Rational functions

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\CancelColor }[0]{} \newcommand {\cancelto }[0]{1} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

Rational functions are functions defined by fractions of polynomials.

What are rational functions?

A rational function in the variable $x$ is a function 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.

Which of the following are rational functions?

$f(x) = 0$ $f(x) = \frac {3x+1}{x^2-4x+5}$ $f(x)=e^x$ $f(x)=\frac {\sin (x)}{\cos (x)}$ $f(x) = -4x^{-3}+5x^{-1}+7-18x^2$ $f(x) = x^{1/2}-x +8$ $f(x)=\frac {\sqrt {x}}{x^3-x}$

All polynomials can be thought of as rational functions.

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 $\begin{align*} &\frac{x^2-3x+2}{x-2}, &&\frac{x^2-3x+2}{x+2}, \\ &\frac{x-2}{x^2-3x+2}, &&\frac{x+2}{x^2-3x+2}. \end{align*}$

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.