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
$$\sqrt{{f}\left({x}\right)}=\mathrm{0}\:\mathrm{has}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{for}\:{c}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$$\mathrm{min}\left({f}\left({x}\right)\right)=\mathrm{0}\:\mathrm{at}\:{f}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{cx}^{\mathrm{2}} +\left(\mathrm{3}{c}^{\mathrm{2}} +\mathrm{1}\right){x}−{c}\left({c}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}={c}+\sqrt[{\mathrm{3}}]{−{c}+\sqrt{\frac{\mathrm{27}{c}^{\mathrm{2}} +\mathrm{1}}{\mathrm{27}}}}−\sqrt[{\mathrm{3}}]{{c}+\sqrt{\frac{\mathrm{27}{c}^{\mathrm{2}} +\mathrm{1}}{\mathrm{27}}}} \\ $$
Answered by mr W last updated on 17/Jan/22
$${f}\left({x}\right)=\left\{\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)\right\}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${f}\left({x}\right)_{{min}} =\mathrm{0}\:{when}\: \\ $$$$\frac{\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{c}−{x}}−\left({c}−{x}\right)=\mathrm{0} \\ $$$$\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\left({c}−{x}\right)^{\mathrm{2}} \\ $$$${c}+{x}=\left({c}−{x}\right)^{\mathrm{3}} \\ $$$$\left({c}−{x}\right)^{\mathrm{3}} +\left({c}−{x}\right)−\mathrm{2}{c}=\mathrm{0} \\ $$$${let}\:{t}={c}−{x} \\ $$$${t}^{\mathrm{3}} +{t}−\mathrm{2}{c}=\mathrm{0} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}+{c}}−\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}−{c}} \\ $$$$\Rightarrow{x}={c}−\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}+{c}}+\sqrt[{\mathrm{3}}]{\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}−{c}} \\ $$