Determine the domain of the following functions:
 $$\displaystyle f(x)=\sqrt{x+3}$$
 $$\displaystyle f(x)=\sqrt{x^2+3x+2}$$
 $$\displaystyle f(x)=\sqrt[3]{\frac{x3}{2x}}$$
See development and solution
Development:

In this case we have an irrational function. Since the degree of the root is even we must verify that what we have inside the root is greater than or equal to zero. $$$x+3 \geq 0 \Rightarrow x\geq 3$$$ Therefore $$Dom (f) = [3, +\infty)$$.

We have an irrational function of even degree. We must verify that its interior is positive or $$0$$. For this, we solve the inequality: $$$x^2+3x+2 \geq 0$$$ $$$x^2+3x+2=0 \Rightarrow x=1,x=2$$$ Therefore, since the parabola has a $$\cup$$ shape we have: $$Dom (f) = (\infty,2] \cup [1, +\infty)$$.
 Finally we have an irrational function of odd degree. Therefore we only need the interior of the root to be positive: $$$2x=0 \Rightarrow x=2$$$ And we have, $$Dom (f) =\mathbb{R}  \{2\}$$.
Solution:
 $$Dom (f) = [3, +\infty)$$
 $$Dom (f) = (\infty,2] \cup [1, +\infty)$$
 $$Dom (f) =\mathbb{R}  \{2\}$$