What is the domain and range of $y=sqrt(x^2 - 2x + 5)$ ?

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Answer 1 · 2016-01-07T09:19:19

domain:

$]-oo,+oo[$

range:
$]0,+oo[$

Domain:

The real conditions for:

$y=sqrt(h(x))$

are:

$h(x)>=0$

then:

$x^2-2x+5>=0$

$x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(2+-sqrt(4-20))/(2)=(2+-sqrt(-16))/(2)=$
$=1+-2i$

Then
$h(x)>0 AAx in RR$

Range:
$lim_(x rarr +-oo) f(x)=lim_(x rarr +-oo)sqrt(x^2-2x+5)=lim_(x rarr +-oo)sqrt(x^2)$

$=lim_(x rarr +-oo)x=+-oo$

Remembering that:

$x^2-2x+5>0 AAx in RR$

Then the range is:

$]0,+oo[$

What is the domain and range of y=sqrt(x^2 - 2x + 5)y=sqrt(x^2 - 2x + 5)?