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Question Number 123396 by bemath last updated on 25/Nov/20

 Find the greatest and the least values  of the function on indicates  interval (−∞,∞)   (i) y = sin x sin 2x .

$$\:{Find}\:{the}\:{greatest}\:{and}\:{the}\:{least}\:{values} \\ $$$${of}\:{the}\:{function}\:{on}\:{indicates} \\ $$$${interval}\:\left(−\infty,\infty\right)\: \\ $$$$\left({i}\right)\:{y}\:=\:\mathrm{sin}\:{x}\:\mathrm{sin}\:\mathrm{2}{x}\:. \\ $$

Answered by TANMAY PANACEA last updated on 25/Nov/20

y=sinx×2sinxcosx  y=2cosx(1−cos^2 x)  y=2c−2c^3   (dy/dc)=2−6c^2   for max/min (dy/dc)=0  6c^2 =2→c=±(1/( (√3)))  (d^2 y/dc^2 )=−12c  ((d^2 y/dc^2 ))_(c=(1/( (√3))))  =((−12)/( (√3)))<0  maximum  max value y_(max) =2×(1/( (√3)))−2×((1/( (√3))))^3   =(2/( (√3)))−(2/(3(√3)))=(4/(3(√3)))  min at c=((−1)/( (√3)))  y_(min) =2×((−1)/( (√3)))−2×((−1)/(3(√3)))=((−6+2)/(3(√3)))=((−4)/(3(√3)))

$${y}={sinx}×\mathrm{2}{sinxcosx} \\ $$$${y}=\mathrm{2}{cosx}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right) \\ $$$${y}=\mathrm{2}{c}−\mathrm{2}{c}^{\mathrm{3}} \\ $$$$\frac{{dy}}{{dc}}=\mathrm{2}−\mathrm{6}{c}^{\mathrm{2}} \\ $$$${for}\:{max}/{min}\:\frac{{dy}}{{dc}}=\mathrm{0} \\ $$$$\mathrm{6}{c}^{\mathrm{2}} =\mathrm{2}\rightarrow{c}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dc}^{\mathrm{2}} }=−\mathrm{12}{c} \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dc}^{\mathrm{2}} }\right)_{{c}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:=\frac{−\mathrm{12}}{\:\sqrt{\mathrm{3}}}<\mathrm{0}\:\:{maximum} \\ $$$${max}\:{value}\:{y}_{{max}} =\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{2}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{3}} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${min}\:{at}\:{c}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${y}_{{min}} =\mathrm{2}×\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{−\mathrm{6}+\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{−\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

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