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Given-f-x-0-x-t-4-t-2-t-2-1-dt-Find-minimum-value-of-f-x-

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Question Number 134391 by bramlexs22 last updated on 03/Mar/21
Given f(x)=∫_0 ^( x) (((t^4 −t^2 )/(t^2 +1))) dt. Find minimum value of f(x).
Givenf(x)=∫0x(t4−t2t2+1)dt.Findminimumvalueoff(x).
Answered by liberty last updated on 03/Mar/21
Given f(x) = ∫_0 ^( x) [ ((t^4 −t^2 )/(t^2 +1))] dt. Find minimum value of f(x). (•) ((df(x))/dx) = ((x^4 −x^2 )/(x^2 +1)) = 0 x^2 (x−1) = 0 → { ((x=0)),((x=1)) :} (••) ((d^2 f(x))/dx^2 )∣_(x=1) = (((4x^3 −2x)(x^2 +1)−2x(x^4 −x^2 ))/((x^2 +1)^2 ))>0 for x=1 so minimum value is f(1) (•••) f(1)=∫_0 ^( 1) ((t^4 −t^2 )/(t^2 +1)) dt f(1)=∫_0 ^( 1) (t^2 −2+ (2/(t^2 +1)))dt f(1)= [(t^3 /3)−2t+2arctan t ]_0 ^1 f(1)= (1/3)−2+2((π/4))= (π/2)−(5/6)
Givenf(x)=∫0x[t4−t2t2+1]dt.Findminimumvalueoff(x).(∙)df(x)dx=x4−x2x2+1=0x2(x−1)=0→{x=0x=1(∙∙)d2f(x)dx2∣x=1=(4x3−2x)(x2+1)−2x(x4−x2)(x2+1)2>0forx=1sominimumvalueisf(1)(∙∙∙)f(1)=∫01t4−t2t2+1dtf(1)=∫01(t2−2+2t2+1)dtf(1)=[t33−2t+2arctant]01f(1)=13−2+2(π4)=π2−56

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