What is the domain and range of $f(x) = (x-1)/(x^2+1)$ ?

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Answer 1 · 2018-03-15T18:44:51

$"Domain":x inRR$
$"Range":f(x)in[-(sqrt(2)+1)/2,(sqrt(2)-1)/2]$

Considering that all real values of $x$ will give a non-zero value for $x^2+1$ , we can say that for $f(x)$ , domain = $x inRR$

For range, we need the maximum and minimum.
$f(x)=(x-1)/(x^2+1)$

$f'(x)=((x^2+1)-2x(x-1))/(x^2+1)^2=(x^2+1-2x^2+2x)/(x^2+1)=(-x^2+2x+1)/(x^2+1)$

The maximum and minimum values occur when $f'(x)=0$

$x^2-2x-1=0$

$x=(2+-sqrt((-2)^2-4(-1)))/2$

$x=(2+-sqrt8)/2=(2+-2sqrt(2))/2=1+-sqrt2$

Now, we input our $x$ values into $f(x)$ :
$(1+sqrt(2)-1)/((1+sqrt(2))^2+1)=(sqrt(2)-1)/2$
$(1-sqrt(2)-1)/((1-sqrt(2))^2+1)=-(sqrt(2)+1)/2$

$f(x)in[-(sqrt(2)+1)/2,(sqrt(2)-1)/2]$

What is the domain and range of f(x) = (x-1)/(x^2+1)f(x) = (x-1)/(x^2+1)?