Rational Function

Quotients of Polynomials

A rational function is a quotient of two polynomials. Its general form is

In this form, $P(x)$ and $Q(x)$ are polynomials, and $Q(x)$ is not the zero polynomial.

The denominator adds one essential rule: every input that makes $Q(x)=0$ must be excluded. That restriction belongs to the function even when a common factor later cancels.

A Rational Model from Area

Suppose Mr. Budi is planning a rectangular chicken coop with area $100\text{ m}^2$. Let its length be $x\text{ m}$ and its width be $w\text{ m}$. Then

The formula is rational because a variable appears in the denominator. In the physical situation, both dimensions must be positive, so the meaningful domain is $x>0$. The condition $x\neq 0$ still includes negative lengths that do not fit the situation.

The model also shows how the two dimensions depend on each other: making the coop longer forces its width to become smaller if the area must stay fixed.

Common Forms

A rational function is a quotient of two polynomials, and a few shapes appear often. The degree of each polynomial decides which shape you get and which features, such as a horizontal asymptote or a hole, are worth looking for.

Reciprocal Form

The basic reciprocal form is

In this form, $k$ is a constant. For example, $f(x)=\frac{5}{x}$ has domain $x\neq 0$.

Linear over Linear Form

When the numerator and denominator are linear,

In this form, the denominator must satisfy $cx+d\neq 0$. One example is $f(x)=\frac{2x+3}{x-1}$, whose domain excludes $x=1$.

Higher Degree Form

Either polynomial may have degree two or higher. For example,

This function is still rational, and its domain excludes $x=3$.

Finding the Domain

To find the real domain of a rational function:

1. Set the original denominator equal to zero.
2. Solve for every value that makes it zero.
3. Exclude those values from the real numbers.

The three steps apply to a rational function whose denominator can be zero:

Its denominator is zero when:

The calculation gives:

Simplifying Without Losing the Domain

You may cancel a common factor only where that factor is nonzero. The simplified expression gives the same output on the original domain, but it does not restore an excluded input.

Removing a Common Power of the Variable

Simplify

The original denominator requires $x\neq 0$. On that domain,

Although the expression $2x$ is defined at $x=0$, the original rational function is not.

Canceling After Factoring

Simplify

First record $x\neq 2$, then factor the difference of squares:

The graph agrees with the line $y=x+2$ except for a hole at $x=2$.

Operations on Rational Expressions

Adding, subtracting, multiplying, and dividing rational expressions follows the rules for fractions. The extra step is the domain, because a factor that cancels from the numerator and the denominator still removes one value from the allowed inputs.

Addition and Subtraction

Use a common denominator and keep every restriction from the original denominators.

Multiplication

Multiply numerators and denominators, then factor before canceling:

The canceled factor $x$ does not remove the original restriction $x\neq 0$.

Division

Dividing by a fraction means multiplying by its reciprocal:

This requires $b\neq 0$ and $d\neq 0$ so both original fractions exist, and $c\neq 0$ so the divisor is not zero.

Exercises

Each problem gives a rational expression and asks for its domain or its simplified form. Find the excluded values first, then simplify.

1. Determine the domain of $f(x)=\frac{x+3}{x^2-9}$.

2. Simplify $g(x)=\frac{x^2-1}{x+1}$ and retain its original restriction.

3. Calculate $\frac{1}{x-1}-\frac{2}{x+1}$ and state its domain restrictions.

4. A car travels $300\text{ km}$ at an average speed of $v\text{ km/h}$. Write travel time as a function of $v$ and state the physically meaningful domain.

Worked Solutions

1. The numerator does not determine the domain. Factor the denominator:

   $$x^2-9=(x-3)(x+3)$$

   $$\begin{aligned} &(x-3)(x+3) = 0 \\ &\Longrightarrow x = 3 \text{ or } x = -3. \end{aligned}$$

   Both values make the original denominator zero, so

   $$D_f=\{x\in\mathbb{R}\mid x\neq -3,\ x\neq 3\}.$$

2. The original denominator gives $x\neq -1$. Factor and cancel only on that domain:

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

   $$=x-1,\qquad x\neq -1.$$

   The simplified formula still has a hole at the excluded input $x=-1$.

3. The original denominators require $x\neq 1$ and $x\neq -1$. Using $(x-1)(x+1)$ as the common denominator gives

   $$\frac{1}{x-1}-\frac{2}{x+1}$$

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

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

   $$=\frac{3-x}{(x-1)(x+1)},\qquad x\neq -1,1.$$

4. Time equals distance divided by speed:

   $$\text{time}=\frac{\text{distance}}{\text{speed}}$$

   $$t(v)=\frac{300}{v}.$$

   Algebraically, $v\neq 0$. For a moving car, speed must be positive, so the physical model uses $v>0$.