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

Transcript

Example 32 Find local maximum and local minimum values of the function f given by f (๐‘ฅ)=3๐‘ฅ4 + 4๐‘ฅ3 โ€“ 12๐‘ฅ2 + 12 f (๐‘ฅ)=3๐‘ฅ4 + 4๐‘ฅ3 โ€“ 12๐‘ฅ2 + 12 Finding fโ€™ (๐’™) fโ€™ (๐‘ฅ)=๐‘‘(3๐‘ฅ4 + 4๐‘ฅ3 โ€“ 12๐‘ฅ2 + 12)/๐‘‘๐‘ฅ fโ€™ (๐‘ฅ)=12๐‘ฅ^3+12๐‘ฅ^2 โ€“ 24๐‘ฅ "+ 0" fโ€™ (๐‘ฅ)=12(๐‘ฅ^3+๐‘ฅ^2โˆ’2๐‘ฅ) fโ€™ (๐‘ฅ)=12๐‘ฅ (๐‘ฅ^2+2๐‘ฅโˆ’๐‘ฅโˆ’2) fโ€™ (๐‘ฅ)=12๐‘ฅ (๐‘ฅ(๐‘ฅ+2)โˆ’1(๐‘ฅ+2)) fโ€™ (๐‘ฅ)=12๐‘ฅ (๐‘ฅโˆ’1)(๐‘ฅ+2) f (๐‘ฅ)=3๐‘ฅ4 + 4๐‘ฅ3 โ€“ 12๐‘ฅ2 + 12 Finding fโ€™ (๐’™) fโ€™ (๐‘ฅ)=๐‘‘(3๐‘ฅ4 + 4๐‘ฅ3 โ€“ 12๐‘ฅ2 + 12)/๐‘‘๐‘ฅ fโ€™ (๐‘ฅ)=12๐‘ฅ^3+12๐‘ฅ^2 โ€“ 24๐‘ฅ "+ 0" fโ€™ (๐‘ฅ)=12(๐‘ฅ^3+๐‘ฅ^2โˆ’2๐‘ฅ) fโ€™ (๐‘ฅ)=12๐‘ฅ (๐‘ฅ^2+2๐‘ฅโˆ’๐‘ฅโˆ’2) fโ€™ (๐‘ฅ)=12๐‘ฅ (๐‘ฅ(๐‘ฅ+2)โˆ’1(๐‘ฅ+2)) fโ€™ (๐‘ฅ)=12๐‘ฅ (๐‘ฅโˆ’1)(๐‘ฅ+2) At ๐’™=๐ŸŽ f โ€™โ€™(0)=12(3(0)^2+2(0)โˆ’2)= 32 (0+0 โˆ’2)= โ€“ 64 < 0 Since fโ€™โ€™(๐‘ฅ)<0 at ๐‘ฅ=0 โˆด ๐‘ฅ is point of local maxima Thus, f(๐‘ฅ) is maximum at ๐‘ฅ=0 At ๐’™=๐Ÿ fโ€™(1)=12(3(1)^2+2(1)โˆ’2)= 12 (3+2โˆ’2) = 36 > 0 Since fโ€™โ€™(๐‘ฅ)>0 at ๐‘ฅ=1 โˆด ๐‘ฅ is point of local minima Thus, f(๐‘ฅ) is minimum at ๐‘ฅ=1 At ๐’™=โˆ’๐Ÿ fโ€™(โˆ’2)=12(3(โˆ’2)^2+2(โˆ’2)โˆ’2)= 12 (3(4)โˆ’4โˆ’2)= 72 > 0 Since fโ€™โ€™(๐‘ฅ)>0 at ๐‘ฅ=โˆ’2 โˆด ๐‘ฅ is point of local minima Thus, f(๐‘ฅ) is minimum at ๐‘ฅ=โˆ’2 Finding local minimum and maximum value Local maximum value of f (๐‘ฅ) at ๐‘ฅ=0 f (0)=3(0)4 + 4(0)3 โ€“ 12(0)2 + 12 = 0 + 0 โ€“ 0 + 12 = 12 Local minimum value of f (๐‘ฅ) at ๐‘ฅ=1 f (1)=3(1)4 + 4(1)3 โ€“ 12(1)2 + 12 = 3 + 4 โ€“ 12 + 12 = 7 Local Minimum value of f (๐‘ฅ) at ๐‘ฅ=โˆ’2 f (โˆ’2)=3(โˆ’2)4 + 4(โˆ’2)3 โ€“ 12(โˆ’2)2 + 12 = 48 โ€“ 32 โ€“ 48 + 12 = โ€“ 20

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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.