Question

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

Short answer

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

Step-by-step solution

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.

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 \( x + 2 \).

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 \( x = -2 \).

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 \( x \).

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 \( x \).

Key concepts

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.