\(f(x) = 0\) \(f(x) = \frac {3x+1}{x^2-4x+5}\) \(f(x)=e^x\) \(f(x)=\frac {\sin (x)}{\cos (x)}\) \(f(x) = -4x^{-3}+5x^{-1}+7-18x^2\) \(f(x) = x^{1/2}-x +8\) \(f(x)=\frac {\sqrt {x}}{x^3-x}\)
All
polynomials can be thought of as rational functions.
Rational functions are functions defined by fractions of polynomials.
1 What are rational functions?
In algebra and calculus, polynomials play the same role as the integers do in arithmetic. We add them, subtract them, multiply them, and factor them. We cannot divide them, however, if we only allow integer results. (Since \(4\) is not a factor of \(7\), the quotient \(\dfrac {7}{4}\) is not an integer.) If we want to be able to divide integers, we end up with rational numbers, which are fractions \(\dfrac {p}{q}\) where \(p\) and \(q\) are integers, and \(q \ne 0\).
The same idea holds for polynomials. We can add them, subtract them, multiply them, and factor them. However, to divide them we have to start considering rational functions.
A rational function in the variable \(x\) is a function which can be written in
the form
\[ f(x) = \frac {p(x)}{q(x)} \]
where \(p\) and \(q\) are polynomial functions. The domain of a rational
function is all real numbers except for where the denominator is equal to zero.
If \(p\) is a polynomial function, then it can be written in the form \(\dfrac {p(x)}{1}\), showing that any polynomial function is also a rational function. Since a rational function has zeroes only where the numerator polynomial has zeroes (as long as the fraction is written in lowest terms), functions like \(\sin (x)\) which have infinitely many zeroes cannot be rational functions.
Which of the following are rational functions?
2 What can the graphs look like?
There is a somewhat wide variation in the graphs of rational functions.
Here we see the the graphs of four rational functions.
Match the curves \(A\), \(B\), \(C\), and \(D\) with the functions
![[Picture]](digInRationalFunctions.online-54d843932ba8b66fc0e51eb2e7418bdb.svg) |
![[Picture]](digInRationalFunctions.online-57ce45fe7493dd7277a11b9ee2d59633.svg) |
![[Picture]](digInRationalFunctions.online-3f1c3b2b42e03473213011a9dfd325f9.svg) |
![[Picture]](digInRationalFunctions.online-953df98568310d56c1ee2ceb8237942b.svg) |
\begin{align*} &\frac {x^2-3x+2}{x-2}, &&\frac {x^2-3x+2}{x+2}, \\ &\frac {x-2}{x^2-3x+2}, &&\frac {x+2}{x^2-3x+2}. \end{align*}
Consider \(\frac {x^2-3x+2}{x-2}\). This function is undefined only at \(x=2\). Of the curves that we see above, \(\answer [given]{D}\) is
undefined exactly at \(x=2\).
Now consider \(\frac {x^2-3x+2}{x+2}\). This function is undefined only at \(x=-2\). The only function above that undefined exactly at \(x=-2\) is curve \(\answer [given]{A}\).
Now consider \(\frac {x-2}{x^2-3x+2}\). This function is undefined at the roots of
\[ x^2-3x+2 = (x-2)(x-1). \]
Hence it is undefined at \(x=2\)
and \(x=1\). It looks like both curves \(B\) and \(C\) would work. Distinguishing between these two
curves is easy enough if we evaluate at \(x=-2\). Check it out. \begin{align*} \eval {\frac {x-2}{x^2-3x+2}}_{x=-2} &= \frac {-2-2}{(-2)^2-3(-2)+2}\\ &= \frac {-4}{4+6+2}\\ &=\frac {-4}{12}. \end{align*}
Since this is negative, we see that \(\frac {x-2}{x^2-3x+2}\) corresponds to curve \(\answer [given]{B}\).
Finally, it must be the case that curve \(\answer [given]{C}\) corresponds to \(\frac {x+2}{x^2-3x+2}\). We should note that if this function is evaluated at \(x=-2\), the output is zero, and this corroborates our work above.