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

Transcript

Example 30 (Method 1) Find all the points of local maxima and local minima of the unction f given by f (๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+6๐‘ฅ+5. f (๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+ 6๐‘ฅ+5 Finding fโ€™ (๐’™) f โ€ฒ(๐‘ฅ)= ๐‘‘(2๐‘ฅ3 โ€“ 6๐‘ฅ2 + 6๐‘ฅ + 5)/๐‘‘๐‘ฅ f โ€ฒ(๐‘ฅ)=6๐‘ฅ2 โ€“12๐‘ฅ+6+0 f โ€ฒ(๐‘ฅ)=6(๐‘ฅ^2โˆ’2๐‘ฅ+1) Putting f โ€ฒ(๐’™)= 0 6(๐‘ฅ^2โˆ’2๐‘ฅ+1)=0 ๐‘ฅ^2โˆ’2๐‘ฅ+1=0 (๐‘ฅ)^2+(1)^2โˆ’2(๐‘ฅ)(1)=0 (๐‘ฅโˆ’1)^2=0 (๐‘ฅโˆ’1)(๐‘ฅโˆ’1)=0 So, ๐‘ฅ=1 is only critical point Hence ๐’™=๐Ÿ is point of inflexion Example 30 (Method 2) Find all the points of local maxima and local minima of the function f given by f (๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+6๐‘ฅ+5. f (๐‘ฅ)=2๐‘ฅ3 โ€“6๐‘ฅ2+ 6๐‘ฅ+5 Finding fโ€™ (๐’™) f โ€ฒ(๐‘ฅ)= ๐‘‘(2๐‘ฅ3 โ€“ 6๐‘ฅ2+ 6๐‘ฅ + 5)/๐‘‘๐‘ฅ f โ€ฒ(๐‘ฅ)=6๐‘ฅ2 โ€“12๐‘ฅ+6+0 f โ€ฒ(๐‘ฅ)=6(๐‘ฅ^2โˆ’2๐‘ฅ+1) Putting f โ€ฒ(๐’™)= 0 6(๐‘ฅ^2โˆ’2๐‘ฅ+1)=0 ๐‘ฅ^2โˆ’2๐‘ฅ+1=0 ๐‘ฅ^2+1^2โˆ’2(๐‘ฅ)(1)=0 (๐‘ฅโˆ’1)^2=0 (๐‘ฅโˆ’1)(๐‘ฅโˆ’1)=0 So, ๐‘ฅ=1 is only critical point Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=6(๐‘ฅ^2โˆ’2๐‘ฅ+1) fโ€™โ€™(๐‘ฅ)=6๐‘‘(๐‘ฅ^2 โˆ’ 2๐‘ฅ + 1)/๐‘‘๐‘ฅ fโ€™โ€™(๐‘ฅ)=6(2๐‘ฅโˆ’2+0) fโ€™โ€™(๐‘ฅ)=12(๐‘ฅโˆ’1) Putting ๐‘ฅ=1 fโ€™โ€™(1)=12(1โˆ’1) = 12 ร— 0 = 0 Thus, second derivative test fails Hence, ๐‘ฅ=1 is neither point of Maxima nor point of Minima โˆด 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.