What is the domain and range of #y = sqrt(x^2 + 2x + 3)#?

1 Answer
MeneerNask
Apr 25, 2016

With radical functions the argument under the root-sign and the outcome are always non-negative (in real numbers).

Explanation:

Domain :
The argument under the root sign must be non-negative:
We 'translate' by completing the square:

#x^2+2x+3=(x^2+2x+1)+2=(x+1)^2+2#

Which is always #>=2# for every value of #x#

So there are no restrictions to #x#:
#x in(-oo,+oo)#

Range:
Since the lowest value the argument can take is #2#, the lowest value of #y=sqrt2#, so:
#y in [sqrt2,+oo)#

Related questions
See all questions in Domain and Range of a Function
Impact of this question
6840 views around the world
You can reuse this answer
Creative Commons License