Given the functions,
1) $$f(x)=x^2-2$$
2) $$f(x)=\sqrt{x+4}$$
3) $$f(x)=\dfrac{1}{x+1}$$
Determine the real domain of each of them.
Development:
1) The first function is a polynomial of second degree. Therefore, $$Dom (f) =\mathbb{R}$$
2) In this case, we need to check that the inside expression is positive, which is: $$x+4\geq 0 \Rightarrow x \geq -4$$.
Therefore, $$Dom (f) = [-4, +\infty)$$.
3) Finally, since it is a rational function we have to verify that the denominator is not zero (since it is not possible to divide by $$0$$): $$$x + 1 = 0$$$ $$$x =-1$$$ Therefore, $$Dom (f)\mathbb{R}- \lbrace-1\rbrace$$
Solution:
1) $$f(x)=x^2-2$$
$$Dom (f) =\mathbb{R}$$
2) $$f(x)=\sqrt{x+4}$$
$$Dom (f) = [-4, +\infty)$$
3) $$f(x)=\dfrac{1}{x+1}$$
$$Dom (f)\mathbb{R}- \lbrace-1\rbrace$$