Last updated at April 19, 2021 by Teachoo

Transcript

Example 33 (Method 1) Find all the points of local maxima and local minima of the function f given by ๐(๐ฅ)=2๐ฅ3 โ6๐ฅ2+6๐ฅ+5. ๐(๐ฅ)=2๐ฅ3 โ6๐ฅ2+6๐ฅ+5 Finding fโ(๐) ๐โ(๐ฅ)=๐(2๐ฅ3 โ 6๐ฅ2 + 6๐ฅ + 5)/๐๐ฅ ๐โ(๐ฅ)=6๐ฅ^2โ12๐ฅ+6+0 ๐โ(๐ฅ)=6(๐ฅ^2โ2๐ฅ+1) ๐โ(๐ฅ)=6((๐ฅ)^2+(1)^2โ2(๐ฅ)(1)) ๐โ(๐ฅ)=๐(๐โ๐)^๐ Putting fโ(๐)=๐ 6(๐ฅโ1)^2=0 (๐ฅโ1)^2=0 ๐ฅโ1=0 So, ๐=๐ is only critical point Hence ๐=๐ 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(๐ฅ)=2๐ฅ3 โ6๐ฅ2+6๐ฅ+5. ๐(๐ฅ)=2๐ฅ3 โ6๐ฅ2+6๐ฅ+5 Finding fโ(๐) ๐โ(๐ฅ)=๐(2๐ฅ3 โ 6๐ฅ2 + 6๐ฅ + 5" " )/๐๐ฅ ๐โ(๐ฅ)=6๐ฅ^2โ12๐ฅ+6++0 ๐โ(๐ฅ)=6(๐ฅ^2โ2๐ฅ+1) ๐โ(๐ฅ)=6((๐ฅ)^2+(1)^2โ2(๐ฅ)(1)) ๐โ(๐ฅ)=๐(๐โ๐)^๐ Putting fโ(๐)=๐ 6(๐ฅโ1)^2=0 (๐ฅโ1)^2=0 So, ๐=๐ is the only critical point Finding fโโ(๐) fโโ(๐ฅ)=6.(๐(๐ฅ โ 1)^2)/๐๐ฅ fโโ(๐ฅ)=6 ร 2(๐ฅโ1) fโโ(๐ฅ) = 12 (๐ฅโ1) Putting ๐=๐ fโโ(๐ฅ)=12(1โ1) = 0 Since fโโ(1) = 0 Hence, ๐ฅ=1 is neither point of Maxima nor point of Minima โด ๐=๐ is Point of Inflexion.

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.