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

Question

6479 views around the world

You can reuse this answer
Creative Commons License

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[#