Question

Find the sum of the local maximum and local minimum values of the function $f(x)=\frac{\tan 3 x}{\tan ^3 x}$ on interval $\left(0, \frac{\pi}{2}\right)$

Answer 1

$y=\frac{\tan 3 x}{\tan ^3 x}=\frac{3 \tan x-\tan ^3 x}{\tan ^3 x\left(1-3 \tan ^2 x\right)}$
$y=\frac{3-\tan ^2 x}{\tan ^2 x\left(1-3 \tan ^2 x\right)}=\frac{3-t}{t(1-3 t)} \text { where } \tan ^2 x=t>0 $
$\left(t-3 t^2\right) y=3-t$
$3 y^2-(1+y) t+3=0$
$\therefore t>0 \Rightarrow D \geq 0 ; \text { Sum of roots }>0 ; \text { Product of roots }>0$
$\text { hence }(1+y)^2-36 y \geq 0 ; \frac{1+y}{3 y}>0 \text { and } \frac{1}{y}>0 \text { hence } y>0 $
$ y^2-34 y-1 \geq 0$
$(y-17)^2 \geq 288 $
$(y-17)^2-(12 \sqrt{2})^2 \geq 04
$(y-17-12 \sqrt{2})(y-17+12 \sqrt{2}) \geq 0$
${[y-(17+12 \sqrt{2})][y-(17-12 \sqrt{2})] \geq 0}$

Hence $y_{\max }=17-12 \sqrt{2} $
$y_{\min }=17+12 \sqrt{2} $
$y_{\max }+y_{\min }=34 \text { which is rational }$