Domain And Range Of Rational Function A Step-by-Step Solution

In mathematics, determining the domain and range of a function is a fundamental task. The domain represents the set of all possible input values (x-values) for which the function is defined, while the range represents the set of all possible output values (y-values) that the function can produce. When dealing with rational functions, which are functions expressed as a ratio of two polynomials, finding the domain and range requires careful consideration of the function's behavior, especially where the denominator might be zero.

This article will provide a step-by-step explanation on how to find the domain and range of the following rational function:

$$f(x) = \frac{(x-1)(x-3)^2(x+1)^2}{(x-2)(x+2)(x-1)(x+3)}$$

We will explore the concepts of removable discontinuities, vertical asymptotes, and horizontal asymptotes to accurately determine the domain and range of this function. Understanding these concepts is crucial for working with rational functions and for their graphical representation.

Understanding Domain and Range

Before diving into the specific function, it's essential to solidify our understanding of domain and range. Let's delve deeper into these concepts:

Domain

The domain of a function is the set of all possible input values (x-values) that will produce a valid output. In simpler terms, it's the set of numbers you can plug into the function without causing any mathematical errors. For rational functions, the most common restriction on the domain arises from the denominator. A rational function is undefined when the denominator equals zero, as division by zero is not allowed in mathematics. Therefore, to find the domain of a rational function, we must identify the values of x that make the denominator zero and exclude them from the set of all real numbers.

For example, consider the function:

$$g(x) = \frac{1}{x-3}$$

The denominator, $x-3$, becomes zero when $x = 3$. Thus, the domain of $g(x)$ is all real numbers except 3, which can be written in interval notation as $(-\infty, 3) \cup (3, \infty)$. This means that you can input any number into the function g(x) except for 3, because plugging in 3 would result in division by zero, which is undefined.

Range

The range of a function is the set of all possible output values (y-values) that the function can produce. Determining the range can sometimes be more challenging than finding the domain. It involves analyzing the function's behavior over its entire domain. For rational functions, the range is influenced by factors such as the presence of horizontal asymptotes, local maxima, and local minima. A horizontal asymptote is a horizontal line that the graph of the function approaches as x approaches positive or negative infinity. The function may or may not intersect its horizontal asymptote.

Consider the function:

$$h(x) = \frac{1}{x^2 + 1}$$

The denominator, $x^2 + 1$, is always positive for any real value of x. The function h(x) will always produce a positive output value. As x approaches positive or negative infinity, h(x) approaches 0. The maximum value of h(x) occurs when x is 0, where h(0) = 1. Therefore, the range of h(x) is $(0, 1]$. This means that the output values of the function will always be between 0 and 1, excluding 0 but including 1.

Step-by-Step Solution

Now, let's apply our understanding of domain and range to the given function:

$$f(x) = \frac{(x-1)(x-3)^2(x+1)^2}{(x-2)(x+2)(x-1)(x+3)}$$

1. Simplify the Function

The first step is to simplify the function by canceling out any common factors in the numerator and the denominator. Notice that the factor $(x-1)$ appears in both the numerator and the denominator. We can cancel this factor out, but it's crucial to remember that this creates a removable discontinuity at $x = 1$. This means that the function is not defined at $x = 1$, but the limit of the function as $x$ approaches 1 exists.

After canceling the common factor, the simplified function becomes:

$$f(x) = \frac{(x-3)^2(x+1)^2}{(x-2)(x+2)(x+3)}, \quad x \neq 1$$

2. Determine the Domain

To find the domain, we need to identify the values of x that make the denominator equal to zero. These values must be excluded from the domain. The denominator of the simplified function is $(x-2)(x+2)(x+3)$. Setting this equal to zero, we get:

$$(x-2)(x+2)(x+3) = 0$$

This equation has three solutions:

• $x - 2 = 0 \implies x = 2$
• $x + 2 = 0 \implies x = -2$
• $x + 3 = 0 \implies x = -3$

Additionally, we must remember the removable discontinuity at $x = 1$. Therefore, the domain of the function is all real numbers except -3, -2, 1, and 2. In interval notation, this can be written as:

$$D: (-\infty, -3) \cup (-3, -2) \cup (-2, 1) \cup (1, 2) \cup (2, \infty)$$

This means the function is defined for all real numbers except for x values of -3, -2, 1, and 2. These values would make the denominator of the function zero, leading to an undefined expression.

