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Reed College Department of Mathematics & Statistics

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Worksheet Rational Functions

1.

For each function below:
  1. Determine the values of x for which the rational expression is undefined. If there is more than one value, enter your answers as a comma separated list. If there is no value that makes the expression undefined enter “NONE”.
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  2. Find the domain of the function. Write your answer in interval notation. Use INF for \(\infty\text{.}\)
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\(f(x) = {\frac{3}{5x-7}}\)
  1. The expression is undefined when \(x =\)
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  2. The domain of \(f\) is
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\(g(x) = {\frac{-4x}{x^{2}+9}}\)
  1. The expression is undefined when \(x =\)
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  2. The domain of \(g\) is
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Answer 1.
\(1.4\)
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Answer 2.
\(\left(-\infty ,1.4\right)\cup \left(1.4,\infty \right)\)
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Answer 3.
\(\text{NONE}\)
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Answer 4.
\(\left(-\infty ,\infty \right)\)
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Solution.
Division by zero is undefined, so to determine where a rational expression is undefined set the denominator = 0.
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For \(f(x)\text{:}\)
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\(5x - 7=0\)
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\(5x = + 7\)
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\(x = {1.4}\)
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The domain is the set of all numbers where the function is NOT undefined. That would be every number except for \(x = {1.4}\text{.}\) In interval notation this is \({\left(-\infty ,1.4\right)\cup \left(1.4,\infty \right)}\text{.}\)
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We can do the same thing for the 2nd function, \(g(x)\text{.}\) Start by setting the denominator to zero:
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\(x^2 + 9 = 0\)
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\(x^2 = -9\)
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There is no number that can be squared to give \(-9\text{.}\) Or if you try to solve for x by taking the square root of both sides, you won’t get a real solution because we can’t take the square root of a negative number. That means there are no values of x that make the expression undefined and the domain is all real numbers or \({\left(-\infty ,\infty \right)}\text{.}\)
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2.

Solve the rational equation. Enter the answer as a reduced fraction. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\(\dfrac{1}{a}+\dfrac{6}{9}=\dfrac{1}{6}\)
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\(a\) =
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Hint.
Try clearing the fractions by multiplying every term by the least common denominator.
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In this problem, LCD = \(6\cdot 9 \cdot a\)
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Answer.
\(-2\)
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3.

Solve the rational equation. Enter the answer as a reduced fraction. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\(\dfrac{11}{5}+\dfrac{1}{11}=\dfrac{2}{v}\)
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\(v\) =
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Answer.
\({\frac{55}{63}}\)
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Solution.
First, clear the fractions by multiplying every term by the LCD. The LCD = \({55v}\text{.}\)
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\(\displaystyle{\frac{11}{5} \cdot {55v} + \frac{1}{11} \cdot {55v} = \frac{2}{v} \cdot {55v}}\)
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\(\displaystyle{{121v+5v} = 110}\)
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\(126v = 110\)
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\(v = {{\frac{55}{63}}}\)
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4.

Solve the rational equation. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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1+\(\dfrac{7}{p}=-\dfrac{10}{p^2}\)
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\(p\) =
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Answer.
\(-2, -5\)
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5.

Solve the rational equation. Enter the answer as a reduced fraction. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\(\dfrac{6}{z-11}-\dfrac{4}{z+11}=\dfrac{5}{z^2-121}\)
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\(z\) =
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Hint.
Clear the fractions by multiplying every term by the LCD.
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The LCD is \((z - 11)(z + 11)\)
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Answer.
\(-{\frac{105}{2}}\)
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Solution.
Clear the fractions by multiplying every term by the LCD.
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The LCD is \((z - 11)(z + 11)\)
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\(\displaystyle{\frac{6}{z-11} \cdot (z-11)(z+11)-\frac{4}{z+11}\cdot (z-11)(z+11)=\frac{5}{z^2-121}\cdot (z-11)(z+11)}\)
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\(6(z+11) - 4(z-11) = 5\)
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\({6z+66-4z+44} = 5\)
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\({2z+110} = 5\)
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\({2z} = -105\)
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\(z = \frac{-105}{{2z}}\)
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Finally, make sure the answer doesn’t give you zero in any of the denominators. In this case, the solution is:
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\(z = {-{\frac{105}{2}}}\)
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6.

