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

1 Answer
Jul 7, 2017

Answer:

#x>=-4# or #[-4, oo)#

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>=-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>=-4# will be our final answer. It can also be written as #[-4, oo)#.