Question Number 85146 by john santu last updated on 19/Mar/20
$$\mathrm{find}\:\mathrm{minimum}\:\&\:\mathrm{maximum}\:\mathrm{value}\: \\ $$$$\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\:,\:−\pi\leqslant\mathrm{x}\leqslant\pi \\ $$
Commented by mr W last updated on 19/Mar/20
$${f}\left({x}\right)=−\left(\mathrm{sin}^{\mathrm{2}} \:{x}−\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=−\left(\mathrm{sin}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${f}_{{max}} =−\frac{\mathrm{1}}{\mathrm{4}}\:{at}\:\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\mathrm{2}},\:{i}.{e}.\:{x}=\frac{\pi}{\mathrm{6}},\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$$${f}_{{min}} =−\frac{\mathrm{5}}{\mathrm{2}}\:{at}\:\mathrm{sin}\:{x}=−\mathrm{1},\:{i}.{e}.\:{x}=−\frac{\pi}{\mathrm{2}} \\ $$
Answered by jagoll last updated on 19/Mar/20
Answered by Rio Michael last updated on 19/Mar/20
$$\mathrm{Maximum}\:\mathrm{value}\:=\:−\mathrm{0}.\mathrm{25} \\ $$$$\mathrm{note}\:\mathrm{at}\:\mathrm{maximum}\:\mathrm{vaue}\:\:{f}^{'} \left({x}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:=\:−\mathrm{2}.\mathrm{5} \\ $$