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let-f-0-1-R-be-given-by-f-x-1-x-1-3-3-1-x-1-3-3-8-1-x-then-max-f-x-x-0-1-min-f-x-x-0-1-is-

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Question Number 178624 by infinityaction last updated on 19/Oct/22
let f:[0,1]→ R be given by f(x) = (((1+x^(1/3) )^3 +(1−x^(1/3) )^3 )/(8(1+x))) then max{f(x): x∈[0,1]}−min{f(x):x∈[0,1]} is
letf:[0,1]→Rbegivenbyf(x)=(1+x13)3+(1−x13)38(1+x)thenmax{f(x):x∈[0,1]}−min{f(x):x∈[0,1]}is
Answered by a.lgnaoui last updated on 20/Oct/22
posons=x^(1/3) =z ; x=z^3 (1+z)^3 +(1−z)^3 =2[(1+z)^2 +(1−z)^2 −(1−z^2 )]=2(1+3z^2 ) f(x)=((1+3z^2 )/(4(1+z^3 )))=((1+3x^(2/3) )/(4(1+x))) (df/dx)=(1/4)[((3(2/3)x^((−1)/3) (1+x)−(1+3x^(2/3) ))/((1+x)^2 ))]=((2x^((−1)/3) −x^(2/3) −1)/(4(1+x)^2 )) ((2/z) −z^2 −1)×(1/(4(1+z^3 )^2 ))=((2−z^3 −z)/(4z(1+z^3 ))) −((z^3 +z−2)/(4z(1+z^3 ))) =−(((z−1)^2 (z+2))/(4z(1+z^3 ))) ⇒[extremum(x=1 f(1)=(1/2)=max(f(x)) ; f(0)=(1/4)=min(f(x))] max(f(x))−min(f(x))=(1/4)
posons=x13=z;x=z3(1+z)3+(1−z)3=2[(1+z)2+(1−z)2−(1−z2)]=2(1+3z2)f(x)=1+3z24(1+z3)=1+3x234(1+x)dfdx=14[323x−13(1+x)−(1+3x23)(1+x)2]=2x−13−x23−14(1+x)2(2z−z2−1)×14(1+z3)2=2−z3−z4z(1+z3)−z3+z−24z(1+z3)=−(z−1)2(z+2)4z(1+z3)⇒[extremum(x=1f(1)=12=max(f(x));f(0)=14=min(f(x))]max(f(x))−min(f(x))=14
Commented by infinityaction last updated on 20/Oct/22
thanks sir
thankssir

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