Problem 70 - A Graphical Approach to College Algebra

Question

Graph each rational function by hand. Give the domain and range, and discuss symmetry. Give the equations of any asymptotes. $$f(x)=\frac{1}{x^{2}+3}$$

Step-by-Step Solution

Answer

Domain: \((-\infty, \infty)\); Range: \((0, 1/3]\); Symmetry: Even; Asymptote: \(y = 0\).

Step 1: Identify the Domain

The domain of a function is all the real numbers for which the function is defined. Since the denominator of \( f(x) = \frac{1}{x^2 + 3} \) is always positive (\( x^2 + 3 > 0 \) for any real \( x \)), \( f(x) \) is defined for all real numbers. Thus, the domain is \( (-\infty, \infty) \).

Step 2: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero. For \( f(x) = \frac{1}{x^2 + 3} \), the denominator \( x^2 + 3 eq 0 \) for any real \( x \). Hence, there are no vertical asymptotes.

Step 3: Find Horizontal Asymptotes

Evaluate the limits of \( f(x) \) as \( x \to \infty \) and \( x \to -\infty \). Since \( x^2 + 3 \to \infty \), we have \( \lim_{{x \to \pm\infty}} f(x) = 0 \). Thus, \( y = 0 \) is the horizontal asymptote.

Step 4: Determine Symmetry

To check for symmetry, evaluate \( f(-x) = \frac{1}{(-x)^2 + 3} = \frac{1}{x^2 + 3} = f(x) \). Therefore, the function is even, indicating symmetry about the y-axis.

Step 5: Identify the Range

The range is the set of possible output values of the function. Since \( \frac{1}{x^2 + 3} \) is always positive and approaches 0 as \( x \to \pm\infty \), the smallest value occurs as \( x \to \infty \) or \( x \to -\infty \). Thus, the range is \( (0, 1/3] \).

Step 6: Graph the Function

Plot a few points to get a sense of the curve: \((0, 1/3)\), \((1, 1/4)\), \((-1, 1/4)\), \((2, 1/7)\), \((-2, 1/7)\). Note the even symmetry and the horizontal asymptote at \( y = 0 \). Sketch the curve, reflecting its decreasing values as \( x \) moves away from zero in both directions.

Key Concepts

Domain and Range

The domain of a rational function is a key aspect to understand because it defines all the possible input values (x-values) that a function can accept. In the function \( f(x) = \frac{1}{x^2 + 3} \), the denominator \( x^2 + 3 \) never equals zero for any real number \( x \). This is because \( x^2 \) is always zero or positive, so \( x^2 + 3 \) is always greater than zero. Therefore, the domain is all real numbers, which we express using the interval notation as \( (-\infty, \infty) \).

On the other hand, the range talks about the possible output values (y-values) of the function. For this function, as \( x \) either increases or decreases without bound, \( \frac{1}{x^2 + 3} \) gets closer and closer to zero but never actually reaches it. Hence, the function's smallest value is just above zero. At \( x = 0 \) the function takes its maximum value of \( \frac{1}{3} \) because the function value decreases as \( x \) moves away from zero. Therefore, the range of this function is \( (0, \frac{1}{3}] \).

Asymptotes

Asymptotes are lines that a graph approaches but never touches. In our rational function \( f(x) = \frac{1}{x^2 + 3} \), we need to explore both vertical and horizontal asymptotes.

Vertical asymptotes occur when the denominator is zero, which causes the function to be undefined at certain x-values. However, because \( x^2 + 3 \) can never be zero, there are no vertical asymptotes for this function.

Horizontal asymptotes guide us on how the function behaves as \( x \) approaches infinity. This happens when the degree of the polynomial in the denominator is higher than that in the numerator—here the numerator is 1 and the denominator is \( x^2 + 3 \). As \( x \to \pm\infty \), \( x^2 + 3 \to \infty \), therefore, \( f(x) \to 0 \). Hence the horizontal asymptote is the line \( y = 0 \). Through this knowledge, we can visualize that as \( x \) increases in positive or negative direction, the function values get smaller and approach zero.

Symmetry Analysis

Symmetry in a function can greatly simplify the graphing process and helps us understand the behavior of the function across different x-values. For the given rational function, we need to check symmetry about the y-axis, x-axis, or the origin.

A function is symmetric about the y-axis if \( f(-x) = f(x) \) for all x in the domain. For \( f(x) = \frac{1}{x^2 + 3} \), we calculate \( f(-x) = \frac{1}{(-x)^2 + 3} = \frac{1}{x^2 + 3} = f(x) \). This means that our function is even, and we see a mirror image of the function on either side of the y-axis.

This kind of symmetry indicates that you only need to graph one-half of the function and can then reflect it across the y-axis to get the other half. Symmetry about the y-axis often simplifies determining points on the graph and understanding function behavior on either side of the y-axis.