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

$$Findthegreatestandtheleastvalues$$$$ofthefunctiononindicates$$$$interval(-\mathrm{\infty },\mathrm{\infty })$$$$\left(i\right)y=\sin x\sin 2x.$$

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

$$y=sinx\times 2sinxcosx$$$$y=2cosx(1-{cos}^{2}x)$$$$y=2c-2c^3$$$$\frac{dy}{dc}=2-6c^2$$$$formax/min\frac{dy}{dc}=0$$$$6c^2=2\to c=\pm \frac{1}{\sqrt{3}}$$$$\frac{d^{2}y}{{dc}^2}=-12c$$$${\left(\frac{d^{2}y}{{dc}^2}\right)}_{c=\frac{1}{\sqrt{3}}}=\frac{-12}{\sqrt{3}}<0maximum$$$$maxvalue{y}_{max}=2\times \frac{1}{\sqrt{3}}-2\times {\left(\frac{1}{\sqrt{3}}\right)}^3$$$$=\frac{2}{\sqrt{3}}-\frac{2}{3\sqrt{3}}=\frac{4}{3\sqrt{3}}$$$$minatc=\frac{-1}{\sqrt{3}}$$$$y_{min}=2\times \frac{-1}{\sqrt{3}}-2\times \frac{-1}{3\sqrt{3}}=\frac{-6+2}{3\sqrt{3}}=\frac{-4}{3\sqrt{3}}$$

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