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Question Number 142167 by iloveisrael last updated on 27/May/21

Find max & min of function f(x)=4tan x+3cot x

$$\:\:\:{Find}\:{max}\:\&\:{min}\:{of}\:{function} \\ $$ $$\:\:{f}\left({x}\right)=\mathrm{4tan}\:{x}+\mathrm{3cot}\:{x} \\ $$

Answered by MJS_new last updated on 27/May/21

−∞<tan x <+∞ ⇒ no absolute min, max g(t)=4t+(3/t) g′(t)=4−(3/t^2 )=0 ⇔ 4t^2 −3=0 ⇔ t=±((√3)/2) g′′(t)=(6/t^3 ) g′′(−((√3)/2))<0 ⇒ g(t) has a local max at (−((√3)/2), −4(√3)) g′′(((√3)/2))>0 ⇒ g(t) has a local min at (((√3)/2), 4(√3)) ⇒ f(x) has local maxima when tan x =−((√3)/2) f(x) has local minima when tan x =((√3)/2)

$$−\infty<\mathrm{tan}\:{x}\:<+\infty\:\Rightarrow\:\mathrm{no}\:\mathrm{absolute}\:\mathrm{min},\:\mathrm{max} \\ $$ $${g}\left({t}\right)=\mathrm{4}{t}+\frac{\mathrm{3}}{{t}} \\ $$ $${g}'\left({t}\right)=\mathrm{4}−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }=\mathrm{0}\:\Leftrightarrow\:\mathrm{4}{t}^{\mathrm{2}} −\mathrm{3}=\mathrm{0}\:\Leftrightarrow\:{t}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ $${g}''\left({t}\right)=\frac{\mathrm{6}}{{t}^{\mathrm{3}} } \\ $$ $${g}''\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)<\mathrm{0}\:\Rightarrow\:{g}\left({t}\right)\:\mathrm{has}\:\mathrm{a}\:\mathrm{local}\:\mathrm{max}\:\mathrm{at}\:\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}},\:−\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$ $${g}''\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)>\mathrm{0}\:\Rightarrow\:{g}\left({t}\right)\:\mathrm{has}\:\mathrm{a}\:\mathrm{local}\:\mathrm{min}\:\mathrm{at}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}},\:\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$ $$\Rightarrow \\ $$ $${f}\left({x}\right)\:\mathrm{has}\:\mathrm{local}\:\mathrm{maxima}\:\mathrm{when}\:\mathrm{tan}\:{x}\:=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ $${f}\left({x}\right)\:\mathrm{has}\:\mathrm{local}\:\mathrm{minima}\:\mathrm{when}\:\mathrm{tan}\:{x}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

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