Question

Find the domain and range of $f\left(x\right)=x^2/(1+x^2)$ (x real ). Is the function one-to-one?

Answer 1

$f\left(x\right)=\frac{x^2}{1+x^2}$

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

$x^2=y(1+x^2)$

$x^2=y+y{x}^2$

$(y-1)x^2+y=0$

$b^2-4ac\Rightarrow 0^2-4(y-1)y\Rightarrow -4y^2+4y$

$-4y^2+4y\ge 0\Rightarrow 0\le y\le 1$

$0\le f\left(x\right)\le 1$

Range$[0,1)$

Domain

$-\infty <x<\infty$

$f\left(x\right)(-\infty ,\infty )$

Let,

$f\left(x\right)=\frac{x^2}{1+x^2}$

$f\left(x_1\right)=1$

$=\frac{1^2}{1+1^2}=\frac{1}{2}$

$f\left(x_2\right)=-1$

$=\frac{(-1{)}^2}{1+(-1{)}^2}=\frac{1}{2}$

Since, $f\left(x_1\right)=f\left(x_2\right)$

function is not one-one.