Definition and key concepts
Definition
Rational function
A rational function is any function of the following form:
q\left(x\right)=\dfrac{f\left(x\right)}{g\left(x\right)}
f\left(x\right) and g\left(x\right) are polynomial functions, provided that g\left(x\right) is not the zero polynomial.
The following is a rational function:
q\left(x\right)=\dfrac{x^2-1}{x^2+1}
The following is an example of a rational function:
f\left(x\right)=\dfrac{1}{x}
The following is a rational function:
q\left(x\right)=\dfrac{x^3-1}{x}
Observe that every polynomial function is a rational function.
The following is a rational function:
q\left(x\right)=x^2-13x=\dfrac{x^2-13x}{1}
A linear function is a special type of polynomial function. Therefore every linear function is a rational function.
The following function is a linear function, therefore it is a rational function as well:
f\left(x\right)=7x-12=\dfrac{7x-12}{1}
Domain and range, excluded values
The domain of a rational function
Let f\left(x\right) and g\left(x\right) be polynomial functions and consider the rational function:
q\left(x\right)=\dfrac{f\left(x\right)}{g\left(x\right)}
Then the domain of q\left(x\right) is all real numbers except the real numbers which make g(x)=0.
Consider the following rational function:
q\left(x\right)=\dfrac{x-1}{x}
The domain of q\left(x\right) is all real numbers except 0.
Excluded values
The numbers which are not in the domain of a rational function are the excluded values of that rational function.
Consider the following rational function:
q\left(x\right)=\dfrac{x}{x-1}
q\left(x\right) has one excluded value, namely 1.
Finding the range of rational function can be tricky and sometimes impossible. Determining when it is impossible to determine the range of a rational function can also be extremely difficult, if not impossible. However, below is an explanation of how to find the range of a rational function which has an inverse.
Calculation rules and simplification of rational expressions
Let a, b and i be real numbers, with a and c nonnegative.
A fraction of the form \dfrac{ab}{ac} can be simplified by canceling the a. The same rules apply when simplifying rational functions.
When x is nonnegative, the rational expression:
\dfrac{2x^3}{4x}
Simplifies as follows:
\dfrac{2x^3}{4x}=\dfrac{x^3}{2x}=\dfrac{x^2}{2}\\
One must find the domain of a rational function before simplifying. Simplifying a rational function first and then finding the domain of the simplified rational function will lead to missing excluded values.
Consider the following rational function:
f\left(x\right)=\dfrac{x^2-1}{x^2+x-2}
f(x) simplifies as follows:
f\left(x\right)=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+2\right)}\\=\dfrac{x+1}{x+2}
However, the domain of f\left(x\right) is all real numbers except 1 and -2 because these are the two numbers which make the original denominator of f\left(x\right) equal to 0.
If \dfrac{a}{b} and \dfrac{c}{d} are two rational numbers, then their sum or difference is found by finding a common denominator and then adding the numerators.
Similar rules apply when adding or subtracting rational functions.
Consider the following rational functions:
- f\left(x\right)=\dfrac{1}{x}
- g\left(x\right)=\dfrac{1}{x-1}
The sum f\left(x\right)+g\left(x\right) is simplified as follows:
f\left(x\right)+g\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x-1}\\=\dfrac{x-1}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}\\=\dfrac{x-1+x}{x\left(x-1\right)}\\=\dfrac{2x-1}{x\left(x-1\right)}
Consider the following rational functions:
- f\left(x\right)=\dfrac{1}{\left(x-1\right)\left(x+1\right)}
- g\left(x\right)=\dfrac{2}{\left(x-1\right)\left(x+2\right)}
A common denominator for these two rational functions is the following polynomial:
\left(x-1\right)\left(x+1\right)\left(x+2\right)
Therefore the difference is computed as follows:
\begin{aligned}f\left(x\right)-g\left(x\right)&=\dfrac{1}{\left(x-1\right)\left(x+1\right)}-\dfrac{2}{\left(x-1\right)\left(x+2\right)} \\ &= \dfrac{x+2}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x-1\right)\left(x+2\right)} \\ &= \dfrac{x+2-2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)} \\ &= \dfrac{-x}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\end{aligned}
Graphic representation
Vertical asymptote
A vertical asymptote of a rational function f\left(x\right) is a vertical line that the graph of f\left(x\right) approaches as the line goes to infinity but never touches.
Consider the following rational function:
f\left(x\right)=\dfrac{x}{\left(x-1\right)\left(x+2\right)}
It has two vertical asymptotes. Their equations are x=1 and x=-2.
The graph of a rational function can have a vertical asymptote around its excluded values.
The domain of the following rational function is all real numbers except 1:
f\left(x\right)=\dfrac{1}{x-1}
The graph of f\left(x\right) is the following:
The graph of f\left(x\right) comes close but never crosses the vertical line with equation x=1. The line x=1 is a vertical asymptote of the rational function.
Removable point
A removable discontinuity of a rational function is a point on the graph which is in the domain of the reduced form of the rational function.
