Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Example 18 (Method 1) Find all the points of local maxima and local minima of the function f given by f (đĽ)=2đĽ3 â6đĽ2+6đĽ+5.f (đĽ)=2đĽ3 â6đĽ2+ 6đĽ+5 Finding fâ (đ) f â˛(đĽ)= đ(2đĽ3 â 6đĽ2 + 6đĽ + 5)/đđĽ f â˛(đĽ)=6đĽ2 â12đĽ+6+0 f â˛(đĽ)=6(đĽ^2â2đĽ+1) Putting f â˛(đ)= 0 6(đĽ^2â2đĽ+1)=0 đĽ^2â2đĽ+1=0 (đĽ)^2+(1)^2â2(đĽ)(1)=0 (đĽâ1)^2=0 So, đ=đ is only critical point Hence đ=đ is point of inflexion Example 18 (Method 2) Find all the points of local maxima and local minima of the function f given by f (đĽ)=2đĽ3 â6đĽ2+6đĽ+5. f (đĽ)=2đĽ3 â6đĽ2+ 6đĽ+5 Finding fâ (đ) f â˛(đĽ)= đ(2đĽ3 â 6đĽ2+ 6đĽ + 5)/đđĽ f â˛(đĽ)=6đĽ2 â12đĽ+6+0 f â˛(đĽ)=6(đĽ^2â2đĽ+1) Putting f â˛(đ)= 0 6(đĽ^2â2đĽ+1)=0 đĽ^2â2đĽ+1=0 đĽ^2+1^2â2(đĽ)(1)=0 (đĽâ1)^2=0 So, đ=đ is only critical point Finding fââ(đ) fââ(đĽ)=6 đ(đĽ^2 â 2đĽ + 1)/đđĽ fââ(đĽ)=6(2đĽâ2+0) fââ(đĽ)=12(đĽâ1) Putting đ=đ fââ(1)=12(1â1) = 12 Ă 0 = 0 Since fââ(1) = 0 Hence, đĽ=1 is neither point of Maxima nor point of Minima â´ đ=đ is Point of Inflexion.

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.