Let f:R-2,6 R be real valued function defined as f(x)=x^2+2x+1x^2-8x+12.

Question

Let $f:R-{2,6}	o R$ be real valued function

defined as $f(x)=\frac{{x}^{2}+2x+1}{{x}^{2}-8x+12}$.

Then range of $f$ is

Accepted Answer

$y=\frac{{x}^{2}+2x+1}{{x}^{2}-8x+12}$

$\Rightarrow (y-1){x}^{2}-(8y+2)x+12y-1=0$

Let $y
e 1$, then $D\ge 0$

$4(4y+1{)}^{2}-4(y-1)(12y-1)\ge 0$

$\Rightarrow 16{y}^{2}+1+8y-(12{y}^{2}-13y+1)\ge 0$

$\Rightarrow 4{y}^{2}+21y\ge 0$

$\Rightarrow y\in (-\infty,-\frac{21}{4})\cup [0,\infty)-{1}$

$y=1$

$\begin{gathered} -8x+12=2x+1 \ x=\frac{11}{10}\, 	herefore I\in R \end{gathered}$

$	herefore $ Range $=(-\infty,-\frac{21}{4}]\cup [0,\infty)$

Solutions · 1