Question

find the domain and range of $x^{2} /1+x^{2}$

Answer 1

Given function is,

$\longrightarrow f(x)=\dfrac{x^2}{1+x^2}$

Here the denominator should not be zero but here the denominator is always positive since $x^2$ is non - negative.

$\displaystyle\longrightarrow x^2\geq0$

$\displaystyle\longrightarrow 1+x^2\geq1$

Thus $f(x)$ can accept all real values of $x$ without any restriction.

Hence the domain is $\mathbb{R.}$

$\displaystyle\longrightarrow\underline{\underline{x\in\mathbb{R}}}$

Let,

$\longrightarrow y=\dfrac{x^2}{1+x^2}$

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

$\longrightarrow y+x^2y=x^2$

$\longrightarrow y+x^2y-x^2=0$

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

Now we have a quadratic equation in $x$ where,

• $a=y-1$

• $b=0$

• $c=y$

Solving it,

$\longrightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\longrightarrow x=\dfrac{\pm\sqrt{0^2-4(y-1)y}}{2(y-1)}$

$\longrightarrow x=\dfrac{\pm\sqrt{-4y(y-1)}}{2(y-1)}$

Here the denominator should not be zero.

$\longrightarrow 2(y-1)\neq0$

Dividing by 2,

$\longrightarrow y-1\neq0$

Adding 1,

$\longrightarrow y\neq1$

$\Longrightarrow y\in\mathbb{R}-\{1\}\quad\quad\dots(1)$

And we know $\sqrt x$ is defined only for $x\geq0.$

$\Longrightarrow -4y(y-1)\geq0$

Dividing by -4, (note the sign change)

$\longrightarrow y(y-1)\leq0$

$\Longrightarrow y\in[0,\ 1]\quad\quad\dots(2)$

Combining (1) and (2),

$\longrightarrow y\in\left[\mathbb{R}-\{1\}\right]\cap[0,\ 1]$

$\longrightarrow\underline{\underline{y\in[0,\ 1)}}$

Or,

$\longrightarrow\underline{\underline{f(x)\in[0,\ 1)}}$

This is the range of the function.