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Question Number 88042 by jagoll last updated on 08/Apr/20

$$\mathrm{find}\max \mathrm{and}\min \mathrm{value}\mathrm{of}$$$$\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\frac{5}{-3\cos \mathrm{x}-4\sin \mathrm{x}}$$

Answered by mr W last updated on 08/Apr/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{5}{-3\cos \mathrm{x}-4\sin \mathrm{x}}$$$$=\frac{1}{-(\frac{3}{5}\cos \mathrm{x}+\frac{4}{5}\sin \mathrm{x})}$$$$=\frac{1}{-(\cos \alpha \cos \mathrm{x}+\sin \alpha \sin \mathrm{x})}with\alpha ={\tan }^{-1}\frac{4}{3}$$$$=\frac{1}{-\cos (\mathrm{x}-\alpha )}$$$$\Rightarrow max.f\left(x\right)\to +\mathrm{\infty }when\cos (x-\alpha )\to 0^{-}$$$$\Rightarrow min.f\left(x\right)\to -\mathrm{\infty }when\cos (x-\alpha )\to 0^{-}$$$$$$$$\Rightarrow localmax.f\left(x\right)=-1when\cos (x-\alpha )=1$$$$\Rightarrow localmin.f\left(x\right)=1when\cos (x-\alpha )=-1$$

Commented by jagoll last updated on 08/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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