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Question Number 183669 by mnjuly1970 last updated on 28/Dec/22
f(x)= (x/(1 + x + x^( 2) )) min_( f) = ?
f(x)=x1+x+x2minf=?
Answered by manolex last updated on 28/Dec/22
f(x)=(1/((1/x)+1+x)) (1/x)+x ≫2 (1/x)+x+1≫3 (1/((1/x)+x+1))≪(1/3) is max l_(x→∞) imf(x)= (1/((1/∞)+1+∞))=(1/(0+1+∞))=(1/∞)=0 correction x⟨0 x+(1/x)≪−2 x+(1/x)+1≪−1 (1/(x+(1/x)+1))≫−1 −1≪f(x)≪(1/3) min=−1
f(x)=11x+1+x1x+x≫21x+x+1≫311x+x+1≪13ismaxlx→∞imf(x)=11∞+1+∞=10+1+∞=1∞=0correctionx⟨0x+1x≪−2x+1x+1≪−11x+1x+1≫−1−1≪f(x)≪13min=−1
Commented by manolex last updated on 28/Dec/22
thanks,Sir W, for correction
thanks,SirW,forcorrection
Answered by mr W last updated on 28/Dec/22
if x=0, f(x)=0 for x≠0: f(x)=(1/(x+(1/x)+1)) for x>0: f(x)=(1/(x+(1/x)+1))≤(1/(2+1))=(1/3) for x<0: f(x)=(1/(x+(1/x)+1))=(1/(−(−x+(1/(−x)))+1))≥(1/(−2+1))=−1 ⇒−1≤f(x)≤(1/3) i.e. max. f(x)=(1/3) and min. f(x)=−1
ifx=0,f(x)=0forx≠0:f(x)=1x+1x+1forx>0:f(x)=1x+1x+1⩽12+1=13forx<0:f(x)=1x+1x+1=1−(−x+1−x)+1⩾1−2+1=−1⇒−1⩽f(x)⩽13i.e.max.f(x)=13andmin.f(x)=−1
Answered by Frix last updated on 28/Dec/22
f′(x)=((1−x^2 )/((1+x+x^2 )^2 )) f′′(x)=−((2(1+3x−x^3 ))/((1+x+x^2 ))) f′(x)=0 ⇒ x_1 =−1∧x_2 =1 f′′(−1)=2>0 ⇒ minimum is −1 at x=−1 f′′(1)=−(2/9) ⇒ maximum is (1/3) at x=1
f′(x)=1−x2(1+x+x2)2f″(x)=−2(1+3x−x3)(1+x+x2)f′(x)=0⇒x1=−1∧x2=1f″(−1)=2>0⇒minimumis−1atx=−1f″(1)=−29⇒maximumis13atx=1
Answered by manxsol last updated on 28/Dec/22
a ≪ (x/(1+x+x^2 ))≪b a≪(x/(1+x+x^2 )) ((a+ax+ax^2 −x)/(1+x+x^2 )) ≪0 ax^2 +(a−1)x+a≪0 △=(a−1)^2 −4a^2 ≫0 (a−1−2a)(a−1+2a)≫0 (−a−1)(3a−1)≫0 (a+1)(3a−1)≪0 −1≪a≪(1/3) min f(x)=−1
a≪x1+x+x2≪ba≪x1+x+x2a+ax+ax2−x1+x+x2≪0ax2+(a−1)x+a≪0△=(a−1)2−4a2≫0(a−1−2a)(a−1+2a)≫0(−a−1)(3a−1)≫0(a+1)(3a−1)≪0−1≪a≪13minf(x)=−1
Commented by mr W last updated on 29/Dec/22
nice approach!
niceapproach!
Commented by manxsol last updated on 29/Dec/22
following his vision, thanks
followinghisvision,thanks

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