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

find max and min value of   function f(x) = (5/(−3cos x−4sin x))

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

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

f(x) = (5/(−3cos x−4sin x))  = (1/(−((3/5)cos x+(4/5)sin x)))  = (1/(−(cos α cos x+sin α sin x))) with α=tan^(−1) (4/3)  = (1/(−cos (x−α)))  ⇒max. f(x) →+∞ when cos (x−α)→0^−   ⇒min. f(x) →−∞ when cos (x−α)→0^−     ⇒local max. f(x)=−1 when cos (x−α)=1  ⇒local min. f(x)=1 when cos (x−α)=−1

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{5}}{−\mathrm{3cos}\:\mathrm{x}−\mathrm{4sin}\:\mathrm{x}} \\ $$$$=\:\frac{\mathrm{1}}{−\left(\frac{\mathrm{3}}{\mathrm{5}}\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{4}}{\mathrm{5}}\mathrm{sin}\:\mathrm{x}\right)} \\ $$$$=\:\frac{\mathrm{1}}{−\left(\mathrm{cos}\:\alpha\:\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{x}\right)}\:{with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}} \\ $$$$=\:\frac{\mathrm{1}}{−\mathrm{cos}\:\left(\mathrm{x}−\alpha\right)} \\ $$$$\Rightarrow{max}.\:{f}\left({x}\right)\:\rightarrow+\infty\:{when}\:\mathrm{cos}\:\left({x}−\alpha\right)\rightarrow\mathrm{0}^{−} \\ $$$$\Rightarrow{min}.\:{f}\left({x}\right)\:\rightarrow−\infty\:{when}\:\mathrm{cos}\:\left({x}−\alpha\right)\rightarrow\mathrm{0}^{−} \\ $$$$ \\ $$$$\Rightarrow{local}\:{max}.\:{f}\left({x}\right)=−\mathrm{1}\:{when}\:\mathrm{cos}\:\left({x}−\alpha\right)=\mathrm{1} \\ $$$$\Rightarrow{local}\:{min}.\:{f}\left({x}\right)=\mathrm{1}\:{when}\:\mathrm{cos}\:\left({x}−\alpha\right)=−\mathrm{1} \\ $$

Commented by jagoll last updated on 08/Apr/20

thank you sir

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

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