Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x) = \frac{x^{2} - 3x - 6}{x^{2} + 2x + 4}$ Then which of the following statements is (are) TRUE?

Let f : R → R be defined by f(x) = (x² - 3x - 6)/(x² + 2x + 4) Then which of the following statements is (are) TRUE? $\text{Given: } f(x) = \dfrac{x^2 - 3x - 6}{x^2 + 2x + 4},\qquad x^2 + 2x + 4 = (x+1)^2 + 3 > 0$ $\text{To find: which of (1)--(4) are TRUE}$ $1.\ \text{plot } f(x)$ $2.\ \text{sign of } f' \Rightarrow \text{increasing/decreasing}$ $3.\ \text{extrema} \Rightarrow \text{range, onto}$ $y = 1 \text{ is a horizontal asymptote } (x \to \pm\infty)$ $f'(x) = \dfrac{(2x-3)(x^2+2x+4) - (x^2-3x-6)(2x+2)}{(x^2+2x+4)^2}$ $f'(x) = \dfrac{5x(x+4)}{(x^2+2x+4)^2}$ $f'(x) = 0 \Rightarrow x = 0,\ -4$ $f' > 0\ (x<-4),\quad f' < 0\ (-4<x<0),\quad f' > 0\ (x>0)$ $\text{increasing on } (-\infty,-4)\cup(0,\infty);\quad \text{decreasing on } (-4,0)$ $(-2,-1) \subset (-4,0) \Rightarrow f \text{ decreasing} \Rightarrow \text{(1) TRUE}$ $(1,2) \subset (0,\infty) \Rightarrow f \text{ increasing} \Rightarrow \text{(2) TRUE}$ $f(0) = \dfrac{-6}{4} = -\dfrac{3}{2}\quad \text{(local min)}$ $f(-4) = \dfrac{22}{12} = \dfrac{11}{6}\quad \text{(local max)}$ $\text{Range} = \left[-\dfrac{3}{2},\ \dfrac{11}{6}\right] \neq \left[-\dfrac{3}{2}, 2\right] \Rightarrow \text{(4) FALSE}$ $\text{Range} \neq \mathbb{R} \Rightarrow f \text{ is NOT onto} \Rightarrow \text{(3) FALSE}$ $\boxed{\text{Answer: (1) and (2)}}$