Suppose $f$ and $g$ are functions with domains $A$ and $B$, respectively. Then the new functions $f+g, f-g, f\cdot g$ and $\frac{f}{g}$ are defined as follows:
\begin{align}
(f+g)(x) &= f(x) + g(x) && \text{Domain } A \cap B \\
(f-g)(x) &= f(x) - g(x) && \text{Domain } A \cap B\\
(f\cdot g)(x) &= f(x)\cdot g(x) && \text{Domain } A \cap B\\
\left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)} && \text{Domain } \{x \in A
\cap B : g(x) \neq 0\}
\end{align}

Simply put, $+, -, \times, \div$ functions together is done by performing the specified operation on the function output. Let's look at some examples.

Let $f(x) = \frac{1}{x-2}$ and $g(x) = x - 3$. Find the following:

- $f+g, f-g, fg, \frac{f}{g}$ and their domains
- $(f+g)(4), \left(\frac{f}{g}\right)(4)$, and $\left(\frac{f}{g}\right)(3)$

- Let's first find the domains and intersect them. $f(x)$ has domain $ A = \{x : x \neq 2\}$ and $g(x)$ has domain $B = \mathbb{R}$. Thus \[A \cap B = \{x : x \neq 2\} = (-\infty, 2) \cup (2, \infty)\] Following the above table, we see that \begin{align} (f+g)(x) &= f(x) + g(x) &&= \frac{1}{x-2} + x - 3 \\ (f-g)(x) &= f(x) - g(x) &&= \frac{1}{x-2} - (x - 3) = \frac{1}{x-2} - x + 3 \\ (f\cdot g)(x) &= f(x)\cdot g(x) &&= \frac{x - 3}{x-2} \\ \left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)} &&= \frac{1}{(x-3)(x-2)} \\ \end{align} The first three functions have the domain $(-\infty, 2) \cup (2, \infty)$ while the last one has domain $(-\infty, 2) \cup (2, 3) \cup (3, \infty)$ because plugging in zero results in division by zero.
- From part 1, the definitions of $f+g$ and $f/g$ give \begin{align} (f+g)(4) &= f(4) + g(4) &&= \frac{1}{4-2} + (4 - 3) & = \frac{3}{2} \\ \left(\frac{f}{g}\right)(4) &= \frac{f(4)}{g(4)} &&= \frac{1}{(4-2)(4 - 3)} &= \frac{1}{2} \\ \end{align} Since we cannot divide by zero, $\left(\frac{f}{g}\right)(3)$ is undefined.

Adding functions together has a nice graphical interpretation. In this example, we see the graph $f(x) + g(x)$ is just the vertical heights of $f(x)$ and $g(x)$ added together.

Function evaluation on a number is straightforward: the preceding example utilized this concept. What if we evaluate a function on another function? This idea is called **function composition.**

Given two functions $f$ and $g$, the composite function $f \circ g$ (also called the composition of $f$ and $g$) is defined by \[(f\circ g)(x) = f(g(x))\]

To evaluate a function composition, you first plug in $g(x)$ into $f(x)$, and wherever there is an $x$ in $f(x)$, substitute the definition of $g(x)$ there.

Suppose $f(x) = x^2$ and $g(x) = x-3$. Find $f\circ g, g \circ f, (f\circ g)(5), (g\circ f)(5)$

We see that \[(f\circ g)(x) = f(g(x)) = f(x-3) = (x-3)^2\] and \[(g\circ f)(x) = g(f(x)) = g(x^2) = x^2 - 3\] Thus $(f\circ g)(5) = (5-3)^2 = 4$ and $(g\circ f)(5) = 5^2 - 3 = 22$.

Ideas:

- This shows that the operation $\circ$ is not commutative, i.e. $f\circ g \neq g \circ f$. Do not mistake $f\circ g$ to mean function multiplication!
- We can also take compositions of three or more functions. For example, $f\circ g \circ h = f(g(h(x)))$.

Find $f\circ g \circ h$ where $f(x) = \frac{x}{x+1}, g(x) = x^{10}, h(x) = x+3$.

\begin{align}
(f\circ g \circ h)(x) &= f(g(h(x)))\\
&= f(g(x+3))\\
&= f((x+3)^{10})\\
&= \frac{(x+3)^{10}}{(x+3)^{10}+1}
\end{align}

Given $F(x) = \sqrt[4]{x + 9}$, find functions $f$ and $g$ such that $F = f\circ g$.

We can think of this function as the $\sqrt[4]{\cdot}$ as the outside function, and $x + 9$ as the inside function.

Since $f\circ g$ is the desired composition, let $g(x) = x+9$ and $f(x) = \sqrt[4]{x}$. Then \[(f\circ g)(x) = f(g(x)) = f(x+9) = \sqrt[4]{x+9} = F(x)\]

Since $f\circ g$ is the desired composition, let $g(x) = x+9$ and $f(x) = \sqrt[4]{x}$. Then \[(f\circ g)(x) = f(g(x)) = f(x+9) = \sqrt[4]{x+9} = F(x)\]