Rational Functions and Domain Restrictions, precalculus lesson

Split a 600 dollar cabin rental evenly among $x$ friends and each person pays $600/x$ dollars. Quotients like this one, a polynomial divided by a polynomial, show up whenever one quantity is shared out or compared with another. You can evaluate them at almost every input, but the few inputs that make the denominator zero need careful bookkeeping.

Key terms

Rational function

Definition A rational function is a function that can be written as

r(x) = \frac{p(x)}{q(x)},

where $p$ and $q$ are polynomials and $q$ is not the zero polynomial. Every polynomial is a rational function, since it can be written with $q(x) = 1$ . When no domain is stated, the domain is the natural domain of the formula, described below.

In plain words A rational function is a fraction with a polynomial on top and a polynomial on the bottom.

Example The functions

\begin{gathered} f(x) = \frac{3x - 1}{x^2 + 4} \\[0.8em] \text{and} \\[0.4em] g(x) = \frac{x^3 - 2x}{x - 5} \end{gathered}

are rational functions. So is $h(x) = 2 + \frac{1}{x}$ , because combining the terms over a common denominator gives $h(x) = \frac{2x + 1}{x}$ . What counts is whether the function can be written as one quotient of polynomials, not how the formula looks at first. The function $\frac{\sqrt{x}}{x + 1}$ is not rational, because no rewriting turns $\sqrt{x}$ into a quotient of polynomials.

Numerator and denominator

Definition In the quotient $\frac{p(x)}{q(x)}$ the polynomial $p$ is the numerator and $q$ is the denominator. For an input $a$ with $q(a) \ne 0$ the value is $r(a) = \frac{p(a)}{q(a)}$ , and it equals $0$ exactly when $p(a) = 0$ . When $q(a) = 0$ the quotient is undefined, whatever $p(a)$ is. This includes the case $p(a) = 0$ , because the expression $\frac{0}{0}$ has no value.

In plain words The top decides where the function is zero and the bottom decides where it is undefined. If both are zero at the same input, the bottom wins and the function is undefined there.

Example Take

f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}.

The numerator is zero at $x = 3$ and at $x = -3$ , and the denominator is zero at $x = 3$ and at $x = -1$ . At $x = -3$ the denominator is $9 + 6 - 3 = 12$ , so $f(-3) = \frac{0}{12} = 0$ . At $x = 3$ both parts are $0$ , so $f(3)$ is undefined, not $0$ . The value $f(-1)$ is undefined as well.

Original domain

Definition The original domain of a rational function $r(x) = \frac{p(x)}{q(x)}$ is its natural domain, read from the formula exactly as given, before any simplification,

\{x \in \mathbb{R} \mid q(x) \ne 0\}.

The real numbers left out are the excluded values. Since $q$ is not the zero polynomial, it has at most $\deg q$ real zeros, so a rational function has only finitely many excluded values.

In plain words Before you touch the formula, find every real number that makes the denominator zero and remove it. What is left is the domain, and later algebra cannot change it.

Example For

g(x) = \frac{x + 2}{x^2 - 5x + 6}

the denominator factors as $(x - 2)(x - 3)$ , so the excluded values are $2$ and $3$ and the original domain is

(-\infty, 2) \cup (2, 3) \cup (3, \infty).

For $k(x) = \frac{x}{x^2 + 1}$ the denominator is at least $1$ for every real $x$ , so nothing is excluded and the domain is $\mathbb{R}$ .

Factors

Definition A polynomial $c$ of positive degree is a factor of a polynomial $p$ when $p = c \cdot s$ for some polynomial $s$ . A common factor of $p$ and $q$ is a polynomial of positive degree that is a factor of both. By the Factor Theorem, $x - a$ is a factor of $q$ exactly when $q(a) = 0$ , so every excluded value $a$ of $\frac{p}{q}$ shows up as a linear factor $x - a$ of the denominator. Every nonconstant polynomial with real coefficients is a constant times a product of linear factors and quadratic factors that have no real zeros.

In plain words Factoring the top and the bottom shows at a glance where each one is zero and which pieces they share.

Example In the quotient

\frac{x^3 - 4x}{x^2 + x - 6}

the numerator factors as $x(x - 2)(x + 2)$ and the denominator as $(x - 2)(x + 3)$ . The common factor is $x - 2$ . The excluded values $2$ and $-3$ come straight from the linear factors of the denominator.

Cancellation

Definition Suppose $p = c \cdot P$ and $q = c \cdot Q$ for a common factor $c$ . At every $x$ with $q(x) \ne 0$ both $c(x)$ and $Q(x)$ are nonzero, so

\frac{p(x)}{q(x)} = \frac{c(x) \, P(x)}{c(x) \, Q(x)} = \frac{P(x)}{Q(x)}.

Cancelling every common factor, as many times as it appears in both, gives the reduced form $\frac{P}{Q}$ , in which $P$ and $Q$ share no factor of positive degree. Only factors cancel. A term that is added to or subtracted from something in the numerator or the denominator cannot be cancelled.

In plain words You may divide the top and the bottom by the same nonzero factor, just as $\frac{6}{15} = \frac{2}{5}$ after dividing out $3$ . Factor first, then cancel whole factors, never single terms.