3. Analyze for Asymptotes

Asymptotes are lines that the graph of a function approaches but never quite touches. They provide valuable information about the function's behavior as x approaches certain values or infinity. There are three main types of asymptotes:

• Vertical Asymptotes: These occur at the values of x where the denominator of the simplified function is zero and the numerator is non-zero. Vertical asymptotes indicate that the function approaches infinity (or negative infinity) as x approaches these values.
• Horizontal Asymptotes: These describe the behavior of the function as x approaches positive or negative infinity. The horizontal asymptote is determined by comparing the degrees of the numerator and denominator polynomials.
• Oblique (Slant) Asymptotes: These occur when the degree of the numerator is exactly one greater than the degree of the denominator. We won't delve into oblique asymptotes in this specific example, but it's worth noting their existence.

Vertical Asymptotes

From the domain analysis, we know that the function is undefined at $x = -3$, $x = -2$, and $x = 2$. Examining the simplified function:

$$f(x) = \frac{(x-3)^2(x+1)^2}{(x-2)(x+2)(x+3)}$$

We see that the numerator is not zero at these values. Therefore, we have vertical asymptotes at $x = -3$, $x = -2$, and $x = 2$.

Horizontal Asymptotes

To find the horizontal asymptote, we compare the degrees of the numerator and the denominator. The degree of the numerator is 4 (since $(x-3)^2$ is degree 2 and $(x+1)^2$ is degree 2, their product is degree 4). The degree of the denominator is 3 (a product of three linear factors). Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, there might be an oblique asymptote, but as mentioned earlier, we won't focus on those here.

4. Determine the Range

Finding the range of a rational function can be more complex than finding the domain. It often requires a combination of analytical techniques and graphical analysis. Here's a breakdown of the process for this function:

1. Consider the behavior near vertical asymptotes: As x approaches the vertical asymptotes at $x = -3$, $x = -2$, and $x = 2$, the function will approach either positive or negative infinity. This suggests that the range will likely include large positive and negative values.
2. Analyze the function's behavior as x approaches infinity: Since there is no horizontal asymptote, the function will either increase or decrease without bound as x approaches positive or negative infinity. The leading term of the numerator is $x^4$ and the leading term of the denominator is $x^3$. Thus, as x approaches infinity, f(x) will behave like $x^4 / x^3 = x$, indicating that the function will approach both positive and negative infinity.
3. Identify any local maxima or minima: To find local maxima and minima, you would typically take the derivative of the function, set it equal to zero, and solve for x. Then, you would plug these x-values back into the original function to find the corresponding y-values. However, finding the derivative of this function is quite complex and beyond the scope of this step-by-step solution. We will rely on the other pieces of information to make an educated determination of the range.
4. Consider the removable discontinuity: The removable discontinuity at $x = 1$ does not directly affect the range, but it's important to remember that the function does not take on the value at $x = 1$.

Based on the analysis above, we can conclude that the function can take on a wide range of values, including very large positive and negative numbers. Without explicitly finding local maxima and minima, it is reasonable to assume that the range of the function is all real numbers. Therefore, the range is:

$$R: (-\infty, \infty)$$

5. Final Answer

In conclusion, the domain and range of the function

$$f(x) = \frac{(x-1)(x-3)^2(x+1)^2}{(x-2)(x+2)(x-1)(x+3)}$$

are:

• Domain: $D: (-\infty, -3) \cup (-3, -2) \cup (-2, 1) \cup (1, 2) \cup (2, \infty)$
• Range: $R: (-\infty, \infty)$

Therefore, the correct answer is:

A. $D:(-\infty,-3) \cup(-3,-2) \cup(-2,1) \cup(1,2) \cup(2, \infty) R:(-\infty, \infty)$

Importance of Domain and Range

Understanding the domain and range of a function is crucial for several reasons:

• Function Definition: The domain specifies the set of inputs for which the function is valid. Without a clear understanding of the domain, you might attempt to evaluate the function at points where it is undefined, leading to incorrect results.
• Graphing: The domain and range provide essential information for graphing functions. Knowing the domain helps you determine the intervals on the x-axis where the function exists, while the range indicates the possible y-values. Asymptotes, which are closely related to the domain, also guide the shape of the graph.
• Modeling Real-World Situations: In many real-world applications, functions are used to model relationships between variables. The domain and range often have physical or practical interpretations. For example, if a function models the height of an object over time, the domain would represent the time interval of interest, and the range would represent the possible heights the object can reach.
• Calculus: The concepts of domain and range are fundamental in calculus. For example, when finding limits, derivatives, and integrals, it's essential to consider the domain of the function to ensure that the operations are valid.

Conclusion

Finding the domain and range of a function, particularly a rational function, requires a systematic approach. This involves simplifying the function, identifying values that make the denominator zero, analyzing asymptotes, and considering the function's behavior as x approaches infinity. While determining the range can sometimes be challenging, understanding the concepts and techniques discussed in this article provides a solid foundation for tackling such problems. By carefully considering these steps, you can accurately determine the domain and range of a rational function, which is essential for understanding its behavior and applications.