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

Transcript

Example 33 (Method 1) Find all the points of local maxima and local minima of the function f given by ๐‘“(๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+6๐‘ฅ+5. ๐‘“(๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+6๐‘ฅ+5 Finding fโ€™(๐’™) ๐‘“โ€™(๐‘ฅ)=๐‘‘(2๐‘ฅ3 โˆ’ 6๐‘ฅ2 + 6๐‘ฅ + 5)/๐‘‘๐‘ฅ ๐‘“โ€™(๐‘ฅ)=6๐‘ฅ^2โˆ’12๐‘ฅ+6+0 ๐‘“โ€™(๐‘ฅ)=6(๐‘ฅ^2โˆ’2๐‘ฅ+1) ๐‘“โ€™(๐‘ฅ)=6((๐‘ฅ)^2+(1)^2โˆ’2(๐‘ฅ)(1)) ๐‘“โ€™(๐‘ฅ)=6(๐‘ฅโˆ’1)^2 Putting fโ€™(๐’™)=๐ŸŽ 6(๐‘ฅโˆ’1)^2=0 (๐‘ฅโˆ’1)^2=0 ๐‘ฅโˆ’1=0 So, ๐‘ฅ=1 Hence ๐‘ฅ=1 is point of inflexion Example 33 (Method 2) Find all the points of local maxima and local minima of the function f given by ๐‘“(๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+6๐‘ฅ+5. ๐‘“(๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+6๐‘ฅ+5 Finding fโ€™(๐’™) ๐‘“โ€™(๐‘ฅ)=๐‘‘(2๐‘ฅ3 โˆ’ 6๐‘ฅ2 + 6๐‘ฅ + 5" " )/๐‘‘๐‘ฅ ๐‘“โ€™(๐‘ฅ)=6๐‘ฅ^2โˆ’12๐‘ฅ+6++0 ๐‘“โ€™(๐‘ฅ)=6(๐‘ฅ^2โˆ’2๐‘ฅ+1) ๐‘“โ€™(๐‘ฅ)=6((๐‘ฅ)^2+(1)^2โˆ’2(๐‘ฅ)(1)) ๐‘“โ€™(๐‘ฅ)=6(๐‘ฅโˆ’1)^2 Putting fโ€™(๐’™)=๐ŸŽ 6(๐‘ฅโˆ’1)^2=0 (๐‘ฅโˆ’1)^2=0 ๐‘ฅโˆ’1=0 ๐‘ฅ=1 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=6(๐‘ฅโˆ’1)^2 fโ€™โ€™(๐‘ฅ)=6.(๐‘‘(๐‘ฅ โˆ’ 1)^2)/๐‘‘๐‘ฅ fโ€™โ€™(๐‘ฅ)=6 ร— 2(๐‘ฅโˆ’1) fโ€™โ€™(๐‘ฅ) = 12 (๐‘ฅโˆ’1) Putting ๐‘ฅ=1 fโ€™โ€™(๐‘ฅ)=12(1โˆ’1) = 0 โˆด Second derivative test fails โ‡’ ๐‘ฅ=1 is neither point of maxima nor point of minima Thus, ๐’™=๐Ÿ is point of inflexion

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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.