$- 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$ $\Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$ $\Rightarrow 2{x^2} - 4x + 5 \ge 0$ & $- 4x \le 1$ $x \in R$ & $x \ge - {1 \over 4}$ So domain is $\left[ { - {1 \over 4},\infty } \right)$