Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=\begin{cases} \dfrac{6x+\sin x}{2x+\sin x} & \text{if } x \neq 0, \\ \dfrac{7}{3} & \text{if } x=0 \end{cases}$ Then which of the following statements is (are) TRUE?

$f(x) = \dfrac{6x+\sin x}{2x+\sin x}\ (x\ne0)$

$f(0) = \dfrac{7}{3}$

$(1),(2):\ x=0\ \text{max or min}?$

$(3):\ \#\text{local max in } [\pi,6\pi]$

$(4):\ \#\text{local min in } [2\pi,4\pi]$

$1.\ \text{continuity and behaviour near } 0$

$2.\ \text{critical points: solve } \tan x = x$

$3.\ \text{classify and count}$

$\lim_{x\to0}\dfrac{6x+\sin x}{2x+\sin x} = \dfrac{7x}{3x} = \dfrac{7}{3} = f(0)$

$\Rightarrow f \text{ continuous at } 0$

$t=\dfrac{\sin x}{x},\quad f=\dfrac{6+t}{2+t}$

$\dfrac{df}{dt} = \dfrac{-4}{(2+t)^2}<0$

$t<1\ (x\ne0,\ \text{near }0) \Rightarrow f(x)>f(0) \Rightarrow (2)\ \text{TRUE}$

$f'(x) = \dfrac{-4\,t'(x)}{(2+t)^2},\quad t'(x)=\dfrac{x\cos x-\sin x}{x^2}$

$h(x)=x\cos x-\sin x,\quad h'(x)=-x\sin x$

$f'=0 \Leftrightarrow h=0 \Leftrightarrow \tan x=x$

$\tan x=x:\ r_1\in(\pi,\tfrac{3\pi}{2}),\ r_2\in(2\pi,\tfrac{5\pi}{2}),\ r_3\in(3\pi,\tfrac{7\pi}{2})$

$r_4\in(4\pi,\tfrac{9\pi}{2}),\quad r_5\in(5\pi,\tfrac{11\pi}{2})$

$f\uparrow \text{ where } h<0,\qquad f\downarrow \text{ where } h>0$

$r_1,r_3,r_5:\ h\ -\to+ \Rightarrow \text{local MAX}$

$r_2,r_4:\ h\ +\to- \Rightarrow \text{local MIN}$

$[\pi,6\pi]\supset r_1,r_3,r_5 \Rightarrow 3 \text{ maxima} \Rightarrow (3)\ \text{TRUE}$

$\text{minima at } r_2,r_4;\quad r_2\in(2\pi,\tfrac{5\pi}{2})\subset[2\pi,4\pi]$

$r_4\in(4\pi,\tfrac{9\pi}{2})\not\subset[2\pi,4\pi] \Rightarrow 1 \text{ min} \Rightarrow (4)\ \text{TRUE}$

$\text{TRUE: } (2),\ (3),\ (4)$

$\boxed{(2),\ (3),\ (4)}$