What is the domain and range of y=sqrt(x^2 - 2x + 5)y=√x2−2x+5?

1 Answer
Lucio Falabella
Jan 7, 2016

domain:

]-oo,+oo[]−∞,+∞[

range:
]0,+oo[]0,+∞[

Explanation:

Domain:

The real conditions for:

y=sqrt(h(x))y=√h(x)

are:

h(x)>=0h(x)≥0

then:

x^2-2x+5>=0x2−2x+5≥0

x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(2+-sqrt(4-20))/(2)=(2+-sqrt(-16))/(2)=x1,2=−b±√b2−4ac2a=2±√4−202=2±√−162=
=1+-2i=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[

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