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Question Number 87171 by naka3546 last updated on 03/Apr/20

$$Givenf\left(x\right)=2{\sin }^{2}x-\sin x+1,0⩽x⩽2\pi$$$$Findmaximumandminumumvalue$$$$off\left(x\right)withoutdifferential.$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{for}\sin \mathrm{x}=1\Rightarrow {\mathrm{y}}_1=2-1+1=2$$$$\mathrm{for}\sin \mathrm{x}=-1\Rightarrow {\mathrm{y}}_2=2+1+1=4$$$$\mathrm{for}\sin \mathrm{x}=\frac{1}{4}\Rightarrow {\mathrm{y}}_3=\frac{1}{8}-\frac{1}{4}+1$$$${\mathrm{y}}_3=\frac{1}{8}+\frac{3}{4}=\frac{7}{8}$$$$\max =4\mathrm{\&}\min =\frac{7}{8}$$

Commented by mr W last updated on 03/Apr/20

$$f\left(x\right)=2({\sin }^{2}x-2\times \frac{1}{4}\sin x+\frac{1}{4^2})+1-2\times \frac{1}{4^2}$$$$=2{(\sin x-\frac{1}{4})}^2+\frac{7}{8}$$$$min.=\frac{7}{8}$$$$max.=2{(-1-\frac{1}{4})}^2+\frac{7}{8}=4$$

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