Domain of a function

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.

  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, $Dom (f) = [-4, +\infty)$.

  1. 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, $Dom (f)\mathbb{R}- \lbrace-1\rbrace$
  1. $f(x)=x^2-2$

$$Dom (f) =\mathbb{R}$$

  1. $f(x)=\sqrt{x+4}$

$$Dom (f) = [-4, +\infty)$$

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

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

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