Question

Find the domain and range of $f\left(x\right)=\frac{x}{x^2-2x-3}$

Answer 1

Equate the denominator $x^2-2x-3=0$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2-3x+x-3=0$

$\Rightarrow x(x-3)+(x-3)=0$

$\Rightarrow (x+1)(x-3)=0$

$\Rightarrow x=-1,3$

$\therefore$ When x=-1 or 3. $x^2-2x-3$ is not defined.

Hence, domain = set of all real rolls except -1 & 3.

$D_f=(-50,-1)\cup (-1,3)\cup (3,\infty )$

Set $f\left(x\right)=y\Rightarrow y=\frac{x}{x^2-2x-3}$

$\Rightarrow y(x^2-2x-3)=x$

$\Rightarrow y{x}^2-(2y+1)x-3y=0$

solutions are real when $b^2-4ac\ge 0$

$\Rightarrow (2y+1{)}^2-4\times y\times -3⩾0$

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

$\Rightarrow 4y^2+4y+1+12y\ge 0$

$\Rightarrow 4y^2+16y+1\ge 0$

$\Rightarrow 16[{(y+\frac{1}{3})}^2-\frac{1}{64}+\frac{1}{16}]\ge 0$

$\Rightarrow (y+\frac{1}{8}{)}^2+\frac{3}{64}\ge 0$ $\forall y\ge 0$

$\therefore yϵR(=)$ & $\left(x\right)ϵR$

so the range is R.