Function Composition

Assume that we have a function that defines the relationship between $a$ and $b$ and another function that expresses the relationship between $b$ and $c$. We can use function composition to create a brand new function using these functions to express a relationship between $a$ and $c$. This is one of the main ways to combine functions together.

Algebraic Operations

Besides function composition, we can combine functions using algebraic operations. Given the functions $f(x)$ and $g(x)$, we can combine them using these operations...

1. $(f + g)(x) = f(x) + g(x)$
2. $(f - g)(x) = f(x) - g(x)$
3. $(fg)(x) = f(x)g(x)$
4. $(\frac{f}{g})(x) = \frac{f(x)}{g(x)} \:\text{where}\: g(x) \neq 0$

An example of combining function using algebraic operation includes...

$\underline{Example}$

Let $f(x) = x - 1$ and $g(x) = x^2 - 1$, find $(f - g)(x)$.

$(f - g)(x) = f(x) - g(x)$

$(f - g)(x) = (x - 1) - (x^2 - 1)$

$(f - g)(x) = x - 1 - x^2 + 1$

$(f - g)(x) = -x^2 + x$

Composition of Functions

As discussed before, another way to combine functions is through function composition. Given the functions $f(x)$ and $g(x)$, the function composition is $(f \circ g)(x) = f(g(x))$. The left side of the equation is read as $f$ composed with $g$ at $x$, and the right side of the equation is read as $f$ of $g$ of $x$.

The way $(f \circ g)(x)$ works is that the $x$ is the input of $g$ and the output of $g$ becomes the input of $f$. So, to evaluate $(f \circ g)(x)$ for a specific value of $x$, we substitute the value into $g(x)$ and solve for $g(x)$. After this we substitute the resulting value of $g(x)$ into $f(x)$ and solve for $f(x)$. This is the solution for $(f \circ g)(x)$.

On the other hand, to evaluate $(f \circ g)(x)$ for all values of $x$ we substitute the expression for $g(x)$ into $f(x)$ and simplify.

$\underline{Example}$

Let $f(x) = 3x + 1$ and $g(x) = 3 - x$, find $f(g(x))$ and $g(f(x))$.

$f(g(x)) = 2(g(x)) + 1 = 2(3 - x) + 1 = 6 - 2x + 1 = -2x + 7$

$g(f(x)) = 3 - (f(x)) = 3 - (2x + 1) = 3 - 2x - 1 = -2x + 2$

$(f \circ g)(x) = -2x + 7$ and $(g \circ f)(x) = -2x + 2$

Domain

The domain of $f(g(x))$ is the set of inputs $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$.

To determine the domain of $f(g(x))$, we first find the domain of $g$ and $f$. We then exclude any inputs of $x$ for which $g(x)$ is not in the domain of $f$. The remaining domain of $g$ is the domain of the entire composition.

$\underline{Example}$

Find the domain of $(f \circ g)(x)$ where $f(x) = \frac{5}{x - 1}$ and $g(x) = \frac{4}{3x - 2}$.

The domain of $g(x)$ is all real numbers except for when $3x - 2 = 0$ which is $x = \frac{2}{3}$.

The domain of $f(x)$ is all real numbers except for when $x - 1 = 0$ which is $x = 1$.

We need to find out when $g(x) = 1$ and exclude it from the domain.

$\frac{4}{3x - 2} = 1 \to 4 = 3x - 2 \to 6 = 3x \to x = 2$

This means our domain is all real numbers excluding $x = \frac{2}{3}, 2$. In other words, $(-\infty, \frac{2}{3}) \cup (\frac{2}{3}, 2) \cup (2, \infty)$.

Decomposition

In some cases, we can decompose a complicated function into a composition of two simpler functions. Note that there are various compositions a function can have. An example of a decomposition is...

$\underline{Example}$

Write $f(x) = \frac{4}{3 - \sqrt{4 + x^2}}$ as the composition of two functions.

Let $h(x) = \sqrt{4 + x^2}$ and $g(x) = \frac{4}{3 - x}$. Then $f(x) = (g \circ h)(x) = g(h(x))$.