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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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\(f(x) = {\frac{2}{-7x-2}}\)
  1. The expression is undefined when \(x =\)
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  2. The domain of \(f\) is
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\(g(x) = {\frac{2x}{x^{2}+36}}\)
  1. The expression is undefined when \(x =\)
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  2. The domain of \(g\) is
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Answer 1.
\(-0.285714\)
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Answer 2.
\(\left(-\infty ,-0.285714\right)\cup \left(-0.285714,\infty \right)\)
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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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\(-7x - 2=0\)
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\(-7x = + 2\)
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\(x = {-0.285714}\)
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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 \(x = {-0.285714}\text{.}\) In interval notation this is \({\left(-\infty ,-0.285714\right)\cup \left(-0.285714,\infty \right)}\text{.}\)
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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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\(x^2 + 36 = 0\)
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\(x^2 = -36\)
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There is no number that can be squared to give \(-36\text{.}\) 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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\(\dfrac{1}{a}+\dfrac{2}{3}=\dfrac{1}{2}\)
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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 = \(2\cdot 3 \cdot a\)
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Answer.
\(-6\)
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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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\(\dfrac{8}{7}+\dfrac{1}{8}=\dfrac{4}{v}\)
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\(v\) =
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Answer.
\({\frac{224}{71}}\)
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Solution.
First, clear the fractions by multiplying every term by the LCD. The LCD = \({56v}\text{.}\)
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\(\displaystyle{\frac{8}{7} \cdot {56v} + \frac{1}{8} \cdot {56v} = \frac{4}{v} \cdot {56v}}\)
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\(\displaystyle{{64v+7v} = 224}\)
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\(71v = 224\)
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\(v = {{\frac{224}{71}}}\)
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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+\(\dfrac{9}{p}=-\dfrac{20}{p^2}\)
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\(p\) =
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Answer.
\(-5, -4\)
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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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\(\dfrac{7}{z-11}-\dfrac{2}{z+11}=\dfrac{5}{z^2-121}\)
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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.
\(-{\frac{94}{5}}\)
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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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\(\displaystyle{\frac{7}{z-11} \cdot (z-11)(z+11)-\frac{2}{z+11}\cdot (z-11)(z+11)=\frac{5}{z^2-121}\cdot (z-11)(z+11)}\)
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\(7(z+11) - 2(z-11) = 5\)
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\({7z+77-2z+22} = 5\)
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\({5z+99} = 5\)
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\({5z} = -94\)
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\(z = \frac{-94}{{5z}}\)
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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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\(z = {-{\frac{94}{5}}}\)
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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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\(\dfrac{v-12}{v^2-6v+8} = \dfrac{2}{v-2} - \dfrac{7}{v-4}\)
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\(v\) =
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Answer.
\(3\)
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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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\(\dfrac{v-12}{v^2-6v+8} \cdot {\left(v-2\right)\mathopen{}\left(v-4\right)} = \dfrac{2}{v-2} \cdot {\left(v-2\right)\mathopen{}\left(v-4\right)} - \dfrac{7}{v-4}\cdot {\left(v-2\right)\mathopen{}\left(v-4\right)}\)
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\({v-12} = {2\mathopen{}\left(v-4\right)-7\mathopen{}\left(v-2\right)}\)
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\({v-12} = {2v-8-7v+14}\)
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\({v-12} = {-5v+6}\)
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\(6v = 18\)
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\(v = 3\)
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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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\(v = 3\)
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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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\(\dfrac{b+4}{4b}+\dfrac{b}{24}=\dfrac{1}{b}\)
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\(b\) =
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Answer.
\(-6\)
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8.

Solve the following equation for \(p\text{.}\)
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\(\displaystyle{{\frac{q+3}{p-2}} = 4/5}\)
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\(p\) =
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Answer.
\(\frac{5q+23}{4}\)
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9.

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

Solve the following equation for \(r\text{.}\)
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\(\displaystyle{{\frac{C}{r}} = {2\pi }}\)
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\(r\) =
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Answer.
\(\frac{C}{2\pi }\)
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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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\(1+\dfrac{-4}{c}=-\dfrac{-45}{c^2}\)
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\(c\) =
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Answer.
\(9, -5\)
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Solution.
The LCD is \(c^2\text{.}\) We can clear the fractions by multiplying every term by the LCD:
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\(1 \cdot c^2 +\dfrac{-4}{c} \cdot c^2 =-\dfrac{-45}{c^2} \cdot c^2\)
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\(c^2 - 4c = + 45\)
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\(c^2 - 4c - 45 = 0\)
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\((c -9)(c + 5) = 0\)
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\(c = 9, c = -5\)
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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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\({\frac{r}{r-1}+3} = {\frac{1}{r-1}}\)
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\(r\) =
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Answer.
\(\text{NONE}\)
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Solution.
Multiply every term by the least common denominator. The LCD is \({r-1}\text{.}\)
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\(\displaystyle{\left( {r-1} \right) \left( {\frac{r}{r-1}+3} \right) = \left( {r-1} \right) \cdot {\frac{1}{r-1}}}\)
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\({r+3\mathopen{}\left(r-1\right)} = 1\)
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\({r+3r-3} = 1\)
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\({4r-3} = 1\)
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\(4r = 4\)
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\(r = 1\)
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Notice that if we plug \(r = 1\) 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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\(r = NONE\)
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13.

Let \(\displaystyle{ f(x)={-\frac{6}{x+8}} }\text{.}\) 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.
\(\left(-\infty ,-8\right)\cup \left(-8,\infty \right)\)
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Answer 2.
\(\text{x=}\)
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Answer 3.
\(-8\)
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Answer 4.
\(\text{y=}\)
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Answer 5.
\(0\)
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14.

Let \(\displaystyle{ f(x)={\frac{x^{2}+7}{x^{2}-6x+5}} }\text{.}\) 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.
\(\left(-\infty ,1\right)\cup \left(1,5\right)\cup \left(5,\infty \right)\)
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Answer 2.
\(\text{x=}\)
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Answer 3.
\(5, 1\)
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Answer 4.
\(\text{y=}\)
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Answer 5.
\(1\)
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15.

Let \(\displaystyle{ f(x)={\frac{x^{2}-25}{x-5}} }\text{.}\) Identify the removable discontinuity.
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  • Removable Discontinuity: \(x=\)
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Answer.
\(5\)
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16.

Let \(\displaystyle{ f(x)={\frac{x^{2}-4x-21}{x-7}} }\text{.}\) Identify the removable discontinuity.
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  • Removable Discontinuity: \(x=\)
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Answer.
\(7\)
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17.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{x + {8}}{x + {7}} > 0 }\)
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Answer.
\(\left(-\infty ,-8\right)\cup \left(-7,\infty \right)\)
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18.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{{4}x}{x - {5}} \lt {3} }\)
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Answer.
\(\left(-15,5\right)\)
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19.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{{3}x}{x - {8}} > {2} }\)
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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.
\(\left(-\infty ,-16\right)\cup \left(8,\infty \right)\)
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20.

Solve the rational inequality. Enter the answer in interval notation.
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\(\displaystyle{ \dfrac{1}{{3}} - \dfrac{ {12} }{x^2} \le \dfrac{ {3} }{x} }\)
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Answer.
\(\left[-3,0\right)\cup \left(0,12\right]\)
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