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Question Number 85146 by john santu last updated on 19/Mar/20

$$\mathrm{find}\mathrm{minimum}\mathrm{\&}\mathrm{maximum}\mathrm{value}$$$$\mathrm{of}\mathrm{function}$$$$\mathrm{f}\left(\mathrm{x}\right)=-\sin {}^2\mathrm{x}+\sin \mathrm{x}-\frac{1}{2},-\pi ⩽\mathrm{x}⩽\pi$$

Commented by mr W last updated on 19/Mar/20

$$f\left(x\right)=-({\sin }^{2}x-\sin x+\frac{1}{4})-\frac{1}{4}$$$$=-{(\sin x-\frac{1}{2})}^2-\frac{1}{4}$$$$f_{max}=-\frac{1}{4}at\sin x=\frac{1}{2},i.e.x=\frac{\pi }{6},\frac{5\pi }{6}$$$$f_{min}=-\frac{5}{2}at\sin x=-1,i.e.x=-\frac{\pi }{2}$$

Answered by jagoll last updated on 19/Mar/20

Answered by Rio Michael last updated on 19/Mar/20

$$\mathrm{Maximum}\mathrm{value}=-0.25$$$$\mathrm{note}\mathrm{at}\mathrm{maximum}\mathrm{vaue}{f}^{{}^{\prime }}\left(x\right)=0$$$$\mathrm{minimum}\mathrm{value}=-2.5$$

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