Question Number 194837 by mr W last updated on 16/Jul/23
$${for}\:{x}>\mathrm{0}\:{find}\:{the}\:{minimum}\:{of}\:{the} \\ $$$${function}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{5}}{{x}}. \\ $$
Answered by Frix last updated on 16/Jul/23
$${f}'\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}=\pm\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${f}\left(\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \right)=\mathrm{4}×\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$
Answered by mr W last updated on 16/Jul/23
$${an}\:{other}\:{way}: \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{5}}{\mathrm{3}{x}}+\frac{\mathrm{5}}{\mathrm{3}{x}}+\frac{\mathrm{5}}{\mathrm{3}{x}} \\ $$$$\:\:\:\:\:\:\:\:\geqslant\mathrm{4}\left({x}^{\mathrm{3}} ×\frac{\mathrm{5}}{\mathrm{3}{x}}×\frac{\mathrm{5}}{\mathrm{3}{x}}×\frac{\mathrm{5}}{\mathrm{3}{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\mathrm{4}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\Rightarrow{minimum}=\mathrm{4}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${when}\:{x}^{\mathrm{3}} =\frac{\mathrm{5}}{\mathrm{3}{x}},\:{i}.{e}.\:{x}=\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$