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

Question

3543 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-23T23:10:05

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 $x=-1$ .

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