Question Number 123396 by bemath last updated on 25/Nov/20
$$\:{Find}\:{the}\:{greatest}\:{and}\:{the}\:{least}\:{values} \\ $$$${of}\:{the}\:{function}\:{on}\:{indicates} \\ $$$${interval}\:\left(−\infty,\infty\right)\: \\ $$$$\left({i}\right)\:{y}\:=\:\mathrm{sin}\:{x}\:\mathrm{sin}\:\mathrm{2}{x}\:. \\ $$
Answered by TANMAY PANACEA last updated on 25/Nov/20
$${y}={sinx}×\mathrm{2}{sinxcosx} \\ $$$${y}=\mathrm{2}{cosx}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right) \\ $$$${y}=\mathrm{2}{c}−\mathrm{2}{c}^{\mathrm{3}} \\ $$$$\frac{{dy}}{{dc}}=\mathrm{2}−\mathrm{6}{c}^{\mathrm{2}} \\ $$$${for}\:{max}/{min}\:\frac{{dy}}{{dc}}=\mathrm{0} \\ $$$$\mathrm{6}{c}^{\mathrm{2}} =\mathrm{2}\rightarrow{c}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dc}^{\mathrm{2}} }=−\mathrm{12}{c} \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dc}^{\mathrm{2}} }\right)_{{c}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:=\frac{−\mathrm{12}}{\:\sqrt{\mathrm{3}}}<\mathrm{0}\:\:{maximum} \\ $$$${max}\:{value}\:{y}_{{max}} =\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{2}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{3}} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${min}\:{at}\:{c}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${y}_{{min}} =\mathrm{2}×\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{−\mathrm{6}+\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{−\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$