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) $$f(x)=\dfrac{1}{x+1}$$

$$Dom (f)\mathbb{R}- \lbrace-1\rbrace$$