Problem 69
Question
Graph each rational function by hand. Give the domain and range, and discuss symmetry. Give the equations of any asymptotes. $$f(x)=\frac{1}{x^{2}+2}$$
Step-by-Step Solution
Verified Answer
Domain: (-∞, ∞); Range: (0, 1/2]; Symmetry: even; Asymptote: y=0.
1Step 1: Analyze the Function and Find Its Domain
The given function is \[f(x) = \frac{1}{x^2 + 2}. \]To find the domain, we need to ensure that the denominator is never zero since division by zero is undefined. The expression \[x^2 + 2\]will never be zero for any real number \(x\) because \(x^2\) is always non-negative and 2 is positive. Thus, \[x^2 + 2 \geq 2.\]Therefore, the domain is all real numbers, which can be expressed as \((-\infty, \infty)\).
2Step 2: Determine the Range
Since \(x^2 + 2\) is always positive (and in fact, always greater than or equal to 2), the function \[f(x) = \frac{1}{x^2 + 2}\]is always positive and less than or equal to \(\frac{1}{2}\). Thus, the range of the function is \((0, \frac{1}{2}].\)
3Step 3: Identify Asymptotes
This function does not have vertical asymptotes because the denominator \(x^2 + 2\) is never zero for real \(x\). However, as \(x\) approaches infinity or negative infinity, \(f(x)\) approaches zero. Therefore, there is a horizontal asymptote at \[y = 0.\]
4Step 4: Check for Symmetry
To determine if the function has symmetry, we can check for even or odd symmetry. A function \(f(x)\) is even if \(f(-x) = f(x)\) and odd if \(f(-x) = -f(x)\). Calculate \(f(-x)\):\[f(-x) = \frac{1}{(-x)^2 + 2} = \frac{1}{x^2 + 2} = f(x).\]Since \(f(-x) = f(x)\), the function is even and symmetric about the y-axis.
5Step 5: Sketch the Graph
On a coordinate plane, plot the general shape of the function based on the domain, range, and asymptotes. The graph will take on a bell-shaped curve that flattens out as it moves away from the y-axis, never crossing \(x = 0\) and approaching \(y = 0\) as \(x\) approaches infinity or negative infinity. This graph will be symmetric with respect to the y-axis.
Key Concepts
Domain and RangeAsymptotesSymmetry
Domain and Range
Understanding the domain and range of a rational function is crucial to grasp its complete behavior. The **domain** of a function is the set of all possible input values (x-values) that the function can accept. For our function, \[f(x) = \frac{1}{x^2 + 2},\]the denominator will never equal zero, since \[x^2 + 2\]is never less than 2. This means there are no restrictions on the x-values.
The domain, therefore, is all real numbers, expressed as \((-\infty, \infty)\).
The **range** represents all possible output values (y-values) that the function can produce. Since the denominator \[x^2 + 2\]is always greater than or equal to 2, \[f(x)\]is always a positive value less than or equal to \(\frac{1}{2}\). Consequently, the range can be expressed as \((0, \frac{1}{2}]\). This detail indicates that the values of the function will never reach zero. Understanding both domain and range helps in sketching and analyzing the function's graph effectively.
The domain, therefore, is all real numbers, expressed as \((-\infty, \infty)\).
The **range** represents all possible output values (y-values) that the function can produce. Since the denominator \[x^2 + 2\]is always greater than or equal to 2, \[f(x)\]is always a positive value less than or equal to \(\frac{1}{2}\). Consequently, the range can be expressed as \((0, \frac{1}{2}]\). This detail indicates that the values of the function will never reach zero. Understanding both domain and range helps in sketching and analyzing the function's graph effectively.
Asymptotes
Asymptotes are lines that a graph approaches but never touches. They can be horizontal, vertical, or even oblique. For the function \[f(x) = \frac{1}{x^2 + 2},\]there are no **vertical asymptotes** because the denominator never becomes zero.
However, a **horizontal asymptote** exists at \(y = 0\). As x approaches positive or negative infinity, the function values approach zero, indicating the horizontal asymptote. This behavior demonstrates that the graph will get closer and closer to the line \(y = 0\) but never actually touch or cross it. Recognizing asymptotes is essential as they provide critical insights into the behavior of rational functions at extreme ends.
However, a **horizontal asymptote** exists at \(y = 0\). As x approaches positive or negative infinity, the function values approach zero, indicating the horizontal asymptote. This behavior demonstrates that the graph will get closer and closer to the line \(y = 0\) but never actually touch or cross it. Recognizing asymptotes is essential as they provide critical insights into the behavior of rational functions at extreme ends.
Symmetry
Symmetry in functions refers to how they reflect across an axis or the origin. A function can be symmetric about the y-axis (even symmetry) or the origin (odd symmetry).
To determine this, we calculate \[f(-x)\]for our function:\[f(-x) = \frac{1}{(-x)^2 + 2} = \frac{1}{x^2 + 2} = f(x).\]Since \[f(-x) = f(x),\]this function is even, meaning it possesses symmetry about the y-axis.
Understanding symmetry can make graphing the function easier because it tells us about the relationship between different parts of the function's graph. For this function, recognizing its even symmetry implies that its left and right sides mirror each other, thus simplifying its graphing and analysis.
To determine this, we calculate \[f(-x)\]for our function:\[f(-x) = \frac{1}{(-x)^2 + 2} = \frac{1}{x^2 + 2} = f(x).\]Since \[f(-x) = f(x),\]this function is even, meaning it possesses symmetry about the y-axis.
Understanding symmetry can make graphing the function easier because it tells us about the relationship between different parts of the function's graph. For this function, recognizing its even symmetry implies that its left and right sides mirror each other, thus simplifying its graphing and analysis.
Other exercises in this chapter
Problem 69
Solve each rational inequality by hand. Do not use a calculator. $$2-\frac{5}{x}+\frac{2}{x^{2}} \geq 0$$
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Complete the following. (a) Simplify the given expression so that it does not have negative exponents. (b) Set the expression from part (a) equal to 0 and solve
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Solve each rational inequality by hand. Do not use a calculator. $$\frac{1}{x-1}+\frac{1}{x+1}>\frac{3}{4}$$
View solution Problem 70
Complete the following. (a) Simplify the given expression so that it does not have negative exponents. (b) Set the expression from part (a) equal to 0 and solve
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