Calculus/Rational functions

Template:Calculus/Top Nav Rational function is "any function which can be defined by a rational fraction, i.e. an algebraic fraction such that both the numerator and the denominator are polynomials".

It can be proved that sum, product, and quotient (except division by the zero polynomial which will cause the function to be undefined) of rational functions are rational functions.

<quiz display=simple> {Example. Define $f_1(x)=x^2/x$, $f_2(x)=x$, $g(x)=\pi$, and $h(x)=\tan x$.}

{Select all rational functions in the following options. |type="[]"} + $f_1(x)$

+ $f_2(x)$

+ $g(x)$ || The value of $g(x)$ is irrational does not imply that $g(x)$ is not rational function. Indeed, it is constant function and thus is rational function.

- $h(x)$ || We cannot express this function as the fraction of two polynomials. Even if we express $h(x)$ as $\tan x/1$, the numerator is || still not a polynomial.

{Is $f_1(x)=f_2(x)$? |type="()"} - yes || What are their domains? + no || Although $x^2/x=x$, their domains are different, because domain of $f_1(x)$ is set of all nonzero real numbers, and domain of $f_2(x)$ is set of all real numbers. Therefore, they are not equal.

{Select all possible expressions of $g(x)$ in the form of $P(x)/Q(x)$ in which $P(x),Q(x)$ are polynomial functions. |type="[]"} - $g(x)$ is not rational function. Therefore, there are no possible expressions. + $g(x)=3\pi /3$ + $g(x)=\pi /1$ - $g(x)=\pi x/x$ || The function $x\mapsto \pi x/x$ does not have the same domain as $g(x)$. Therefore, we cannot express in this way. - $g(x)=0\pi /0$ || $0/0$ is undefined. Therefore, this is an invalid expression. + $g(x)={\pi }^2/\pi$ </quiz>

Examples

The rational function $f(x)=\frac{x^3-2x}{2(x^2-5)}$ is not defined at $x^2=5\iff x=\pm \sqrt{5}$. It is asymptotic to $y=\frac{x}{2}$, i.e. gets closer and closer to $y=\frac{x}{2}$, as $x$ approaches positive or negative infinity.

The rational function $f(x)=\frac{x^2+2}{x^2+1}$ is defined for all real numbers, but not for all complex numbers, since if $x$ were a square root of $-1$ (i.e. the imaginary unit or its negative), then formal evaluation would lead to division by zero: $f(i)=\frac{i^2+2}{i^2+1}=\frac{-1+2}{-1+1}=\frac{1}{0}$, which is undefined.

Every polynomial function $f(x)=P(x)$ is a rational function with $Q(x)=1$. A function that cannot be written in this form, such as $f(x)=\sin (x)$, is not a rational function. The adjective "irrational" is not generally used for functions.

Sketch a graph of a rational function

(1)Let's sketch the graph of $y=\frac{1}{x}$.
First, we must avoid $x=0$ because anything can not be divided by 0. Thus x should not be 0 in the equation. Now we just plug in some values of x. The result is as follows:

$x=1$	$y=1$
$x=2$	$y=\frac{1}{2}$
$x=3$	$y=\frac{1}{3}$
$x=-3$	$y=-\frac{1}{3}$
$x=-2$	$y=-\frac{1}{2}$
$x=-1$	$y=-1$

As x get large the function itself gets smaller and smaller. Here is the graph of $\frac{1}{x}$.

References

• Wikipedia article on rational function
• Paul's Online Math Notes"Rational Functions"