Question

Find the domain and range of the real function $\frac{x+2}{x^2-8x-4}$

Answer 1

Let $f\left(x\right)=\frac{x+2}{x^2-8x-4}=y\to 1$

If $x^2-8x-4=0$

$x=\frac{8\pm \sqrt{64+16}}{2}=\frac{8\pm 4\sqrt{5}}{2}$

$=4\pm 2\sqrt{5}$

At $x=4\pm 2\sqrt{5},x^2-8x-4=0\therefore f\left(x\right)$ is not defined

Domain of $f\left(x\right)$ is $R-\{4-2\sqrt{5},4+2\sqrt{5}\}$

$1\Rightarrow y(x^2-8x-4)=x+2$

$y{x}^2-(1+8y)x-(4y+2)=0$

As x is always real, Discriminant of above equation must be $\ge 0$

${(1+8y)}^2+4y(4y+2)\ge 0$

$64y^2+16y+1+16y^2+8y\ge 0$

$80y^2+24y+1\ge 0$

Factorize above equation,

$y=\frac{-24\pm \sqrt{526-320}}{160}$

$=\frac{-1}{4}$ or $\frac{-1}{20}$

$\therefore (4y+1)(20y+1)\ge 0$

$\therefore y\le \frac{-1}{20}$ or $y\ge \frac{-1}{4}$

$\therefore$ Range of $f\left(x\right)=(-\infty ,\frac{-1}{20}]\bigcup [\frac{-1}{4},\infty )$