Question

Let $f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$ be a function from $R$ into $R$.Determine the range of $f$

Answer 1

Given:$f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$

Domain$=x$

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

We find different values of $\frac{x^2}{1+x^2}$ for different values of $x$

Value of $x$	value of $\frac{x^2}{1+x^2}$
$-2$	$\frac{{(-2)}^2}{1+{(-2)}^2}=\frac{4}{1+4}=\frac{4}{5}=0.8$
$-1$	$\frac{{(-1)}^2}{1+{(-1)}^2}=\frac{1}{1+1}=\frac{1}{2}=0.5$
$0$	$\frac{{\left(0\right)}^2}{1+0}=0$
$1$	$\frac{{\left(1\right)}^2}{1+{\left(1\right)}^2}=\frac{1}{1+1}=\frac{1}{2}=0.5$
$2$	$\frac{{\left(2\right)}^2}{1+{\left(2\right)}^2}=\frac{4}{1+4}=\frac{4}{5}=0.8$

Thus, value will always lie between $0$ and $1$

$\therefore$ the value of range of $y=\frac{x^2}{1+x^2}$ is always positive

Also, it lies between $0$ and $1$

Hence,Range is any positive integer between $0$ and $1$ with $0$ included.

Thus, Range of $f=[0,1)$