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Chapter 3: Problem 9

Determine the domain of each function. (a) \(y=-5 x+1\) (b) \(y=1 /(-5 x+1)\) (c) \(y=\sqrt{-5 x+1}\) (d) \(y=\sqrt[3]{-5 x+1}\)

Short Answer

Expert verified
(a): All real numbers, (b): \((-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)\), (c): \((-\infty, \frac{1}{5}]\), (d): All real numbers.

Step by step solution

01

Analyze the Linear Function

The function given is a linear function: \(y = -5x + 1\). Linear functions are defined for all real numbers, which means there are no restrictions on the domain. Therefore, the domain of this function is the set of all real numbers, represented as \((-\infty, \infty)\).
02

Analyze the Rational Function

The function given is \(y = \frac{1}{-5x + 1}\). Rational functions are undefined where the denominator is zero. To find where the denominator is zero, set \(-5x + 1 = 0\) and solve for \(x\). This gives \(x = \frac{1}{5}\). Therefore, the domain excludes \(x = \frac{1}{5}\), and it is \((-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)\).
03

Analyze the Square Root Function

For the function \(y = \sqrt{-5x + 1}\), the expression inside the square root must be non-negative, i.e., \(-5x + 1 \geq 0\). Solving gives \(x \leq \frac{1}{5}\). Therefore, the domain of this function is \((-\infty, \frac{1}{5}]\).
04

Analyze the Cube Root Function

The function \(y = \sqrt[3]{-5x + 1}\) involves a cube root, which is defined for all real numbers because cube roots can be taken of both positive and negative numbers as well as zero. Therefore, the domain of this function is \((-\infty, \infty)\).

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Key Concepts

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

Linear Function
A linear function is one of the simplest types of functions. It resembles a straight line when graphed on a Cartesian coordinate system. The general form is given by \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Linear functions are defined for all real numbers. This means when asked about the domain of a linear function, the answer is always all real numbers, denoted as
  • \((-\infty, \ \infty)\)
For example, in the linear function \(y = -5x + 1\), whatever value you choose for \(x\), there will always be a corresponding \(y\). Thus, no numbers are excluded from the domain.
Rational Function
Rational functions can appear more complicated, as they are expressed as the ratio of two polynomials. A typical form is \(y = \frac{P(x)}{Q(x)}\), where \(Q(x)\) should not be zero.
To determine the domain of a rational function, it's important to identify where the denominator, \(Q(x)\), equals zero and exclude those values from the domain.
For example, in the rational function \(y = \frac{1}{-5x + 1}\), solving \(-5x + 1 = 0\) reveals \(x = \frac{1}{5}\). This means \(x = \frac{1}{5}\) is excluded from the domain. So, the domain becomes
  • \((-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)\)
Remember, for rational functions, discontinuities like these make the calculation of domains a bit more complex.
Square Root Function
Square root functions involve finding the square root of a function within the equation, and are written in the form \(y = \sqrt{f(x)}\). The key factor to remember is that the expression under the square root must be non-negative.
As such, to determine the domain, solve the inequality \(f(x) \geq 0\). In the case of \(y = \sqrt{-5x + 1}\), solving \(-5x + 1 \geq 0\) gives \(x \leq \frac{1}{5}\). Therefore, the domain
  • \((-\infty, \frac{1}{5}]\)
Square root functions can often involve more restrictive domains to ensure all expression values are validly squared, compared to linear or cube root functions.
Cube Root Function
Cube root functions, expressed as \(y = \sqrt[3]{f(x)}\), are particularly flexible in terms of domain. Unlike square roots, there are no requirements for the expression under a cube root to be non-negative.
This means cube root functions are defined for all real numbers, much like linear functions.
For the function \(y = \sqrt[3]{-5x + 1}\), every number for \(x\) is valid. Thus, the domain is
  • \((-\infty, \infty)\)
This makes cube root functions versatile in their domain applicability, allowing for negative, positive, and zero values under the root.

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Most popular questions from this chapter

Refer to the following table. The left-hand column of the table lists four errors to avoid in working with function notation. In each case, use the function \(f(x)=x^{2}-1\) and give a numerical example showing that the expressions on each side of the equation are not equal. $$f(a+b)=f(a)+f(b)$$

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.) \(F(x)=x^{2}\) (a) \(x_{0}=0.9\) (b) \(x_{0}=1\) (c) \(x_{0}=1.1\)

Let \(g(x)=\sqrt{x}-3\) and \(f(x)=x-1\) (a) Sketch a graph of \(g\). Specify the domain and range. (b) Sketch a graph of \(f .\) Specify the domain and range. (c) Compute \((f \circ g)(x)\). Graph the function \(f \circ g\) and spec. ify its domain and range. (d) Find a formula for \(g[f(x)]\). Which values of \(x\) are acceptable inputs here? That is, what is the domain of \(g \circ f ?\) (e) Use the results of part (d) to sketch a graph of the function \(g \circ f\).

Use a graphing utility to graph each function and then apply the horizontal line test to see whether the function is one-to-one. $$y=2 x^{3}+x^{2}$$

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.) \(G(x)=x^{2}+0.25\) (a) \(x_{0}=0.4\) (b) \(x_{0}=0.5\) (c) \(x_{0}=0.6\)
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