




Ex 6.3
Last updated at Dec. 16, 2024 by Teachoo
Transcript
Ex 6.3,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) đ (đĽ) = â(đĽ â 1)^2+10 f(đĽ)=â(đĽâ1)^2+10 Finding fâ (x) Diff w.r.t đĽ fâ(đĽ)=đ(â(đĽâ1)^2+10)/đđĽ fâ(đĽ) = â2(đĽâ1)(đ(đĽâ1)/đđĽ)+0 fâ(đĽ) = â2(đĽâ1)(1â0) + 0 fâ(đĽ)=â2(đĽâ1) Putting fâ(đ)=đ â2(đĽâ1)=0 (đĽâ1)=0 đĽ=1 Hence, đĽ=1 is point of Maxima & No point of Minima Thus, f(đĽ) has maximum value at đĽ=1 Putting x=1 in f(x) f(đĽ)=â(đĽâ1)^2+10 f(1)=â(1â1)^2+10 = 0 + 10 = 10 Maximum value of f(đĽ) is 10 There is no minimum value of f(đ) Ex 6.3,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) đ (đĽ) = â (đĽ â 1)^2+10 f(đĽ)=â(đĽâ1)^2+10 Finding fâ(x) Diff w.r.t đĽ fâ(đĽ)=đ(â(đĽâ1)^2+10)/đđĽ fâ(đĽ) = â2(đĽâ1)(đ(đĽâ1)/đđĽ)+0 fâ(đĽ) = â2(đĽâ1)(1â0) + 0 fâ(đĽ)=â2(đĽâ1) Putting fâ(đ)=đ â2(đĽâ1)=0 (đĽâ1)=0 đĽ=1 Finding fââ(đ) fâ(đĽ)=â2(đĽâ1) Again diff w.r.t đĽ fââ(đĽ)=đ(â2(đĽ â 1))/đđĽ =â2 đ(đĽ â 1)/đđĽ =â2(1â0) =â2 Since fââ(đĽ) < 0 for đĽ=1 Hence, f(đĽ) has Maximum value at đĽ=1 Finding maximum value of f(đ) f(đĽ)=â(đĽâ1)^2+10 Putting đĽ=1 f(đĽ) =â(1â1)^2+10 = 0 + 10 = 10 Maximum value of f(đ) is 10 There is no minimum value of f(đ)