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Example 29 - Chapter 6 Class 12 Application of Derivatives
Last updated at Jan. 7, 2020 by Teachoo
Transcript
Example 29 (Method 1) Find all points of local maxima and local minima of the function f given by ๐ (๐ฅ)=๐ฅ3 โ 3๐ฅ + 3. ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Finding ๐^โฒ (๐) ๐โฒ(๐ฅ)= 3๐ฅ^2 โ 3+0 ๐โฒ(๐ฅ)= 3(๐ฅ^2โ1) Putting ๐โฒ(๐)= 0 3(๐ฅ^2โ1)=0 ๐ฅ^2โ1=0 (๐ฅโ1)(๐ฅ+1)=0 Thus ๐ฅ = 1 , โ1 are only critical points Finding maximum & minimum value of ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Minimum value ๐(1)=(1)3 โ3(1)+3= 1 โ 3 + 3 = 1 Maximum value ๐(โ1)=(โ1)3 โ3(โ1)+3= โ1 +3 + 3 = 5 Example 29 (Method 2) Find all points of local maxima and local minima of the function f given by ๐ (๐ฅ) = ๐ฅ3 โ 3๐ฅ + 3. ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Finding ๐^โฒ (๐) ๐โฒ(๐ฅ)= 3๐ฅ^2 โ 3+0 ๐โฒ(๐ฅ)= 3(๐ฅ^2โ1) Putting ๐โฒ(๐)= 0 3(๐ฅ^2โ1)=0 ๐ฅ^2โ1=0 (๐ฅโ1)(๐ฅ+1)=0 So, x = 1 & x = โ1 Finding ๐โฒโฒ(๐) ๐^โฒ (๐ฅ)=3(๐ฅ^2โ1) ๐^โฒโฒ (๐ฅ)=3 ๐(๐ฅ^2 โ 1)/๐๐ฅ ๐^โฒโฒ (๐ฅ)=3(2๐ฅโ0) ๐^โฒโฒ (๐ฅ)=6๐ฅ Putting ๐=โ๐ ๐^โฒโฒ (๐ฅ) = 6(โ1) = โ6<0 ๐ฅ = โ1 is point of local maxima Putting ๐=๐ ๐^โฒโฒ (๐ฅ) = 6(1) = 6>0 ๐ฅ = 1 is point of local minima Finding maximum & minimum value of ๐(๐ฅ)=๐ฅ3 โ 3๐ฅ+3 Minimum value at x = 1 ๐(1)=(1)3 โ3(1)+3= 1 โ 3 + 3 = 1 Maximum value at x = โ1 ๐(โ1)=(โ1)3 โ3(โ1)+3= โ1 + 3 + 3 = 5
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Chapter 6 Class 12 Application of Derivatives
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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.