Question Number 163080 by tounghoungko last updated on 03/Jan/22
$$\:\:\:\:\:\:\:{F}\left({x}\right)=\:\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\left.\begin{matrix}{{local}\:{maximum}}\\{{absolut}\:{maximum}}\end{matrix}\right\}\:=? \\ $$
Answered by mahdipoor last updated on 03/Jan/22
$${F}\:'=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\frac{−\mathrm{1}}{\mathrm{3}}} \left(\mathrm{2}{x}−\mathrm{4}\right)=\mathrm{0}\:{or}\:\nexists \\ $$$$\Rightarrow{x}=\mathrm{2}\:,\:\mathrm{4}\:,\:\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{F}\:\left(\pm\infty\right)=\infty\:}\\{{F}\:\left(\mathrm{2}\right)=\left(\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \:\:\:{LM}\:\left({Local}\:{Max}\right)}\\{{F}\:\left(\mathrm{4}\right)=\mathrm{0}\:\:\:\:\:\:\:\:\:\:{LM}}\\{{F}\:\left(\mathrm{0}\right)=\mathrm{0}\:\:\:\:\:\:\:\:\:\:{LM}}\end{cases} \\ $$$$ \\ $$