Solve the rational equation. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\(\dfrac{v-10}{v^2-6v+8} = \dfrac{4}{v-2} - \dfrac{5}{v-4}\)
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\(v\) =
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Answer.
\(\text{NONE}\)
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Solution.
The LCD is \({\left(v-2\right)\mathopen{}\left(v-4\right)}\text{.}\) Multiply every term by the LCD to clear the fractions:
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\(\dfrac{v-10}{v^2-6v+8} \cdot {\left(v-2\right)\mathopen{}\left(v-4\right)} = \dfrac{4}{v-2} \cdot {\left(v-2\right)\mathopen{}\left(v-4\right)} - \dfrac{5}{v-4}\cdot {\left(v-2\right)\mathopen{}\left(v-4\right)}\)
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\({v-10} = {4\mathopen{}\left(v-4\right)-5\mathopen{}\left(v-2\right)}\)
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\({v-10} = {4v-16-5v+10}\)
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\({v-10} = {-1v+\left(-6\right)}\)
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\(2v = 4\)
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\(v = 2\)
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Finally, make sure your solution doens’t give give zeros in any of the denominators. This means \(v \ne 4\) and \(v \ne 2\text{.}\)
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\(v = NONE\)
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7.

Solve the rational equation. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\(\dfrac{b+3}{3b}+\dfrac{b}{15}=\dfrac{1}{b}\)
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\(b\) =
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Answer.
\(-5\)
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8.

Solve the following equation for \(p\text{.}\)
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\(\displaystyle{2/q+1/5 = {\frac{3}{p}}}\)
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\(p\) =
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Answer.
\(\frac{15q}{10+q}\)
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9.

Solve the following equation for \(w\text{.}\)
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\(\displaystyle{{\frac{a+3}{w-1}} = 1/2}\)
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\(w\) =
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Answer.
\(2a+7\)
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10.

Solve the following equation for \(r\text{.}\)
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\(\displaystyle{{\frac{I}{r}} = {P}}\)
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\(r\) =
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Answer.
\(\frac{I}{P}\)
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11.

Solve the rational equation. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\(1+\dfrac{-1}{y}=-\dfrac{-12}{y^2}\)
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\(y\) =
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Answer.
\(4, -3\)
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Solution.
The LCD is \(y^2\text{.}\) We can clear the fractions by multiplying every term by the LCD:
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\(1 \cdot y^2 +\dfrac{-1}{y} \cdot y^2 =-\dfrac{-12}{y^2} \cdot y^2\)
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\(y^2 - 1y = + 12\)
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\(y^2 - 1y - 12 = 0\)
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\((y -4)(y + 3) = 0\)
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\(y = 4, y = -3\)
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12.

Solve the rational equation. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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\({11-\frac{6p}{p+3}} = {\frac{18}{p+3}}\)
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\(p\) =
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Answer.
\(\text{NONE}\)
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Solution.
Multiply every term by the least common denominator. The LCD is \({p+3}\text{.}\)
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\(\displaystyle{\left( {p+3} \right) \left( {11-\frac{6p}{p+3}} \right) = \left( {p+3} \right) \cdot {\frac{18}{p+3}}}\)
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\({11\mathopen{}\left(p+3\right)-6p} = 18\)
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\({11p-6p+33} = 18\)
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\({5p+33} = 18\)
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\(5p = -15\)
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\(p = -3\)
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Notice that if we plug \(p = -3\) into the orignal equation, we get get zeros in the denominator. Since division by zero is undefined, this equation has no solution.
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\(p = NONE\)
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13.

Let \(\displaystyle{ f(x)={-\frac{9}{x-10}} }\text{.}\) Find the domain, vertical asymptote(s), and horizontal asymptote.
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If the video does not work, click here to watch on YouTube.
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Answer 1.
\(\left(-\infty ,10\right)\cup \left(10,\infty \right)\)
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Answer 2.
\(\text{x=}\)
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Answer 3.
\(10\)
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Answer 4.
\(\text{y=}\)
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Answer 5.
\(0\)
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14.

