If f(x)=1/(4x^2 + 2x + 1);x∈R, then find the maximum value of f(x).
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CBSE Class 12 Sample Paper for 2024 Boards
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CBSE Class 12 Sample Paper for 2024 Boards
Last updated at Aug. 14, 2023 by Teachoo
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
f(đĽ)=1/(4đĽ^2 + 2đĽ + 1) Finding fâ(đ) fâ(đĽ)= ((1)^â˛ " " (4đĽ^2 + 2đĽ + 1)" â " (ă4đĽ^2 + 2đĽ + 1)ă^â˛ (1))/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) fâ(đĽ)= (0 (4đĽ^2 + 2đĽ + 1)" â " (8đĽ + 2)(1))/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) fâ(đĽ)= ("â" (8đĽ + 2) )/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) Putting fâ(đ)=đ ("â" (8đĽ + 2) )/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) = 0 -(8x + 2) = 0 8x + 2 = 0 8x = -2 â(8x + 2) = 0 8x + 2 = 0 8x = â2 x = (â2)/8 x = (âđ)/đ Finding fââ(đ) fâ(đĽ)=("â" (8đĽ + 2) )/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) " " Differentiating again w.r.t x fââ(x) =â((8đĽ + 2)^â˛ (ă4đĽ^2 + 2đĽ + 1)ă^2â((ă4đĽ^2+2đĽ+1)ă^2 )^â˛ (8đĽ + 2))/(((ă4đĽ^2 + 2đĽ + 1)ă^2 )^2 ) fââ(x) =â(8(ă4đĽ^2 + 2đĽ + 1)ă^2 â 2(4đĽ^2 + 2đĽ + 1)(8đĽ + 2)(8đĽ + 2))/(4đĽ^2 + 2đĽ + 1)^4 fââ(x) =â(8(ă4đĽ^2 + 2đĽ + 1)ă^2 â 2(4đĽ^2 + 2đĽ + 1)(8đĽ + 2)(8đĽ + 2))/(4đĽ^2 + 2đĽ + 1)^4 fââ(x) =â(8(ă4đĽ^2 + 2đĽ + 1)ă^2 â 2(4đĽ^2 + 2đĽ + 1) (8đĽ + 2)^2)/(4đĽ^2 + 2đĽ + 1)^4 fââ (âđ/đ) = â(8(ă4(â1/4)^2+ 2(â1/4) + 1)ă^2 â 2(4(â1/4)^2+ 2(â1/4)+ 1) (8(â1/4)+ 2)^2)/(4(â1/4)^2+ 2(â1/4)+ 1)^4 fââ (âđ/đ) = â(8(ă4(â1/4)^2+ 2(â1/4) + 1)ă^2 â 2(4(â1/4)^2+ 2(â1/4)+ 1) (â2 + 2)^2)/(4(â1/4)^2+ 2(â1/4)+ 1)^4 fââ (âđ/đ) = â(8(3/4)^2â0)/(3/4)^4 = â8/(3/4)^2 fââ (âđ/đ) < 0 Since fââ (âđ/đ) < 0 , đĽ = âđ/đ is point of local maxima Putting đĽ = âđ/đ , we can calculate maximum value f(đĽ) =1/(4đĽ^2+2đĽ+1) f(âđ/đ)=1/(4(â1/4)^2+ 2(â1/4)+ 1) =1/(4(1/16)+ 2(â1/4)+ 1) =1/(1/4 â 2/4+ 1) = 4/(1 â2+ 4) = đ/đ