# How do you find the domain of f(x)=sqrt(x+4)?

Jul 7, 2017

$x \ge - 4$ or $\left[- 4 , \infty\right)$

#### Explanation:

In order to find the domain of $\sqrt{x + 4}$, we need to understand what domain is. A domain is, in essence, any real number $x$ that produces a real number $y$.

So, looking at $\sqrt{x + 4}$, we must ask at what value $x$ does the function (equation) stop producing a number that is real? In other words, not an irrational number.

We know that the square root of a negative number produces a non-real number, thus using the definition of domain we will simply find when does $x$ stop giving us real $y$ values.

The first step to solving this now is to look at the $x + 4$ itself and disregard the square root. Let's set $x + 4$ to equal zero so: $x + 4 = 0$. Then subtracting 4 from both sides, we discover that $x = - 4$.

Plug $x = - 4$ into sqrt(x+4, so $\sqrt{- 4 + 4}$. This will be $\sqrt{0}$ which is basically 0. At $x = - 4$, the function still gives us a real number. What if we plug in $x = - 5$? We get $\sqrt{- 1}$ which will result in an irrational number.

Discovering this, we can now conclude that the domain of $x$ starts at $x = - 4$. Thus we will get $x \ge - 4$.

Now, let's look at $\sqrt{x + 4}$ and ask does it ever hit zero if we go towards the positive $x$-axis? No, it doesn't. Due to this, $x \ge - 4$ will be our final answer. It can also be written as $\left[- 4 , \infty\right)$.