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 $x^2+4$]
Best Answer
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 $4x = 0 \implies x = 0$.