$Ifx\in Rand1\le \frac{3x^2-7x+8}{x^2+1}\le 2,thenthe\min imumvaluesofxarereapectively$

$\begin{array}{l}Given1\le \frac{3x^2-7x+8}{x^2+1}\le 2\\ Let1\le \frac{3x^2-7x+8}{x^2+1}\Rightarrow x^2+1\le 3x^2-7x+8\\ \Rightarrow 2x^2-7x+7\ge 0\\ and\frac{3x^2-7x+8}{x^2+1}\le 2\\ \Rightarrow 3x^2-7x+8\le 2x^2+2\\ \Rightarrow x^2-7x+6\le 0\\ \Rightarrow (x-1)(x-6)\le 0\\ \therefore 1\le x\le 6\end{array}$