Find the domain and range of the real function, defined by $f(x)=\frac{x^{2}}{\left(1+x^{2}\right)}$ Show that f is many - one.
For domain $\left(1+x^{2}\right) \neq 0$
$\Rightarrow x^{2} \neq-1$
$\Rightarrow \mathrm{dom}(\mathrm{f})=\mathrm{R}$
For the range of $x$ :
$\Rightarrow y=\frac{x^{2}+1-1}{x^{2}+1}=1-\frac{1}{x^{2}+1}$
$y_{\min }=0($ when $x=0)$
$y_{\max }=1(w h e n x=\infty)$
$\therefore$ range of $f(x)=[0,1)$
For many one the lines cut the curve in 2 equal valued points of y therefore the function f(x) $=\frac{x^{2}}{x^{2}+1}$ is many - one.
Ans:
$\operatorname{dom}(f)=R$
range $(f)=[0,1)$
function $f(x)=\frac{x^{2}}{x^{2}+1}$ is many - one.
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