Domain of $(f \circ g)(x)$ when $f(x) = x^2 + 9$ and $g(x) = \sqrt{x + 3}$ I am in need of some help. Right now I am doing some test corrections for my Honors High School Pre-calc class. This class is Algebra-based calculus. So the question is: 


  Find $(f\circ g)(x)$ and its domain when $f(x)=x^2+9$ and $g(x)=\sqrt{x+3}$. 


I solved $(f\circ g)(x)$ to get $x+12$. So I understand how to get $(f\circ g)(x)$, but I do not understand how to get the domain. According to my test, it says that answer for the domain is $[-3,\infty)$. When I graph it on Desmos.com and my TI-84 Plus I get a straight line, so I really don't know how they got that answer. I would really appreciate it if someone could help me! Thanks in advance!
The domain of this function is indeed $[-3, \infty)$, assuming $\sqrt[]{x}$ is undefined when $x&lt;0$.

This is because $$(f\circ g)(x) = \left[\sqrt[]{x+3}\right]^2+3.$$ Notice that when $x &lt; -3$, we are taking the square root of a negative number, which is undefined. Therefore, the only values of $x$ for which $(f\circ g)(x)$ is defined are those satisfying $x \geq -3$.
Let $D(f)$ be a domain of the function $f$. 

Thus, by definition we have:
$$D(f\circ g)=D(g)\cap\{x|g(x)\in D(f)\}$$
In our case 
$$f(g(x))=\left(\sqrt{x+3}\right)^2+9,$$
$$D(f)=\mathbb R$$ and $$D(g)=[-3,+\infty).$$

Thus, the domain is $[-3,+\infty)$.