How do you find the domain and range of $f(x) = 1/(1+x^2)$ ?

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Answer 1 · 2018-04-06T19:04:36

The domain is $x in RR$ . The range is $y in (0,1]$

The denominator is $=1+x^2$

$AA x in RR$ , $1+x^2>0$

Therefore,

The domain of $f(x)$ is $x in RR$

To determine the range, proceed as follows

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

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

$y+yx^2=1$

$yx^2=1-y$

$x^2=(1-y)/y$

$x=sqrt((1-y)/y)$

The range of $f(x)$ is the domain of $x$

$((1-y)/y)>0$

$y in RR _+^("*")$

$1-y>=0$

$y<=1$

Therefore,

The range is $y in (0,1]$

graph{1/(1+x^2) [-11.25, 11.25, -5.625, 5.625]}

How do you find the domain and range of f(x) = 1/(1+x^2)f(x) = 1/(1+x^2)?