In Functions, we looked at how to determine if a given mapping is a function. We saw that we could look at these when we have mappings defined as sentences, graphs and tables. Here we will look at determining the domain when given a function in function notation.

**Determine the domain of a function**

We will now determine the domain of the function

\begin{align*}

\frac{\sqrt{x+4}}{x-2}.

\end{align*}

In order to do this, we will need to look at properties of domains of functions.

**Domain properties**

Suppose that \(f\) and \(g\) are functions with domains \(dom(f)\) and \(dom(g)\). We then have that

- \(dom(f+g)=dom(f) \cap dom(g)\).
- \(dom(f*g)=dom(f) \cap dom(g)\)
- \(dom(f/g)=dom(f) \cap dom(g) \cap \left\{x : g(x) \neq 0 \right\}\).
- \(dom(f(g(x)))=dom(g) \cap \left\{x: g(x) \in dom(f)\right\}\).

Using this, we will note that we can determine the domain of our original function by writing it in pieces. That is, if we let \(f(x)=x+4\), \(g(x)=x-2\) and \(h(x)=\sqrt{x}\), we find that our original function is indeed

\begin{align*}

\frac{\sqrt{x+4}}{x-2}=\frac{h(f(x))}{g(x)}.

\end{align*}

Therefore, the domain of top would be \(dom(f) \cap \left\{x: f(x) \in dom(h)\right\}\) and the domain of bottom would be \(dom(g)\). Hence, the domain of our entire function would

\begin{align*}

dom(f) \cap \left\{x: f(x) \in dom(h)\right\} \cap dom(g) \cap \left\{x: g(x) \neq 0 \right\}.

\end{align*}

We will now find each of these.

**Domains of pieces**

We have the following,

- \(f(x)=x+4\) is a polynomial, therefore, the \(dom(f)=\mathbb{R}\).
- \(g(x)=x-2\) is also a polynomial, so \(dom(g)=\mathbb{R}\).
- Since \(g(x)=x-2\), we note that \(g(x) \neq 0\) as long as \(x \neq 2\).
- \(h(x)=\sqrt{x}\) is not defined for \(x < 0\), therefore, \(dom(h)=\left\{x : x \geq 0\right\}\).
- In order to have \(f(x) \in dom(h)\), we must have \(x+4 \geq 0\), that is \(x \geq -4\).

If we now combine all of these, we note that we must have

\begin{align*}

&x \in \mathbb{R} \text{ and } (x \geq -4) \\

&\text{ and } x \in \mathbb{R} \text{ and } (x \neq 2).

\end{align*}

Therefore, we must have \(x \geq -4\) and \(x \neq 2\), so the domain of our function is \([-4,2) \cup (2, \infty)\).

**Conclusion**

By combining functions through products, quotients, and compositions, we were able to construct the given function using simpler functions. Since we could find the domain of these simpler functions, we could use the properties of domains we showed in class in order to find the domain of our goal function. In a similar way, we can build up to find the domains of a large number of functions.

If you look like to look at a more detailed description of functions, including topics of one-to-one, onto, and inverses, please see the post and video at Bijective mappings.

As always, I hope you enjoyed the post and learned something along the way. For more help with Calculus, visit the posts available here. Also, make sure to like the post and share on Social Media.