Define the domain of $f(x)=\sqrt{x^2+4}$ I am to define the domain of $f(x)=\sqrt{x^2+4}$.
I arrived at $[-2,\infty)$ whereas the textbook solution is $(-\infty,\infty)$.
To arrive at my solution I set the radicand to be greater than or equal to zero:
$x^2+4\ge0$
$x+2\ge0$ # square root of each side
$x\ge-2$
Thus I get the domain as $[-2,\infty)$.
Why is the domain actually $(-\infty, \infty)$?
[edit I had a typo the radicand is $x^2+4$]
 A: As you said, the radicand must be greater than or equal to zero.  Since the square of any real number is at least zero, $x^2 + 4 \geq 4 > 0$ for every real number $x$, which implies that the domain is indeed $(-\infty, \infty)$.
In your calculation, you should have had
\begin{align*}
x^2 + 4 & \geq 0\\
x^2 & \geq -4
\end{align*}
which is true for every real number $x$.
It is not true that $\sqrt{x^2 + 4} = x + 2$.  Notice that if $x = 1$, then $\sqrt{x^2 + 4} = \sqrt{1 + 4} = \sqrt{5}$ while $x + 2 = 1 + 2 = 3$.  Squaring $x + 2$ yields
\begin{align*}
(x + 2)^2 & = (x + 2)(x + 2)\\
          & = x(x + 2) + 2(x + 2)\\
          & = x^2 + 2x + 2x + 4\\
          & = x^2 + 4x + 4
\end{align*}
Hence, $\sqrt{x^2 + 4} = x + 2$ is only true when $4x = 0 \implies x = 0$.
A: It must be $$x^2-4\geq 0$$ and this is $$(x-2)(x+2)\geq 0$$
and $$x^2+4\geq 0$$ for all real $x$
A: One should be knowing that square of any real number is non-negative i.e.
$$x^2\ge0\quad (\forall \ \ \ x\in \mathbb R)$$
$$x^2+4\ge0+4$$
$$x^2+4\ge4$$
$$\sqrt{x^2+4}\ge\sqrt4$$
$$\sqrt{x^2+4}\ge2$$
$$f(x)\ge2$$
The above function is defined for all real values of $x$ hence its domain is 
$$x\in (-\infty, \infty)\ \ \ $$ 
A: Here is how to find the domain: ask yourself "What values can x not take?"
In this question, the expression under the square root cannot be less than zero, because the square root of a negative number is not defined yet at this level of mathematics (it will be later).
So for what values of x will produce a less than 0 value under the square root? Those values cannot be part of the domain. The rest, however can.
Can you take it from here?
A: The correct domain of $\sqrt{x^2-4}$ is $(-\infty, -2] \cup [2, \infty)$. More generally, the domain of $\sqrt{x^2-a}$ is $(-\infty, -\sqrt{a}] \cup [\sqrt{a}, \infty)$ for any positive real number $a$.
For $\sqrt{x^2+a}$, e.g. $\sqrt{x^2+4}$, the domain is all the real numbers.
A: The domain of $f(x) = \sqrt{x^{2} + 4}$ is all of $\mathbb{R}$ because every real value $x \in \mathbb{R}$ produces a real corresponding value of $f(x)$. 
In other words, there is no real $x$ that causes us to take the square root of a negative number.
