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Chapter 3: Problem 23

Determine the domain and the range of each function. (a) \(f(x)=\frac{x+3}{x-5}\) (b) \(F(x)=\frac{x^{3}+3}{x^{3}-5}\)

Short Answer

Expert verified
(a) Domain: \(x \neq 5\), Range: \(y \neq 1\). (b) Domain: \(x \neq \sqrt[3]{5}\), Range: \(y \neq 1\).

Step by step solution

01

Finding the Domain of f(x)

To find the domain of the function \(f(x)=\frac{x+3}{x-5}\), we need to determine the values of \(x\) for which the function is defined. Since division by zero is undefined, we set the denominator equal to zero and solve for \(x\): \(x-5 = 0\). This gives us \(x = 5\). Therefore, the domain of \(f(x)\) is all real numbers except \(x = 5\), which can be expressed as \(x \in \mathbb{R}, x eq 5\).
02

Finding the Range of f(x)

To find the range of \(f(x)=\frac{x+3}{x-5}\), we determine the values that \(y=\frac{x+3}{x-5}\) can take. By rearranging and solving for \(x\) in terms of \(y\), we get \(x = \frac{5y+3}{y-1}\). Since \(y = 1\) makes the denominator zero and thus is undefined, the range of \(f(x)\) is all real numbers except \(y = 1\). Thus, the range is \(y \in \mathbb{R}, y eq 1\).
03

Finding the Domain of F(x)

For the function \(F(x)=\frac{x^{3}+3}{x^{3}-5}\), we again need to avoid division by zero. Set the denominator equal to zero: \(x^3 - 5 = 0\). Solving for \(x\), we get \(x^3 = 5\), which gives \(x = \sqrt[3]{5}\). Therefore, the domain of \(F(x)\) is all real numbers except \(x = \sqrt[3]{5}\). So, it can be expressed as \(x \in \mathbb{R}, x eq \sqrt[3]{5}\).
04

Finding the Range of F(x)

For the range, let's examine \(y=\frac{x^3+3}{x^3-5}\). Solving for \(x\) in terms of \(y\), we have \(x^3 = \frac{5y+3}{y-1}\). If \(y=1\), the expression becomes undefined as it leads to division by zero. Thus, the range of \(F(x)\) is all real numbers except \(y = 1\), meaning \(y \in \mathbb{R}, y eq 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Functions
Rational functions are a type of function expressed as the ratio of two polynomials. They often take the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The behavior and characteristics of rational functions significantly depend on these polynomials, particularly the denominator since division by zero is undefined.

This leads to the critical concept of identifying values that make the denominator zero, as these are the points the function is not defined, restricting the domain. For example, in the function \( f(x) = \frac{x+3}{x-5} \), the denominator becomes zero at \( x = 5 \), excluding that point from the domain. Understanding the structure of rational functions is essential for analyzing their domain and range.
Function Properties
When examining rational functions, certain properties help us understand their domain and range.

  • Domain: Includes all real numbers except those that make the denominator zero. We identify these points by setting the denominator equal to zero and solving for \( x \).
  • Range: Describes the possible output values \( f(x) \) can take. Sometimes, this requires expressing \( x \) in terms of \( y \) and finding values that make the function undefined, often involving setting the equation \( y = f(x) \) and solving for \( x \).
Consider the rational function \( F(x) = \frac{x^3+3}{x^3-5} \). Here, the domain excludes \( x = \sqrt[3]{5} \) as it makes the denominator zero. Similarly, for the range, the function becomes undefined when \( y = 1 \) as it leads to zero in the new denominator when rearranged. Building these concepts, you can navigate through complex rational functions more effectively.
Algebraic Manipulation
Algebraic manipulation is a useful tool when dealing with rational functions, especially in determining the range.

  • Express the function \( y = f(x) \) in terms of \( x \) by rearranging the original function's equation. For instance, from \( y = \frac{x+3}{x-5} \), you rearrange to get \( x = \frac{5y+3}{y-1} \).
  • Determine the values that make rearranged expressions undefined, often indicating points that are excluded from the range, like \( y = 1 \) in the example.
This technique helps identify critical points that are not obvious at first glance, enabling a deeper understanding of the function's behavior across its domain. Mastery of algebraic manipulation is key in overcoming challenges posed by complex rational functions in mathematics.

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Most popular questions from this chapter

(a) Complete the given table. $$\begin{array}{l}x \quad x^{2} \quad(x-1)^{2} \quad(x+1)^{2} \\\\\hline 0 \\\1 \\\2 \\\3 \\\\-1 \\\\-2 \\\\-3 \\\\\hline\end{array}$$ (b) Using the results in the table, graph the functions \(y=x^{2}, y=(x-1)^{2},\) and \(y=(x+1)^{2}\) on the same set of axes. How are the graphs related?

Use the given function and compute the first six iterates of each initial input \(x_{0}\). In cases in which a calculator answer contains four or more decimal places, round the final answer to three decimal places. (However, during the calculations, work with all of the decimal places that your calculator affords.) \(g(x)=2 x+1\) (a) \(x_{0}=-2\) (b) \(x_{0}=-1\) (c) \(x_{0}=1\)

Let \(f(x)=2 x+3 .\) Find values for \(a\) and \(b\) such that the equation \(f(a x+b)=x\) is true for all values of \(x\) Hint: Use the fact that if two polynomials (in the variable \(x\) ) are equal for all values of \(x\), then the corresponding coefficients are equal.

Use the given function and compute the first six iterates of each initial input \(x_{0}\). In cases in which a calculator answer contains four or more decimal places, round the final answer to three decimal places. (However, during the calculations, work with all of the decimal places that your calculator affords.) \(f(x)=2 x\) (a) \(x_{0}=1\) (b) \(x_{0}=0\) (c) \(x_{0}=-1\)

Suppose that an oil spill in a lake covers a circular area and that the radius of the circle is increasing according to the formula \(r=f(t)=15+t^{1.65},\) where \(t\) represents the number of hours since the spill was first observed and the radius \(r\) is measured in meters. (Thus when the spill was first discovered, \(t=0 \mathrm{hr}\), and the initial radius was \(\left.r=f(0)=15+0^{1.65}=15 \mathrm{~m} .\right)\) (a) Let \(A(r)=\pi r^{2},\) as in Example \(5 .\) Compute a table of values for the composite function \(A \circ f\) with \(t\) running from 0 to 5 in increments of \(0.5 .\) (Round each output to the nearest integer.) Then use the table to answer the questions that follow in parts (b) through (d). (b) After one hour, what is the area of the spill (rounded to the nearest \(10 \mathrm{~m}^{2}\) )? (c) Initially, what was the area of the spill (when \(t=0\) )? Approximately how many hours does it take for this area to double? (d) Compute the average rate of change of the area of the spill from \(t=0\) to \(t=2.5\) and from \(t=2.5\) to \(t=5\). Over which of the two intervals is the area increasing faster?
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