Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (v) đ (đĽ)=đĽ3 â6đĽ2+9đĽ+15Putting fâ(đ)=đ 3(đĽ^2â4đĽ+3)=0 đĽ^2â4đĽ+3=0 đĽ^2â3đĽâđĽ+3=0 đĽ(đĽâ3)â1(đĽâ3)=0 (đĽâ1)(đĽâ3)=0 So, x = 1 & x = 3 Finding fââ(đ) fâ(đĽ)=3(đĽ^2â4đĽ+3) fââ(đĽ)=đ(3(đĽ^2 â 4đĽ+3))/đđĽ = 3(2đĽâ4+0) = 6đĽâ12 Putting đ=đ in fââ(đ) fââ(1)=6(1)â12 = 6 â 12 = â 6 < 0 Since fââ(đĽ)<0 when đĽ=1 â đĽ=1 is point of local maxima â´ f(đĽ) is maximum at đ=đ Maximum value of f(đĽ) at đĽ = 1 f(đĽ)=đĽ^3â6đĽ^2+9đĽ+15 f(1)=(1)^3â6(1)^2+9(1)+15 = 1 â 6 + 9 + 15 = 19 Putting đ=đ in fââ(x) fââ(đĽ)=6đĽâ12 fââ(3)=6(3)â12 = 18 â 12 = 6 > 0 Since fââ(đĽ)>0 when đĽ=3 â đĽ=3 is point of local minima â´ f(đĽ) is minimum at đ=đ Minimum value of f(đĽ) at đĽ = 3 f(đĽ)=đĽ^3â6đĽ^2+9đĽ+15 f(3)=(3)^3â6(3)^2+9(3)+15 = 27 â 54 + 27 + 15 = 15