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 (i) f (đĽ) = (2đĽ â 1)^2 + 3 Square of number cant be negative It can be 0 or greater than 0 đ(đĽ)=(2đĽâ1)^2+3 Hence, Minimum value of (2đĽâ1)^2 = 0 Minimum value of (2đĽâ1^2 )+3 = 0 + 3 = 3 Also, there is no maximum value of đĽ â´ There is no maximum value of f(x) Ex 6.3, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (đĽ) = (2đĽ â 1)^2+3Finding fâ(x) f(đĽ)=(2đĽâ1)^2+3 fâ(đĽ)= 2(2đĽâ1) Putting fâ(đ)=đ 2(2đĽâ1)=0 2đĽ â 1 = 0 2đĽ = 1 đ = đ/đ Thus, x = 1/2 is the minima Finding minimum value f(đĽ)=(2đĽâ1)^2+3 Putting đĽ = 1/2 f(1/2)=(2 Ă 1/2â1)^2+3= (1â1)^2+3= 3 â´ Minimum value = 3 There is no maximum value Ex 6.3, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) đ (đĽ)= (2đĽ â 1)^2 + 3Double Derivative Test f(đĽ)=(2đĽâ1)^2+3 Finding fâ(đ) fâ(đĽ)=2(2đĽâ1)^(2â1) = 2(2đĽâ1) Putting fâ(đ)=đ 2(2đĽâ1)=0 (2đĽâ1)=0 2đĽ = 0 + 1 đ = đ/đ Finding fââ(đ) fâ(đĽ)=2(2đĽâ1) fâ(đĽ) = 4đĽ â 2 fââ(đĽ)= 4 fââ (đ/đ) = 4 Since fââ (đ/đ) > 0 , đĽ = 1/2 is point of local minima Putting đĽ = 1/2 , we can calculate minimum value f(đĽ) = (2đĽâ1)^2+3 f(1/2)= (2(1/2)â1)^2+3= (1â1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value