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.
Be careful when you are subtracting and multiplying functions together. If you are multiplying by $\geq 2$ terms, do not forget parenthesis.
When you cancel a common factor in a rational function, you must exclude the zero of that factor from the domain.
Such a factor is called a hole.
Redefine the rational function $f(x) = \dfrac{x^2 + 2x + 1}{x+1}$ with the domain exclusion.
Let $f(x) = x^2 + x$ and $g(x) = -x^2 - x$. Find the following:
$f+g, f-g, fg, \frac{f}{g}$ and their domains
$\left(\frac{f}{g}\right)(0)$
Let's first find the domains and intersect them. Both functions have domain $\mathbb{R}$.
Following the above table, we see that
\begin{align}
(f+g)(x) &= f(x) + g(x) &&= x^2 + x + -x^2 - x = 0 \\
(f-g)(x) &= f(x) - g(x) &&= x^2 + x - (-x^2 - x) = x^2 + x + x^2 + x = 2x^2 + 2x \\
(f\cdot g)(x) &= f(x)\cdot g(x) &&= (x^2 + x)(-x^2 - x) = (x^2 + x)(-x^2) + (x^2 + x)(-x) = -x^4 - x^3 - x^3 - x^2 = -x^4 - 2x^3 - x^2 \\
\left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)} &&= \frac{x^2 + x}{-x^2 -x} = \frac{x(x+1)}{-x(x + 1)} = \frac{1}{-1} = -1 , \ x \neq 0, x \neq -1 \\
\end{align}
To be clear, \[\dfrac{f(x)}{g(x)} = -1 , x \neq 0, x \neq -1\] Notice the domain exclusion (caused by cancelling factors) where your input $x$ cannot be $0$ nor $-1$.
Since $\mathbb{R} \cap \mathbb{R} = \mathbb{R}$, all functions have domain $\mathbb{R}$.
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 Composition
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 $f(x) = \frac{x}{x+1}, g(x) = x^{10}, h(x) = x+3$.
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 \[(f\circ g)(x) = f(g(x)) = f(x+9) = \sqrt[4]{x+9} = F(x)\]