Last updated at May 6, 2021 by Teachoo

Transcript

Example 32 Find local maximum and local minimum values of the function f given by f (๐ฅ)=3๐ฅ4 + 4๐ฅ3 โ 12๐ฅ2 + 12f (๐ฅ)=3๐ฅ4 + 4๐ฅ3 โ 12๐ฅ2 + 12 Finding fโ (๐) fโ (๐ฅ)=๐(3๐ฅ4 + 4๐ฅ3 โ 12๐ฅ2 + 12)/๐๐ฅ fโ (๐ฅ)=12๐ฅ^3+12๐ฅ^2 โ 24๐ฅ "+ 0" fโ (๐ฅ)=12(๐ฅ^3+๐ฅ^2โ2๐ฅ) fโ (๐ฅ)=12๐ฅ(๐ฅ^2+๐ฅโ2) fโ (๐ฅ)=12๐ฅ (๐ฅ^2+2๐ฅโ๐ฅโ2) fโ (๐ฅ)=12๐ฅ (๐ฅ(๐ฅ+2)โ1(๐ฅ+2)) fโ (๐ฅ)=๐๐๐ (๐โ๐)(๐+๐) Putting fโ (๐)=๐ 12๐ฅ (๐ฅโ1)(๐ฅ+2)=0 ๐ฅ (๐ฅโ1)(๐ฅ+2)=0 So, ๐=๐,๐ฅ=๐,& ๐ฅ=โ๐ Finding fโโ(๐) f โ(๐ฅ)=12(๐ฅ^3+๐ฅ^2โ2๐ฅ) f โโ(๐ฅ)=12๐(๐ฅ^3 + ๐ฅ^2 โ 2๐ฅ)/๐๐ฅ f โโ(๐ฅ)=๐๐(๐๐^๐+๐๐โ๐) At ๐=๐ f โโ(0)=12(3(0)^2+2(0)โ2)= 32 (0+0 โ2)= โ 64 < 0 Since fโโ(๐ฅ)<0 at ๐ฅ=0 โด ๐ฅ = 0 is point of local maxima Thus, f(๐ฅ) is maximum at ๐ฅ=0 At ๐=๐ fโโ(1)=12(3(1)^2+2(1)โ2)= 12 (3+2โ2) = 36 > 0 Since fโโ(๐ฅ)>0 at ๐ฅ=1 โด ๐ฅ = 1 is point of local minima Thus, f(๐ฅ) is minimum at ๐ฅ=1 At ๐=โ๐ fโโ(โ2)=12(3(โ2)^2+2(โ2)โ2)= 12 (12โ4โ2)= 72 > 0 Since fโโ(๐ฅ)>0 at ๐ฅ=โ2 โด ๐ฅ = โ2 is point of local minima Thus, f(๐ฅ) is minimum at ๐ฅ=โ2 Finding local minimum and maximum value fโ (๐ฅ)=๐๐๐ (๐โ๐)(๐+๐) Local maximum value of f (๐ฅ) at ๐ฅ=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 (๐ฅ) at ๐ฅ=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 (๐ฅ) at ๐ฅ=โ2 f (โ2)=3(โ2)4 + 4(โ2)3 โ 12(โ2)2 + 12 = 48 โ 32 โ 48 + 12 = โ 20

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.