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  1. Chapter 6 Class 12 Application of Derivatives
  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 & the local minimum values, as the case may be: (i) f (๐‘ฅ)=๐‘ฅ2 f(๐‘ฅ)=๐‘ฅ^2 Finding fโ€™(x) fโ€™(x) = 2x Putting fโ€™(x) = 0 2x = 0 x = 0 Finding fโ€™โ€™(x) fโ€™(x) = 2x Differentiating again fโ€™โ€™(x) = 2 Since fโ€™โ€™(x) > 0 for x = 0 So, f(x) is minimum at x = 0 Minimum value of f(x) at x = 0 f(x) = x2 f(0) = 02 = 0 So, Minimum value of f(x) = 0 There is no maximum value 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: (ii) ๐‘”(๐‘ฅ)=๐‘ฅ3 โ€“3๐‘ฅ ๐‘”(๐‘ฅ)=๐‘ฅ3 โ€“3๐‘ฅ Finding gโ€™(๐’™) gโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3 โˆ’ 3๐‘ฅ)/๐‘‘๐‘ฅ gโ€™(๐‘ฅ)=3๐‘ฅ^2โˆ’3 Putting gโ€™(๐’™)=๐ŸŽ 3๐‘ฅ^2โˆ’3=0 3๐‘ฅ^2=3 ๐‘ฅ^2=3/3 ๐‘ฅ^2=1 ๐‘ฅ=ยฑ1 So, x = 1 & x = โ€“1 Finding gโ€™โ€™(๐’™) gโ€™(๐‘ฅ)=3๐‘ฅ^2โˆ’3 gโ€™โ€™(๐‘ฅ)=๐‘‘(3๐‘ฅ^2โˆ’3)/๐‘‘๐‘ฅ = 6๐‘ฅโˆ’0 = 6๐‘ฅ Putting ๐’™=๐Ÿ in gโ€™โ€™(x) gโ€™โ€™(1)=6(1)= 6 > 0 Thus, gโ€™โ€™(๐‘ฅ)>0 when ๐‘ฅ=1 โ‡’ ๐‘ฅ=1 is point of local minima & g(๐‘ฅ) is minimum at ๐‘ฅ=1 Local minimum value g(๐‘ฅ)=๐‘ฅ^3โˆ’3๐‘ฅ g(1)=(1)^3โˆ’3(1) =1โˆ’3 =โˆ’๐Ÿ Putting ๐’™=โˆ’๐Ÿ in gโ€™โ€™(x) gโ€™โ€™(โˆ’1)=6(โˆ’1)= โ€“6 < 0 Thus, gโ€™โ€™(๐‘ฅ)<0 when ๐‘ฅ=โˆ’1 โ‡’ ๐‘ฅ=โˆ’1 is point of local maxima & g(๐‘ฅ) is maximum at ๐‘ฅ=โˆ’1 Local minimum value g(๐‘ฅ)=๐‘ฅ^3โˆ’3๐‘ฅ g(โˆ’1)=(โˆ’1)^3โˆ’3(โˆ’1) =โˆ’1+3 =๐Ÿ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.