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How do you find the domain and range for #y = sqrt(x^2 + 2x + 3)#?

1 Answer

May 23, 2015

Note that

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

#=(x+1)^2+2 >= 2 AA x in RR#

So #sqrt(x^2+2x+3)# is defined #AA x in RR# and the domain is the whole of #RR#.

Also we see that the range is #{y in RR: y >= 2}#. The minimum value #y=2# occurs when #(x+1) = 0#, i.e. when #x=-1#.

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