Question

Find the domain of the function f of x equals square root of the quantity twelve x plus twenty four

Answer 1

\[\Large f(x)=\sqrt{12x+24}\]
Hmm so what values would give us trouble?
Any time we're trying to take the square root of a negative number we'd have a problem, right?

So let's set up an inequality to express this.\[\Large 12x+24\ge0\]This inequality basically is saying, the stuff under the square root must be greater than or equal to 0.

Now solve for x to find your domain! :)

Answer 2

So it would x is equal to or greater to negative 2?

Answer 3

Yes very good! plugging x=-2 into our function gives us,\[\Large f(-2)=\sqrt{12(-2)+24}=\sqrt{0}\]
If we plugged any x value more negative in, it would give us a problem.
Good job \c:/

Answer 4

Oh alright, thank you! I have one more question, I've been hitting a snag on this one Find the domain of the function f of x equals the cube root of the quantity x plus three, minus one.

Answer 5

\[\Large f(x)=\sqrt[3]{x+3}-1\]

Domain is x-values.
So with these types of problems, we're asking ourselves, "Are there any x-values that would cause a problem when I try to plug them in?"

We have to mostly be careful with `square` roots and dividing by x. Stuff like that.

So in this problem, are there any trouble spots?
Maybe the root is confusing you a little.

Answer 6

Here's an example:\[\Large \sqrt{-4}\]This is bad!

But if we have something like this,\[\Large \sqrt[3]{-8} \qquad=\qquad -2\]No problems!
We can take the `odd` root of a negative number.

Answer 7

So we don't actually have anything to worry about with this particular problem, our domain is all real numbers, no restrictions on our x-values.