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Question Number 167098 by mr W last updated on 06/Mar/22

$$findthemaximumof$$$$f\left(x\right)=\sin x+\cos x+\sin x\cos x$$

Answered by cortano1 last updated on 06/Mar/22

$$\sin \mathrm{x}+\cos \mathrm{x}=\sqrt{1+\sin 2\mathrm{x}}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{1+\sin 2\mathrm{x}}+\frac{1}{2}\sin 2\mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{\cos 2\mathrm{x}}{\sqrt{1+\sin 2\mathrm{x}}}+\cos 2\mathrm{x}=0$$$$\Rightarrow \cos 2\mathrm{x}+\cos 2\mathrm{x}\sqrt{1+\sin 2\mathrm{x}}=0$$$$\Rightarrow \cos 2\mathrm{x}(1+\sqrt{1+\sin 2\mathrm{x}})=0$$$$\Rightarrow \cos 2\mathrm{x}=0\Rightarrow \mathrm{x}=\frac{\pi }{4}$$$$\max \mathrm{f}\left(\mathrm{x}\right)=\sqrt{2}+\frac{1}{2}$$

Answered by cortano1 last updated on 06/Mar/22

$$\mathrm{for}\mathrm{minimum}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sin \mathrm{x}+\cos \mathrm{x}+\sin \mathrm{x}\cos \mathrm{x}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sin \mathrm{x}+\cos \mathrm{x}+\frac{1}{2}\sin 2\mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\cos \mathrm{x}-\sin \mathrm{x}+\cos 2\mathrm{x}=0$$$$\cos \mathrm{x}-\sin \mathrm{x}+\cos {}^2\mathrm{x}-\sin {}^2\mathrm{x}=0$$$${(\cos \mathrm{x}+\frac{1}{2})}^2-{(\sin \mathrm{x}+\frac{1}{2})}^2=0$$$$\Rightarrow (\cos \mathrm{x}+\sin \mathrm{x}+1)(\cos \mathrm{x}-\sin \mathrm{x})=0$$$$\Rightarrow \{\begin{array}{l}\cos \mathrm{x}+\sin \mathrm{x}=-1\Rightarrow 1+\sin 2\mathrm{x}=1\\ \cos \mathrm{x}-\sin \mathrm{x}=0\end{array}$$$$\Rightarrow \{\begin{array}{l}\max =\sqrt{2}+\frac{1}{2}\\ \min =-1\end{array}\Rightarrow -1⩽\mathrm{f}\left(\mathrm{x}\right)⩽\sqrt{2}+\frac{1}{2}$$

Answered by mahdipoor last updated on 06/Mar/22

$$sinx+cosx=u=\sqrt{2}sin(45+x)$$$$sinx.cosx=0.5(u^2-1)$$$$f\left(x\right)=f\left(u\right)=0.5(u^2+2u-1)=0.5{(u+1)}^2-1$$$$\Rightarrow manoff=f\left(maxofu\right)=f\left(\sqrt{2}\right)=$$$$0.5{(\sqrt{2}+1)}^2-1=\sqrt{2}+0.5$$

Commented by mr W last updated on 06/Mar/22

$$thankssirs!$$

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