Let  $f:R-\{2,6\}\to R$ be real valued function defined as  $f\left(x\right)=\frac{x^2+2x+1}{x^2-8x+12}$.Then range of$f\left(x\right)$ is.

Let  $y=\frac{x^2+2x+1}{x^2-8x+12}$

By cross multiplying 

$$\begin{gathered} \begin{array}{l}y{x}^2-8xy+12y-x^2-2x-1=0\\ x^2(y-1)-x(8y+2)+(12y-1)=0\\ Case1,y\ne 1\\ D\ge 0\end{array} \\ \begin{array}{l}\Rightarrow {\left(8y-2\right)}^2-4(y-1)(12y-1)\ge 0\\ \Rightarrow y(4y+21)\ge 0\end{array} \end{gathered}$$

$\begin{array}{l}y\in (-\infty ,\frac{-21}{4}]\cup [0,\infty )-\left\{1\right\}\\ Case2,y=1\\ x^2+2x+1=x^2-8x+12\\ 10x=11\\ x=\frac{11}{10}So,ycanbe1\end{array}$