Domain and range of $f(x) = -1 + \sqrt{x+2}$ 
$f(x) = -1 + \sqrt{x+2}$

$$Dom(f(x)) = \{x|x\geq-2\}$$
$$Ran(f(x)) = \{y|y\geq-1\}$$
To see that the range is, all of the set above, it needs to be shown that for $y\geq-1$ there exists a real number x such that, 
$$y = -1 +\sqrt{x+2}$$
Solving for x we have, 
$$y+1 = \sqrt{x+2} \\
x+2 = (y+1)^2 \\
x = (y+1)^2 - 2 $$
this shows that the domain is the all the set $\{y|y\geq -1\}$
Is this correct? 
Also, is any of my notation incorrect or can it be improved?
Thanks
 A: One step missing in the end is to check that $f((y+1)^2-2) = y$ when $y\geq -1$, as Siong Thye Goh points out. This is important, since if $y<-1$, $f((y+1)^2-2) = -y-2$, actually!
To be more clear what I mean, you basically wrote: "if $y = -1 + \sqrt{x+2}$ has a solution for some $y$, then that solution is $x = (y+1)^2-2$", but you didn't comment that in the case $y<-1$ there is no solution and you didn't check that in the case $y\geq -1$ there really exists a solution (as I wrote in the 1st paragraph).
Thus, your phrasing is not as clear as it could be. 
Here's one alternative approach to be more precise. As soon as you wrote $\sqrt{\,\cdot\,}$, it is implicit that by this you mean a bijective function $[0,\infty)\to [0,\infty)$, inverse to $x\mapsto x^2\colon [0,\infty)\to [0,\infty)$.
Let's define $g(x) = -1 + x$, $h(x) = \sqrt x$ and $k(x) = x + 2$. Then, $f = g\circ h\circ k$ and the range of $f$ is
$$f([-2,\infty)) = g(h(k([-2,\infty)))= g(h([0,\infty))) = g([0,\infty)) = [-1,\infty).$$
A: Your working is fine. 
It should be clear that $(y+1)^2-2 \ge -2$ and it is in the domain and if $y \ge -1$, then we have.
$$-1 + \sqrt{(y+1)^2-2+2}=y.$$
Now the question is what if $y<-1$? Is it possible to have a preimage? We know it doesn't because $$-1+\sqrt{x+2}\ge -1$$
