Question

Find the maximum and minimum values of $f\left(x\right)=6x+3\cot x$, where $x\in (0,\pi )$

Answer 1

Given:

$f\left(x\right)=6x+3\cot x$

Differentiating both sides with respect to x:

$f^{\prime }\left(x\right)=6-3cose{c}^{2}x=0$

$\Rightarrow cose{c}^{2}x=2$

$\Rightarrow \sin x=\frac{1}{\sqrt{2}}$

$\therefore x=\frac{\pi }{4},\frac{3\pi }{4}$

$\therefore f\left(\frac{\pi }{4}\right)=6.\frac{\pi }{4}+3\cot \frac{\pi }{4}=\frac{3}{2}[\pi +2]$

$\therefore f\left(\frac{3\pi }{4}\right)=6.\frac{3\pi }{4}+3\cot \frac{3\pi }{4}=\frac{3}{2}[3\pi -2]$

$\therefore f_{max}=\frac{3}{2}[3\pi -2]$

$\therefore f_{min}=\frac{3}{2}[\pi +2]$ $\left(Ans\right)$