Question

If f (x) =$\frac{x^2-x}{x^2+2x}$, find the domain of $f\left(x\right)$. Show that f is one-one.

Answer 1

The function $f\left(x\right)$ is defined only when denominator $\ne 0$

$\Rightarrow x^2+2x\ne 0\Rightarrow x(x+2)\ne 0$

$\to x\ne 0,-2$

So for$x=(0,-2)f\left(x\right)$is not defined. Thus,

$D_f:Domainof\left(x\right)=R(0,-2)$

A function is one-one only when $f\left(a\right)=f\left(b\right)\iff a=b$

Consider any $a,bϵD_f$ such that:

$f\left(a\right)=f\left(b\right)$

$\Rightarrow \frac{a^2-a}{a^2+2a}=\frac{b^2-b}{b^2+2b}$

$\Rightarrow \frac{a(a-1)}{a(a+2)}=\frac{b(b-1)}{b(b+2)}[asa,b\ne 0.0\ne D_f]$

$\Rightarrow (a-1)(b+2)=(b-1)(a+2)$

$\Rightarrow ab-b+2a-2=ab-a+2b-2$

$\Rightarrow 3a=3b$

$\Rightarrow a=b\therefore f\left(x\right)$ is one one as $f\left(a\right)=f\left(b\right)\Rightarrow a=b.$