Consider the followiong rational function:
f\left(x\right)=\dfrac{x}{x\left(x+1\right)}
It has a removable discontinuity at x=0 because the rational function simplifies as follows:
f\left(x\right)=\dfrac{x}{x\left(x+1\right)}=\dfrac{1}{x+1}
and 0 is in the domain of the reduced rational function but not the original rational function.
The graph of f\left(x\right) will look like the following:
Observe that the graph has a hole at x=0. For this reason, removable discontinuities are sometimes called holes.
Also observe that the graph of f\left(x\right) still has a vertical asymptote of x=-1. This is because the factor \left(x+1\right) in the denominator cannot be canceled with a factor in the numerator.
The inverse of a rational function
Definition
The graphs of some rational functions pass the horizontal line test. Therefore it is possible to find the inverse function of some rational functions.
Consider the rational function:
f\left(x\right)=\dfrac{1}{x}
The graph of f\left(x\right) passes the horizontal line test. Therefore f\left(x\right) has an inverse function.
When a rational function f\left(x\right) has an inverse, it is possible to find the inverse by solving the equation for y:
x=f\left(y\right)
Consider the following function:
f\left(x\right)=\dfrac{1}{x}
The inverse of f\left(x\right) is found by solving the following equation for y :
x=\dfrac{1}{y}\\xy=1\\y=\dfrac{1}{x}
Therefore the inverse of y=\dfrac{1}{x} is itself.
Domain and range
Suppose f\left(x\right) is a function which has inverse f^{-1}\left(x\right). Then:
- The domain of f^{-1}\left(x\right) is the range of f\left(x\right).
- The range of f^{-1}\left(x\right) is the domain of f\left(x\right).
Consider the rational function:
f\left(x\right)=\dfrac{1}{x}
The inverse of f\left(x\right) is itself. Therefore the domain and range of f\left(x\right) are the same and they are both:
\left(-\infty, 0\right)\cup \left(0,\infty\right)
It is now possible to find the range of a rational function using its inverse.
Let's find the range of the following function:
y=\dfrac{1}{x-1}
The range of a rational function is the domain of its inverse. Let's first find the inverse function.
To find the inverse we solve the following equation for y :
x=\dfrac{1}{y-1}\\\left(y-1\right)x=1\\y-1=\dfrac{1}{x}\\y=\dfrac{1}{x}+1\\y=\dfrac{1+x}{x}
The domain of this rational function is all real numbers except 0.
Therefore the range of the function y=\dfrac{1}{x-1} is all real numbers except 0.
Properties and graphic representation
Suppose f\left(x\right) is a function which has the inverse f^{-1}\left(x\right). Then the graph of f^{-1}\left(x\right) is obtained by reflecting the graph of f\left(x\right) across the line y=x.
In the following example, the graph of a function f\left(x\right) is given in blue and the graph of f^{-1}\left(x\right) is given in red.
Consider the following rational function:
f\left(x\right)=\dfrac{1}{x-1}
The inverse function of f\left(x\right) is the following function:
f^{-1}\left(x\right)=\dfrac{1+x}{x}
In the following image the graph of f\left(x\right) is in blue and the graph of f^{-1}\left(x\right) is given in red.
Rational equations
Rational equation
A rational equation is any equation of the following form:
\dfrac{f\left(x\right)}{g\left(x\right)}=0
f\left(x\right) and g\left(x\right) being polynomial functions.
The following is a rational equation:
\dfrac{x}{x-1}=0
The following is a rational equation:
\dfrac{x^2-2x+1}{x^{17}+11}=0
If \dfrac{a}{b} is a rational number then \dfrac{a}{b}=0 if and only if a=0.
This principal allows us to solve rational equations. Suppose q\left(x\right)=\dfrac{f\left(x\right)}{g\left(x\right)} is a rational function and f\left(x\right) and g\left(x\right) are polynomial functions. Then a real number a is a solution to the rational equation q\left(x\right)=0 if and only if the following two conditions hold:
- f\left(a\right)=0
- g\left(a\right)\not= 0
Consider the following rational equation:
\dfrac{x}{x+1}=0
The numerator is 0 if and only if x=0.
Observe that the denominator of the rational function evaluated at 0 is not 0.
Therefore there is one solution to the rational equation:
x=0
Consider the following equation:
\dfrac{x-1}{x+2}=\dfrac{x-1}{x+1}
The equation is solved as follows:
\dfrac{x-1}{x+2}=\dfrac{x-1}{x+1}\\\dfrac{x-1}{x+2}-\dfrac{x-1}{x+1}=0\\\dfrac{\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}=0\\\dfrac{x^2-1-\left(x^2+x-2\right)}{\left(x+2\right)\left(x-1\right)}=0\\\dfrac{1-x}{\left(x+2\right)\left(x-1\right)}=0
The only possible solution to the rational equation is x=1 since 1 is the only real number which makes the numerator 0.
However, the number 1 also makes the denominator of the rational function 0 and so we cannot include it as part of our solutions.
Therefore there are no solutions to the original equation.