Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) f (đĽ) = 9đĽ2+12đĽ+2Finding fâ(x) f (đĽ)=9đĽ2+12đĽ+2 Diff. w.r.t đĽ fâ(đĽ)=18đĽ+12 fâ(đĽ)=6(3đĽ+2) Putting fâ(đ)=đ 6(3đĽ+2)=0 3đĽ+2=0 3đĽ=â2 đĽ=(â2)/( 3) Hence đĽ=(â2)/3 is point of minima of f(đĽ) Finding minimum value of f(đĽ) at đĽ=(â2)/3 f(đĽ)=9đĽ^2+12đĽ+2 Putting đĽ=(â2)/3 f(đĽ)=9((â2)/3)^2+12((â2)/3)+2=9(4/3)â12(2/3)+2=â2 Thus, Minimum value of f(đ)=âđ There is no maximum value Ex 6.3, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (đĽ) = 9đĽ2+12đĽ+2Finding fâ(đ) f (đĽ)=9đĽ2+12đĽ+2 Diff. w.r.t đĽ fâ(đĽ)=đ(9đĽ^2 + 12đĽ + 2)/đđĽ fâ(đĽ)=18đĽ+12 fâ(đĽ)=6(3đĽ+2) Putting fâ(đ)=đ 6(3đĽ+2)=0 3đĽ+2=0 3đĽ=â2 đĽ=(â2)/3 Finding fââ(đ) fâ(đĽ)= 6(3đĽ+2) Again diff w.r.t đĽ fââ(đĽ)=đ(6(3đĽ+2))/đđĽ fââ(đĽ)=6 đ(3đĽ + 2)/đđĽ fââ(đĽ)=6(3+0) fââ(đĽ)=6(3) fââ(đĽ)=18 So, fââ((â2)/3)=18 Since fââ(đĽ)>0 is for đĽ=(â2)/3 đĽ=(â2)/3 is point of local minima Finding minimum value Putting đĽ=(â2)/3 in f(đĽ) f (đĽ)=9đĽ2+12đĽ+2 f ((â2)/3)=9((â2)/3)^2+12((â2)/3)+2 =9(4/9)+12((â2)/3)+2 =4â8+2 =â2 Hence, minimum value = â2 There is no maximum value