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Question Number 183669 by mnjuly1970 last updated on 28/Dec/22
f(x)= (x/(1 + x + x^( 2) )) min_( f) = ?
$$ \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\:\frac{{x}}{\mathrm{1}\:+\:{x}\:+\:{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:{min}_{\:{f}} \:=\:? \\ $$
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
$${f}\left({x}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}} \\ $$$$\frac{\mathrm{1}}{{x}}+{x}\:\:\gg\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{x}}+{x}+\mathrm{1}\gg\mathrm{3} \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}}+{x}+\mathrm{1}}\ll\frac{\mathrm{1}}{\mathrm{3}}\:\:\:{is}\:{max}\:\:\: \\ $$$${l}_{{x}\rightarrow\infty} {imf}\left({x}\right)=\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\infty}+\mathrm{1}+\infty}=\frac{\mathrm{1}}{\mathrm{0}+\mathrm{1}+\infty}=\frac{\mathrm{1}}{\infty}=\mathrm{0}\: \\ $$$${correction} \\ $$$${x}\langle\mathrm{0} \\ $$$${x}+\frac{\mathrm{1}}{{x}}\ll−\mathrm{2} \\ $$$${x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\ll−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}\gg−\mathrm{1} \\ $$$$−\mathrm{1}\ll{f}\left({x}\right)\ll\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${min}=−\mathrm{1} \\ $$
Commented by manolex last updated on 28/Dec/22
thanks,Sir W, for correction
$${thanks},{Sir}\:{W},\:{for}\:{correction} \\ $$
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
$${if}\:{x}=\mathrm{0},\:{f}\left({x}\right)=\mathrm{0} \\ $$$${for}\:{x}\neq\mathrm{0}: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}} \\ $$$${for}\:{x}>\mathrm{0}: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}\leqslant\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${for}\:{x}<\mathrm{0}: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}=\frac{\mathrm{1}}{−\left(−{x}+\frac{\mathrm{1}}{−{x}}\right)+\mathrm{1}}\geqslant\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{1}}=−\mathrm{1} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${i}.{e}.\:{max}.\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:{and}\:{min}.\:{f}\left({x}\right)=−\mathrm{1} \\ $$
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
$${f}'\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${f}''\left({x}\right)=−\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{3}{x}−{x}^{\mathrm{3}} \right)}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)} \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1}} =−\mathrm{1}\wedge{x}_{\mathrm{2}} =\mathrm{1} \\ $$$${f}''\left(−\mathrm{1}\right)=\mathrm{2}>\mathrm{0}\:\Rightarrow\:\mathrm{minimum}\:\mathrm{is}\:−\mathrm{1}\:\mathrm{at}\:{x}=−\mathrm{1} \\ $$$${f}''\left(\mathrm{1}\right)=−\frac{\mathrm{2}}{\mathrm{9}}\:\Rightarrow\:\mathrm{maximum}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$
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
$$ \\ $$$${a}\:\:\ll\:\frac{{x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\ll{b} \\ $$$${a}\ll\frac{{x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} } \\ $$$$\frac{{a}+{ax}+{ax}^{\mathrm{2}} −{x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\:\ll\mathrm{0} \\ $$$${ax}^{\mathrm{2}} +\left({a}−\mathrm{1}\right){x}+{a}\ll\mathrm{0} \\ $$$$\bigtriangleup=\left({a}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} \gg\mathrm{0} \\ $$$$\left({a}−\mathrm{1}−\mathrm{2}{a}\right)\left({a}−\mathrm{1}+\mathrm{2}{a}\right)\gg\mathrm{0} \\ $$$$\left(−{a}−\mathrm{1}\right)\left(\mathrm{3}{a}−\mathrm{1}\right)\gg\mathrm{0} \\ $$$$\left({a}+\mathrm{1}\right)\left(\mathrm{3}{a}−\mathrm{1}\right)\ll\mathrm{0} \\ $$$$−\mathrm{1}\ll{a}\ll\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${min}\:{f}\left({x}\right)=−\mathrm{1} \\ $$$$ \\ $$
Commented by mr W last updated on 29/Dec/22
nice approach!
$${nice}\:{approach}! \\ $$
Commented by manxsol last updated on 29/Dec/22
following his vision, thanks
$${following}\:{his}\:{vision},\:{thanks} \\ $$

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