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Example 30 - Chapter 6 Class 12 Application of Derivatives
Last updated at Jan. 7, 2020 by Teachoo
Transcript
Example 30 (Method 1) Find all the points of local maxima and local minima of the unction f given by f (๐ฅ)=2๐ฅ3 โ6๐ฅ2+6๐ฅ+5. f (๐ฅ)=2๐ฅ3 โ6๐ฅ2+ 6๐ฅ+5 Finding fโ (๐) f โฒ(๐ฅ)= ๐(2๐ฅ3 โ 6๐ฅ2 + 6๐ฅ + 5)/๐๐ฅ f โฒ(๐ฅ)=6๐ฅ2 โ12๐ฅ+6+0 f โฒ(๐ฅ)=6(๐ฅ^2โ2๐ฅ+1) Putting f โฒ(๐)= 0 6(๐ฅ^2โ2๐ฅ+1)=0 ๐ฅ^2โ2๐ฅ+1=0 (๐ฅ)^2+(1)^2โ2(๐ฅ)(1)=0 (๐ฅโ1)^2=0 (๐ฅโ1)(๐ฅโ1)=0 So, ๐ฅ=1 is only critical point Hence ๐=๐ 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 (๐ฅ)=2๐ฅ3 โ6๐ฅ2+6๐ฅ+5. f (๐ฅ)=2๐ฅ3 โ6๐ฅ2+ 6๐ฅ+5 Finding fโ (๐) f โฒ(๐ฅ)= ๐(2๐ฅ3 โ 6๐ฅ2+ 6๐ฅ + 5)/๐๐ฅ f โฒ(๐ฅ)=6๐ฅ2 โ12๐ฅ+6+0 f โฒ(๐ฅ)=6(๐ฅ^2โ2๐ฅ+1) Putting f โฒ(๐)= 0 6(๐ฅ^2โ2๐ฅ+1)=0 ๐ฅ^2โ2๐ฅ+1=0 ๐ฅ^2+1^2โ2(๐ฅ)(1)=0 (๐ฅโ1)^2=0 (๐ฅโ1)(๐ฅโ1)=0 So, ๐ฅ=1 is only critical point Finding fโโ(๐) fโ(๐ฅ)=6(๐ฅ^2โ2๐ฅ+1) fโโ(๐ฅ)=6๐(๐ฅ^2 โ 2๐ฅ + 1)/๐๐ฅ fโโ(๐ฅ)=6(2๐ฅโ2+0) fโโ(๐ฅ)=12(๐ฅโ1) Putting ๐ฅ=1 fโโ(1)=12(1โ1) = 12 ร 0 = 0 Thus, second derivative test fails Hence, ๐ฅ=1 is neither point of Maxima nor point of Minima โด ๐=๐ is Point of Inflexion.
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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.