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

Transcript

Example 29 (Method 1) Find all points of local maxima and local minima of the function f given by ๐‘“ (๐‘ฅ)=๐‘ฅ3 โ€“ 3๐‘ฅ + 3. ๐‘“(๐‘ฅ)=๐‘ฅ3 โ€“ 3๐‘ฅ+3 Finding ๐’‡^โ€ฒ (๐’™) ๐‘“โ€ฒ(๐‘ฅ)= 3๐‘ฅ^2 โ€“ 3+0 ๐‘“โ€ฒ(๐‘ฅ)= 3(๐‘ฅ^2โˆ’1) Putting ๐’‡โ€ฒ(๐’™)= 0 3(๐‘ฅ^2โˆ’1)=0 ๐‘ฅ^2โˆ’1=0 (๐‘ฅโˆ’1)(๐‘ฅ+1)=0 Thus ๐‘ฅ = 1 , โ€“1 are only critical points Finding maximum & minimum value of ๐‘“(๐‘ฅ)=๐‘ฅ3 โ€“ 3๐‘ฅ+3 Minimum value ๐‘“(1)=(1)3 โ€“3(1)+3= 1 โ€“ 3 + 3 = 1 Maximum value ๐‘“(โˆ’1)=(โˆ’1)3 โ€“3(โˆ’1)+3= โ€“1 +3 + 3 = 5 Example 29 (Method 2) Find all points of local maxima and local minima of the function f given by ๐‘“ (๐‘ฅ) = ๐‘ฅ3 โ€“ 3๐‘ฅ + 3. ๐‘“(๐‘ฅ)=๐‘ฅ3 โ€“ 3๐‘ฅ+3 Finding ๐’‡^โ€ฒ (๐’™) ๐‘“โ€ฒ(๐‘ฅ)= 3๐‘ฅ^2 โ€“ 3+0 ๐‘“โ€ฒ(๐‘ฅ)= 3(๐‘ฅ^2โˆ’1) Putting ๐’‡โ€ฒ(๐’™)= 0 3(๐‘ฅ^2โˆ’1)=0 ๐‘ฅ^2โˆ’1=0 (๐‘ฅโˆ’1)(๐‘ฅ+1)=0 So, x = 1 & x = โˆ’1 Finding ๐’‡โ€ฒโ€ฒ(๐’™) ๐‘“^โ€ฒ (๐‘ฅ)=3(๐‘ฅ^2โˆ’1) ๐‘“^โ€ฒโ€ฒ (๐‘ฅ)=3 ๐‘‘(๐‘ฅ^2 โˆ’ 1)/๐‘‘๐‘ฅ ๐‘“^โ€ฒโ€ฒ (๐‘ฅ)=3(2๐‘ฅโˆ’0) ๐‘“^โ€ฒโ€ฒ (๐‘ฅ)=6๐‘ฅ Putting ๐’™=โˆ’๐Ÿ ๐‘“^โ€ฒโ€ฒ (๐‘ฅ) = 6(โˆ’1) = โˆ’6<0 ๐‘ฅ = โ€“1 is point of local maxima Putting ๐’™=๐Ÿ ๐‘“^โ€ฒโ€ฒ (๐‘ฅ) = 6(1) = 6>0 ๐‘ฅ = 1 is point of local minima Finding maximum & minimum value of ๐‘“(๐‘ฅ)=๐‘ฅ3 โ€“ 3๐‘ฅ+3 Minimum value at x = 1 ๐‘“(1)=(1)3 โ€“3(1)+3= 1 โ€“ 3 + 3 = 1 Maximum value at x = โˆ’1 ๐‘“(โˆ’1)=(โˆ’1)3 โ€“3(โˆ’1)+3= โ€“1 + 3 + 3 = 5

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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.