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

$$Findmax\mathrm{\&}minoffunction$$ $$f\left(x\right)=4\tan x+3\cot x$$

Answered by MJS_new last updated on 27/May/21

$$-\mathrm{\infty }<\tan x<+\mathrm{\infty }\Rightarrow \mathrm{no}\mathrm{absolute}\min ,\max$$ $$g\left(t\right)=4t+\frac{3}{t}$$ $$g^{\prime }\left(t\right)=4-\frac{3}{t^2}=0\iff 4t^2-3=0\iff t=\pm \frac{\sqrt{3}}{2}$$ $$g^{″}\left(t\right)=\frac{6}{t^3}$$ $$g^{″}(-\frac{\sqrt{3}}{2})<0\Rightarrow g\left(t\right)\mathrm{has}\mathrm{a}\mathrm{local}\max \mathrm{at}(-\frac{\sqrt{3}}{2},-4\sqrt{3})$$ $$g^{″}\left(\frac{\sqrt{3}}{2}\right)>0\Rightarrow g\left(t\right)\mathrm{has}\mathrm{a}\mathrm{local}\min \mathrm{at}(\frac{\sqrt{3}}{2},4\sqrt{3})$$ $$\Rightarrow$$ $$f\left(x\right)\mathrm{has}\mathrm{local}\mathrm{maxima}\mathrm{when}\tan x=-\frac{\sqrt{3}}{2}$$ $$f\left(x\right)\mathrm{has}\mathrm{local}\mathrm{minima}\mathrm{when}\tan x=\frac{\sqrt{3}}{2}$$

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