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.
Redefine the rational function $f(x) = \dfrac{x^3 - x}{x}$ with the domain exclusion.
Creating New Functions with Operations
$+, -, \times, \div$ 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.
Let $f(x) = x^2 + x$ and $g(x) = -x^2 - x$. Find the following:
$f(x)+g(x), f(x)-g(x), f(x)g(x), \frac{f}{g}$ and their domains
$\frac{f(0)}{g(0)}$
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 $-1$.
The first three functions have domain $\mathbb{R}$ because you do not divide or have any square roots (sources of problems).
We have \[\dfrac{f(0)}{g(0)} = \dfrac{0^2 + 0}{-0^2 - 0} = \dfrac{0}{0}\] which 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 \[(f\circ g)(x) = f(g(x))\]
The function to the right of the circle, in this case $g(x)$, needs to be the input for the function to the left of the circle.
Suppose $f(x) = x^2$ and $g(x) = x-3$. Find the functions $f\circ g$ and $g \circ f$.
We see that \[(f\circ g)(x) = f(g(x)) = f(x-3) = (x-3)^2\] and \[(g\circ f)(x) = g(f(x)) = g(x^2) = x^2 - 3\]$.
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 $f(x) = \frac{x}{x+1}, g(x) = x^{10}, h(x) = x+3$.
The concept in these last three composition examples will be used extensively in calculus.
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 \[(f\circ g)(x) = f(g(x)) = f(x+9) = \sqrt[4]{x+9} = F(x)\]
Suppose $F(x) = \sin^3(x)$. Find functions $f$ and $g$ where $F = f \circ g$.
Suppose $F(x) = \sin(x^3)$. Find functions $f$ and $g$ where $F = f \circ g$.