Question Number 43471 by Cheyboy last updated on 11/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18
$${one}\:{is}\:{x}\:{another}\:{is}\:\mathrm{2}−{x} \\ $$$${y}={x}^{\mathrm{3}} +\left(\mathrm{2}−{x}\right)^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}−{x}\right)\left(−\mathrm{1}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}+\mathrm{2}{x} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{6}{x}+\mathrm{2} \\ $$$${for}\:{max}/{min}\:\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{48}}}{\mathrm{2}×\mathrm{3}}=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{13}}}{\mathrm{6}} \\ $$$$ \\ $$$${x}=\frac{\mathrm{2}\sqrt{\mathrm{13}}\:−\mathrm{2}}{\mathrm{6}}=\frac{\sqrt{\mathrm{13}}\:−\mathrm{1}}{\mathrm{3}}\:{and}\frac{−\sqrt{\mathrm{13}}\:−\mathrm{1}}{\mathrm{3}}\leftarrow−{ve}\:{neglected} \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\frac{\sqrt{\mathrm{13}}\:−\mathrm{1}}{\mathrm{3}}} \:=\mathrm{6}\left(\frac{\sqrt{\mathrm{13}}\:−\mathrm{1}}{\mathrm{3}}\right)+\mathrm{2}=\mathrm{2}\sqrt{\mathrm{13}}\:>\mathrm{0} \\ $$$${so}\:{one}\:{number}\:{is}\:\frac{\sqrt{\mathrm{13}}\:−\mathrm{1}}{\mathrm{3}} \\ $$$${snother}\:{is}\:\left(\mathrm{2}−\frac{\sqrt{\mathrm{13}}\:−\mathrm{1}}{\mathrm{3}}\right)=\left(\frac{\mathrm{7}−\sqrt{\mathrm{13}}\:}{\mathrm{3}}\right) \\ $$
Commented by Cheyboy last updated on 11/Sep/18
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{Godbless}\:\mathrm{you} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18
$${you}\:{are}\:{welcome}...{God}\:{bless}\:{you}\:{and}\:{all}... \\ $$