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11.3: Exercises

Last updated

May 2, 2022
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- 11.2: Optional section- Rational functions by hand
- 12: Polynomial and Rational Inequalities

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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Exercise $\PageIndex{1}$

Find the domain, the vertical asymptotes and removable discontinuities of the functions.

$f(x)=\dfrac{2}{x-2}$
$f(x)=\dfrac{x^2+2}{x^2-6x+8}$
$f(x)=\dfrac{3x+6}{x^3-4x}$
$f(x)=\dfrac{(x-2)(x+3)(x+4)}{(x-2)^2(x+3)(x-5)}$
$f(x)=\dfrac{x-1}{x^3-1}$
$f(x)=\dfrac{2}{x^3-2x^2-x+2}$

Answer

domain $D = \mathbb{R} − \{2\}$ , vertical asymptote at $x = 2$ , no removable discontinuities
$D = \mathbb{R}−\{2, 4\}$ , vertical asympt. at $x = 2$ and $x = 4$ , no removable discont.
$D = \mathbb{R}− \{−2, 0, 2\}$ , vertical asympt. at $x = 0$ and $x = 2$ , removable discont. at $x = −2$
$D = \mathbb{R} − \{−3, 2, 5\}$ , vertical asympt. at $x = 2$ and $x = 5$ , removable discont. at $x = −3$
$D = \mathbb{R} − \{1\}$ , no vertical asympt., removable discont. at $x = 1$
$D = \mathbb{R} − \{−1, 1, 2\}$ , vertical asympt. at $x = −1$ and $x = 1$ and $x = 2$ , no removable discont.

Exercise $\PageIndex{2}$

Find the horizontal asymptotes of the functions.

$f(x)=\dfrac{8x^2+2x+1}{2x^2+3x-2}$
$f(x)=\dfrac{1}{(x-3)^2}$
$f(x)=\dfrac{x^2+3x+2}{x-1}$
$f(x)=\dfrac{12x^3-4x+2}{-3x^3+2x^2+1}$

Answer

$y = 4$
$y = 0$
no horizontal asymptote (asymptotic behavior $y = x + 4$ )
$y = −4$

Exercise $\PageIndex{3}$

Find the $x$ - and $y$ -intercepts of the functions.

$f(x)=\dfrac{x-3}{x-1}$
$f(x)=\dfrac{x^3-4x}{x^2-8x+15}$
$f(x)=\dfrac{(x-3)(x-1)(x+4)}{(x-2)(x-5)}$
$f(x)=\dfrac{x^2+5x+6}{x^2+2x}$

Answer

$x$ -intercept at $x = 3$ , $y$ -intercept at $y = 3$
$x$ -intercepts at $x = 0$ and $x = −2$ and $x = 2$ , $y$ -intercept at $y = 0$
$x$ -intercepts at $x = −4$ and $x = 1$ and $x = 3$ , $y$ -intercept at $y = \dfrac{6}{5}$
$x$ -intercept at $x = −3$ (but not at $x = −2$ since $f(−2)$ is undefined), no $y$ -intercept since $f(0)$ is undefined

Exercise $\PageIndex{4}$

Sketch the graph of the function $f$ by using the domain of $f$ , the horizontal and vertical asymptotes, the removable singularities, the $x$ - and $y$ -intercepts of the function, together with a sketch of the graph obtained from the calculator.

$f(x)=\dfrac{6x-2}{2x+4}$
$f(x)=\dfrac{x-3}{x^3-3x^2-6x+8}$
$f(x)=\dfrac{x^4-10x^2+9}{x^2-3x+2}$
$f(x)=\dfrac{x^3-3x^2-x+3}{x^3-2x^2}$

Answer

$D = \mathbb{R} − \{2\}$ , horizontal asympt. $y = 3$ , vertical asympt. $x = −2$ , no removable discont., $x$ -intercept at $x = \dfrac 1 3$ , $y$ -intercept at $y = \dfrac{-1}{2}$ , graph:

$clipboard_ecbbba7ef054ee5a490309ff6ad0fbf4b.png$

$f(x)=\dfrac{x-3}{(x-4)(x-1)(x+2)}$ has domain $D = \mathbb{R} − \{−2, 1, 4\}$ , horizontal asympt. $y = 0$ , vertical asympt. $x = −2$ and $x = 1$ and $x = 4$ , no removable discont., $x$ -intercept at $x = 3$ , $y$ -intercept at $y = \dfrac{-3}{8} = −0.375$ , graph:

$clipboard_e0d6db6e49d48f2bad98ab2b86fb86e1b.png$

$f(x)=\dfrac{(x-3)(x+3)(x-1)(x+1)}{(x-2)(x-1)}$ has domain $D = \mathbb{R} − \{1, 2\}$ , no horizontal asympt., vertical asympt. $x = 2$ , removable discont. at $x = 1$ , $x$ -intercept at $x = −3$ and $x = −1$ and $x = 3$ , $y$ -intercept at $y = \dfrac 9 2 = 4.5$ , graph:

$clipboard_e3ce9e8190cd29dd861506874a33ab2d1.png$

$f(x)=\dfrac{(x-3)(x-1)(x+1)}{x^{2}(x-2)}$ has domain $D = \mathbb{R} − \{0, 2\}$ , horizontal asympt. $y = 1$ , vertical asympt. $x = 0$ and $x = 2$ , no removable discont., $x$ -intercepts at $x = −1$ and $x = 1$ and $x = 3$ , no $y$ -intercept since $f(0)$ is undefined, graph:

$clipboard_e46938e7afbc22f2db2e88901bdcc36c6.png$

Note that the graph intersects the horizontal asymptote $y = 1$ at approximately $x \approx-2.3$ and approaches the asymptote from above

Exercise $\PageIndex{5}$

Find a rational function $f$ that satisfies all the given properties.

vertical asymptote at $x=4$ and horizontal asymptote $y=0$
vertical asymptotes at $x=2$ and $x=3$ and horizontal asymptote $y=5$
removable singularity at $x=1$ and no horizontal asymptote

Answer

for example $f(x)=\dfrac{1}{x-4}$
for example $f(x)=\dfrac{5 x^{2}}{x^{2}-5 x+6}$
for example $f(x)=\dfrac{x^{2}-x}{x-1}$

Exercise 11.3.1\PageIndex{1}

Exercise 11.3.2\PageIndex{2}

Exercise 11.3.3\PageIndex{3}

Exercise 11.3.4\PageIndex{4}

Exercise 11.3.5\PageIndex{5}

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Exercise $\PageIndex{1}$

Exercise $\PageIndex{2}$

Exercise $\PageIndex{3}$

Exercise $\PageIndex{4}$

Exercise $\PageIndex{5}$