Find-x-such-that-f-x-is-minimum-f-x-c-2-x-2-c-x-c-x-2-

Question Number 164462 by ajfour last updated on 17/Jan/22

$${Find}\:{x},\:{such}\:{that}\:{f}\left({x}\right)\:{is}\:{minimum}. \\ $$$${f}\left({x}\right)=\left\{\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)\right\}^{\mathrm{2}} \\ $$

Commented by ajfour last updated on 17/Jan/22

$${did}. \\ $$

Commented by MJS_new last updated on 17/Jan/22

$$\mathrm{no}\:{x}\:\mathrm{on}\:\mathrm{the}\:\mathrm{rhs}…\:\mathrm{please}\:\mathrm{correct} \\ $$

Answered by MJS_new last updated on 17/Jan/22

(√(f(x)))=0 has one solution for c∈R ⇒ min(f(x))=0 at f(x)=0 ⇒ ((√(c^2 −x^2 ))/(c−x))−(c−x)=0 ⇔ x^3 −3cx^2 +(3c^2 +1)x−c(c^2 −1)=0 ⇒ x=c+((−c+(√((27c^2 +1)/(27)))))^(1/3) −((c+(√((27c^2 +1)/(27)))))^(1/3)

$$\sqrt{{f}\left({x}\right)}=\mathrm{0}\:\mathrm{has}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{for}\:{c}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$$\mathrm{min}\left({f}\left({x}\right)\right)=\mathrm{0}\:\mathrm{at}\:{f}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{cx}^{\mathrm{2}} +\left(\mathrm{3}{c}^{\mathrm{2}} +\mathrm{1}\right){x}−{c}\left({c}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}={c}+\sqrt[{\mathrm{3}}]{−{c}+\sqrt{\frac{\mathrm{27}{c}^{\mathrm{2}} +\mathrm{1}}{\mathrm{27}}}}−\sqrt[{\mathrm{3}}]{{c}+\sqrt{\frac{\mathrm{27}{c}^{\mathrm{2}} +\mathrm{1}}{\mathrm{27}}}} \\ $$

Answered by mr W last updated on 17/Jan/22

f(x)={((√(c^2 −x^2 ))/(c−x))−(c−x)}^2 ≥0 f(x)_(min) =0 when ((√(c^2 −x^2 ))/(c−x))−(c−x)=0 (√(c^2 −x^2 ))=(c−x)^2 c+x=(c−x)^3 (c−x)^3 +(c−x)−2c=0 let t=c−x t^3 +t−2c=0 t=(((√(c^2 +(1/(27))))+c))^(1/3) −(((√(c^2 +(1/(27))))−c))^(1/3) ⇒x=c−(((√(c^2 +(1/(27))))+c))^(1/3) +(((√(c^2 +(1/(27))))−c))^(1/3)

$${f}\left({x}\right)=\left\{\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)\right\}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${f}\left({x}\right)_{{min}} =\mathrm{0}\:{when}\: \\ $$$$\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)=\mathrm{0} \\ $$$$\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\left({c}−{x}\right)^{\mathrm{2}} \\ $$$${c}+{x}=\left({c}−{x}\right)^{\mathrm{3}} \\ $$$$\left({c}−{x}\right)^{\mathrm{3}} +\left({c}−{x}\right)−\mathrm{2}{c}=\mathrm{0} \\ $$$${let}\:{t}={c}−{x} \\ $$$${t}^{\mathrm{3}} +{t}−\mathrm{2}{c}=\mathrm{0} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}+{c}}−\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}−{c}} \\ $$$$\Rightarrow{x}={c}−\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}+{c}}+\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}−{c}} \\ $$

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