˆ’3) = 4/๐‘ฅ^3 Putting value ๐‘ฅ=2 in gโ€™โ€™(๐‘ฅ) gโ€™โ€™(๐‘ฅ)=4/(2)^3 = 4/8= 1/2 Since gโ€™โ€™ (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(๐‘ฅ) is g(๐‘ฅ)=๐‘ฅ/2+2/๐‘ฅ Putting ๐‘ฅ=2 g(2)= 2/2+2/2 = 1 + 1 = 2

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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.