Question Number 78890 by gopikrishnan last updated on 21/Jan/20
$${the}\:{local}\:{maximum}\:{value}\:{of}\:{f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{32}{x} \\ $$
Commented by mr W last updated on 21/Jan/20
$${there}\:{is}\:{no}\:{local}\:{maximum},\:{only} \\ $$$${local}\:\left(={global}\right)\:{minimum}\:{at}\:{x}=−\mathrm{2}. \\ $$
Commented by mathmax by abdo last updated on 22/Jan/20
$${f}\left({x}\right)={x}^{\mathrm{4}} \:+\mathrm{32}\:\Rightarrow{f}^{'} \left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} \:+\mathrm{32}\:=\mathrm{4}\left({x}^{\mathrm{3}} \:+\mathrm{8}\right)\:=\mathrm{4}\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\right) \\ $$$${f}^{'} \left({x}\right)\geqslant\mathrm{0}\:\Leftrightarrow{x}\geqslant−\mathrm{2} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:\:\:{decr}\:\:\:{f}\left(−\mathrm{2}\right)\:{incr}\:\:+\infty \\ $$$${no}\:{max}\:{forf}\:\:\:\:{and}\:\:{minf}\left({x}\right)={f}\left(−\mathrm{2}\right)\:=\left(−\mathrm{2}\right)^{\mathrm{4}} −\mathrm{64}\:=\mathrm{16}−\mathrm{64}=−\mathrm{48} \\ $$$$ \\ $$