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Chapter 3: Problem 40

Determine whether the equation defines y as a function of \(x .\) \(y=\frac{3 x-1}{x+2}\)

Short Answer

Expert verified
Yes, it is a function of \( x \).

Step by step solution

01

Understanding the Function

To determine whether the equation defines y as a function of x, we need to check if for every value of x, there is only one unique value of y.
02

Identify the Domain

Identify the values of x for which the equation is defined. This is done by finding the values for which the denominator is not zero. Here, the denominator is \( x + 2 \).
03

Solve for Restrictions

Solve for restrictions. We need to solve the equation \( x + 2 = 0 \) to find where the function is undefined. \( x = -2 \). This shows that the function is undefined at \( x = -2 \).
04

Determine Functionality

For all other values of \( x \), the equation \( y = \frac{3x - 1}{x + 2} \) provides exactly one value of \( y \). Therefore, for every \( x \) not equal to \( -2 \), there is exactly one corresponding \( y \), which implies that \( y \) is a function of \( x \).
05

Conclusion

Since there is exactly one value of \( y \) for every value of \( x \) except for \( x = -2 \), the equation \( y = \frac{3x - 1}{x + 2} \) defines \( y \) as a function of \( x \).

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Key Concepts

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

Domain of a Function
In mathematics, the domain of a function refers to all possible input values (typically represented by x) that the function can accept without leading to undefined operations. For our function, given by the equation \( y = \frac{3x - 1}{x + 2} \), we need to ensure that the denominator never equals zero, as that would make the expression undefined.

By identifying the denominator, \(x + 2\), and setting it equal to zero, we find that \(x = -2\). Therefore, the domain of our function includes all real numbers except \(-2\). It can be expressed as \( (-fty, -2) \bigcup (-2, fty) \). This ensures that for any input value \(x\) in this range, the function will provide a valid output.
Function Definition
A function is formally defined as a relation where each input is associated with exactly one unique output. In other words, for every x-value, there must be one and only one y-value. Let's consider our function \( y = \frac{3x - 1}{x + 2} \)

We need to verify if our function meets this criterion. By examining the equation, it's clear that for each x-value (within the domain), the formula simplifies to provide one unique y-value. Therefore, \( y \) is indeed a function of \( x \). Ensuring that every x maps to one y is key to confirming the function's validity.
Undefined Values
Undefined values in functions often arise when operations involve division by zero or taking the square root of a negative number, among other situations. For the function given \( y = \frac{3x - 1}{x + 2} \), we focus on the denominator, \(x + 2\).

This denominator cannot be zero because division by zero is undefined. Solving \( x + 2 = 0 \), we find \( x = -2 \). At \( x = -2 \), the function is undefined. Avoiding undefined values by excluding them from the domain helps in maintaining the integrity of a function.
Unique Output for Inputs
A fundamental property of functions is that each input must have exactly one output. To verify that our function meets this condition, consider the function \( y = \frac{3x - 1}{x + 2} \).

For any x-value within the domain (excluding \( x = -2 \)), substituting x into the equation yields a specific y-value. For instance, substituting \( x = 1 \) results in \( y = \frac{3(1) - 1}{1 + 2} = \frac{2}{3} \). This unique mapping from x to y confirms that our function correctly assigns one output per input, thus validating its status as a function.

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Most popular questions from this chapter

The period \(T\) (in seconds) of a simple pendulum is a function of its length \(l\) (in feet) defined by the equation $$ T=2 \pi \sqrt{\frac{l}{g}} $$ where \(g \approx 32.2\) feet per second per second is the acceleration due to gravity. (a) Use a graphing utility to graph the function \(T=T(l)\). (b) Now graph the functions \(T=T(l+1), T=T(l+2)\) $$ \text { and } T=T(l+3) $$ (c) Discuss how adding to the length \(l\) changes the period \(T\) (d) Now graph the functions \(T=T(2 l), T=T(3 l)\), and \(T=T(4 l)\) (e) Discuss how multiplying the length \(l\) by factors of 2,3 , and 4 changes the period \(T\)

\(h(x)=x^{2}-2 x\) (a) Find the average rate of change from 2 to 4 . (b) Find an equation of the secant line containing \((2, h(2))\) and \((4, h(4))\)

Use a graphing utility. Graph \(y=x^{3}\). Then on the same screen graph \(y=(x-1)^{3}+2\). Could you have predicted the result?

Use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. \(f(x)=-0.4 x^{4}-0.5 x^{3}+0.8 x^{2}-2 \quad[-3,2]\)

Use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. \(f(x)=x^{3}-3 x^{2}+5 \quad[-1,3]\)
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