Example Since $x^2 + x - 6 = (x - 2)(x + 3)$ and $x^2 - 4 = (x - 2)(x + 2)$ ,

\begin{gathered} \frac{x^2 + x - 6}{x^2 - 4} = \frac{x + 3}{x + 2}, \\[0.8em] x \ne 2, \ x \ne -2. \end{gathered}

Crossing out the $x$ in $\frac{x + 3}{x + 2}$ to get $\frac{3}{2}$ would be wrong, because there $x$ is a term, not a factor. At $x = 1$ the fraction equals $\frac{4}{3}$ , not $\frac{3}{2}$ .

Keeping excluded values after simplifying

Definition Two functions are equal when they have the same domain and the same value at every input in that domain. Suppose cancelling the factor $x - a$ turns $r(x) = \frac{p(x)}{q(x)}$ into $\frac{P(x)}{Q(x)}$ and $Q(a) \ne 0$ . Then the reduced form is defined at $a$ while $r$ is not, so they are different functions, and the correct simplified formula carries the restriction along,

r(x) = \frac{P(x)}{Q(x)}, \qquad x \ne a.

Every excluded value of the original domain stays excluded after simplifying, whether or not the new denominator still shows it.

In plain words Simplifying changes how the formula looks, not which inputs are allowed. Write the lost restriction next to the simplified formula so that it does not quietly disappear.

Example Take

\begin{aligned} r(x) &= \frac{x^2 - x - 6}{x^2 - 4} \\[0.8em] &= \frac{(x - 3)(x + 2)}{(x - 2)(x + 2)}. \end{aligned}

The original domain excludes $2$ and $-2$ . Cancelling $x + 2$ gives

r(x) = \frac{x - 3}{x - 2}, \qquad x \ne -2.

The restriction $x \ne 2$ is still visible in the new denominator, but $x \ne -2$ would vanish without the note. The reduced formula gives $\frac{-5}{-4} = \frac{5}{4}$ at $x = -2$ , while $r(-2)$ is undefined.

Removable hole

Definition Let $r = \frac{p}{q}$ have reduced form $\frac{P}{Q}$ , and let $a$ be an excluded value of $r$ with $Q(a) \ne 0$ . Then $r(x) = \frac{P(x)}{Q(x)}$ for every $x \ne a$ close to $a$ , so near $a$ the graph of $r$ is the graph of $\frac{P}{Q}$ with the single point

\left(a, \frac{P(a)}{Q(a)}\right)

missing. This missing point is a removable hole. As $x$ approaches $a$ from either side, the values $r(x)$ approach $\frac{P(a)}{Q(a)}$ . The hole is called removable because a new function, equal to $r$ on its domain and equal to $\frac{P(a)}{Q(a)}$ at $a$ , fills it. That new function is not $r$ . If instead $Q(a) = 0$ , which happens when $x - a$ appears more often in the denominator than in the numerator, there is no hole at $a$ but a vertical asymptote.

In plain words A factor that cancels completely leaves one missing point, while the curve just to its left and right is unbroken. Draw it as an open dot at the height the reduced form gives.

Example For the function $r$ of the previous section the factor $x + 2$ cancels completely, and the reduced form equals $\frac{5}{4}$ at $-2$ . So the graph has a removable hole at $(-2, \frac{5}{4})$ . Nearby values, rounded to four decimals, close in on $1.25$ from both sides.

$x$	$-2.1$	$-2.01$	$-2$	$-1.99$	$-1.9$
$r(x)$	$1.2439$	$1.2494$	undefined	$1.2506$	$1.2564$

In calculus such a point is called a removable discontinuity, and the new function that fills it is a continuous extension.

Check yourself

Find the original domain, the reduced form and any holes of the function below.

f(x) = \frac{x^2 + 3x}{x^2 - 9}

Show the answer

Factoring gives

f(x) = \frac{x(x + 3)}{(x - 3)(x + 3)},

so the original domain excludes $3$ and $-3$ . Cancelling $x + 3$ gives $f(x) = \frac{x}{x - 3}$ for $x \ne -3$ . The reduced form equals $\frac{-3}{-6} = \frac{1}{2}$ at $-3$ , so the graph has a hole at $(-3, \frac{1}{2})$ . At $x = 3$ the reduced denominator is still zero, so there is no hole there.

Does the graph of the function below have a hole at $x = 1$ ?

g(x) = \frac{x - 1}{x^2 - 2x + 1}

Show the answer

No. The denominator is $(x - 1)^2$ , so only one copy of $x - 1$ cancels and $g(x) = \frac{1}{x - 1}$ . The reduced denominator is still $0$ at $x = 1$ , so the graph has a vertical asymptote there, not a hole. Here both formulas exclude only $1$ , so they are the same function and no extra note is needed.

The graph of the function below looks like the line $y = x - 2$ . Does it have an $x$ -intercept?

h(x) = \frac{x^2 - 4x + 4}{x - 2}

Show the answer

No. For $x \ne 2$ the function simplifies to $h(x) = x - 2$ , so its graph is the line $y = x - 2$ with a hole at $(2, 0)$ . The line meets the $x$ -axis only at $x = 2$ , which is not in the domain of $h$ , so the graph of $h$ never meets the $x$ -axis.

Rational Functions and Domain Restrictions

Key terms

Rational function

Numerator and denominator

Original domain

Factors

Cancellation

Keeping excluded values after simplifying

Removable hole

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus