Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Example 20 Find local maximum and local minimum values of the function f given by f (đĽ)=3đĽ4 + 4đĽ3 â 12đĽ2 + 12f (đĽ)=3đĽ4 + 4đĽ3 â 12đĽ2 + 12 Finding fâ (đ) fâ (đĽ)=đ(3đĽ4 + 4đĽ3 â 12đĽ2 + 12)/đđĽ fâ (đĽ)=12đĽ^3+12đĽ^2 â 24đĽ "+ 0" fâ (đĽ)=12(đĽ^3+đĽ^2â2đĽ) fâ (đĽ)=12đĽ(đĽ^2+đĽâ2) fâ (đĽ)=12đĽ (đĽ^2+2đĽâđĽâ2) fâ (đĽ)=12đĽ (đĽ(đĽ+2)â1(đĽ+2)) fâ (đĽ)=đđđ (đâđ)(đ+đ) Putting fâ (đ)=đ 12đĽ (đĽâ1)(đĽ+2)=0 đĽ (đĽâ1)(đĽ+2)=0 So, đ=đ,đĽ=đ,& đĽ=âđ Finding fââ(đ) f â(đĽ)=12(đĽ^3+đĽ^2â2đĽ) f ââ(đĽ)=12đ(đĽ^3 + đĽ^2 â 2đĽ)/đđĽ f ââ(đĽ)=đđ(đđ^đ+đđâđ) At đ=đ f ââ(0)=12(3(0)^2+2(0)â2)= 32 (0+0 â2)= â 64 < 0 Since fââ(đĽ)<0 at đĽ=0 â´ đĽ = 0 is point of local maxima Thus, f(đĽ) is maximum at đĽ=0 At đ=đ fââ(1)=12(3(1)^2+2(1)â2)= 12 (3+2â2) = 36 > 0 Since fââ(đĽ)>0 at đĽ=1 â´ đĽ = 1 is point of local minima Thus, f(đĽ) is minimum at đĽ=1 At đ=âđ fââ(â2)=12(3(â2)^2+2(â2)â2)= 12 (12â4â2)= 72 > 0 Since fââ(đĽ)>0 at đĽ=â2 â´ đĽ = â2 is point of local minima Thus, f(đĽ) is minimum at đĽ=â2 Finding local minimum and maximum value fâ (đĽ)=đđđ (đâđ)(đ+đ) Local maximum value of f (đĽ) at đĽ=0 f (0)=3(0)4 + 4(0)3 â 12(0)2 + 12 = 0 + 0 â 0 + 12 = 12 Local minimum value of f (đĽ) at đĽ=1 f (1)=3(1)4 + 4(1)3 â 12(1)2 + 12 = 3 + 4 â 12 + 12 = 7 Local Minimum value of f (đĽ) at đĽ=â2 f (â2)=3(â2)4 + 4(â2)3 â 12(â2)2 + 12 = 48 â 32 â 48 + 12 = â 20