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Question Number 83850 by Rio Michael last updated on 06/Mar/20

Find the maximum value of the function f, defined by f(x) = (x/(1+ x^2 )) , x∈R

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:{f},\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:{f}\left({x}\right)\:=\:\frac{{x}}{\mathrm{1}+\:{x}^{\mathrm{2}} }\:,\:{x}\in\mathbb{R} \\ $$

Commented by mathmax by abdo last updated on 06/Mar/20

lim_(x→∞) f(x)=0 and f^′ (x)=((1+x^2 −x(2x))/((1+x^2 )^2 )) =((1−x^2 )/((1+x^2 )^2 )) we have f^′ (x)≥0 ⇔ −1≤x≤1 x −∞ −1 0 1 +∞ f^′ − + + − f 0 decr −(1/2) inc0 inc(1/2) dec 0 ⇒max_(x∈R) f(x)=(1/2)

$${lim}_{{x}\rightarrow\infty} {f}\left({x}\right)=\mathrm{0}\:\:\:\:{and}\:{f}^{'} \left({x}\right)=\frac{\mathrm{1}+{x}^{\mathrm{2}} −{x}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)\geqslant\mathrm{0}\:\Leftrightarrow\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$${x}\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:− \\ $$$${f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:{decr}\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:{inc}\mathrm{0}\:{inc}\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:{dec}\:\:\:\:\:\mathrm{0} \\ $$$$\Rightarrow{max}_{{x}\in{R}} \:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by john santu last updated on 06/Mar/20

(x^2 +1)y−x=0 yx^2 −x+y = 0 ⇒ Δ = 1−4.y^2 ≥0 (2y−1)(2y+1) ≤ 0 −(1/2) ≤ y ≤ (1/2) { ((max = (1/2))),((min = −(1/2))) :}

$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{y}−\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{yx}^{\mathrm{2}} −\mathrm{x}+\mathrm{y}\:=\:\mathrm{0}\: \\ $$$$\Rightarrow\:\Delta\:=\:\mathrm{1}−\mathrm{4}.\mathrm{y}^{\mathrm{2}} \:\geqslant\mathrm{0} \\ $$$$\left(\mathrm{2y}−\mathrm{1}\right)\left(\mathrm{2y}+\mathrm{1}\right)\:\leqslant\:\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\mathrm{y}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\begin{cases}{\mathrm{max}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{min}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

Answered by mr W last updated on 07/Mar/20

f(x)=(x/(1+x^2 ))=(1/(x+(1/x))) ≤(1/2) →max. ≥(1/(−2))=−(1/2) →min.

$${f}\left({x}\right)=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}} \\ $$$$\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\rightarrow{max}. \\ $$$$\geqslant\frac{\mathrm{1}}{−\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\:\rightarrow{min}. \\ $$

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