Let \(\displaystyle{ f(x)={\frac{x^{2}+2}{x^{2}-11x+18}} }\text{.}\) Find the domain, vertical asymptote(s), and horizontal asymptote.
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Hint.
A vertical asymptote of a graph is a vertical line \(x=a\) where the graph tends toward positive or negative infinity as the inputs approach \(a\text{.}\)
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A horizontal asymptote of a graph is a horizontal line \(y=b\) where the graph approaches the line as the inputs increase or decrease without bound.
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Steps to identifying vertical and horizontal asymptotes of rational functions:
  1. Factor the numerator and denominator.
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  2. Note any restrictions in the domain of the function.
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  3. Reduce the expression by canceling common factors in numerator and denominator if any.
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  4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
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  5. Note any restrictions in the domain where asymptotes do not occur. Thes are removable discontinuities, or ‘holes’ in the graph.
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  6. The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator:
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  • if the degree of the numerator is less than the degree of the denominator then the horizontal asymptote is \(y=0\)
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  • if the degree of the numerator is greater than the degree of the denominator by one then there is no horizontal asymptote and there is a slant asymptote.
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  • if the degree of the numerator is equal to the degree of the denominator then the horizontal asymptote is \(y=\frac{a}{b}\text{,}\) the ratio of leading coefficients.
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Answer 1.
\(\left(-\infty ,2\right)\cup \left(2,9\right)\cup \left(9,\infty \right)\)
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Answer 2.
\(\text{x=}\)
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Answer 3.
\(2, 9\)
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Answer 4.
\(\text{y=}\)
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Answer 5.
\(1\)
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15.

Let \(\displaystyle{ f(x)={\frac{x^{2}-16}{x+4}} }\text{.}\) Identify the removable discontinuity.
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  • Removable Discontinuity: \(x=\)
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Answer.
\(-4\)
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16.

Let \(\displaystyle{ f(x)={\frac{x^{2}-8x-20}{x+2}} }\text{.}\) Identify the removable discontinuity.
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  • Removable Discontinuity: \(x=\)
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Answer.
\(-2\)
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17.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{x + {6}}{x + {8}} > 0 }\)
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Answer.
\(\left(-\infty ,-8\right)\cup \left(-6,\infty \right)\)
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18.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{{4}x}{x - {6}} \lt -{2} }\)
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Answer.
\(\left(2,6\right)\)
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19.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{{6}x}{x - {8}} > -{2} }\)
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Hint.
It may be helpful to rewrite the inequality so that you have a single rational expression isolated and zero on the other side of the inequality.
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Example: Solve the rational inequality \(\dfrac{x}{x-2} >3\)
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Solution:
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\(\displaystyle{\begin{aligned} \dfrac{x}{x-2} \amp >3\\ \dfrac{x}{x-2} -3\amp >0\\ \dfrac{x}{x-2}-\dfrac{3(x-2)}{x-2}\amp >0\\ \dfrac{-2x+6}{x-2}\amp >0 \end{aligned} }\)
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The rational expression on the left side of this inequality is undefined when \(x=2\) and is equal to zero when \(x=3\text{.}\)
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We now want to find when the rational expression is positive and when it is negative.
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Looking at the interval \((-\infty, 2)\text{,}\) we can see that those \(x\) values make the rational expression \(\dfrac{-2x+6}{x-2}\) negative.
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Looking at the interval \((2, 3)\text{,}\) we can see that those \(x\) values make the rational expression \(\dfrac{-2x+6}{x-2}\) positive.
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Looking at the interval \((2, \infty)\text{,}\) we can see that those \(x\) values make the rational expression \(\dfrac{-2x+6}{x-2}\) negative.
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Since we are looking for\(\dfrac{-2x+6}{x-2}>0\text{,}\) or where the rational expression is positive, then we can see the solution is \((2,3)\text{.}\)
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Answer.
\(\left(-\infty ,2\right)\cup \left(8,\infty \right)\)
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20.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{1}{{5}} - \dfrac{ {15} }{x^2} \le \dfrac{ {2} }{x} }\)
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Answer.
\(\left[-5,0\right)\cup \left(0,15\right]\)
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