Question

Determine which of the values (a) through (d), if any, must be excluded from the domain of the variable in each expression. (a) \(x=0\) (b) \(x=1\) (c) \(x=0\) (d) \(x=-1\) \(\frac{x^{2}+1}{x}\)

Short answer

Exclude values \(x = 0\) from (a) and (c).

Step-by-step solution

Understand the expression

The given expression is \(\frac{x^2 + 1}{x}\). This is a rational function which means it has a numerator and a denominator.

Identify potential domain restrictions

A rational function is undefined when its denominator is zero. Therefore, the value(s) that make the denominator zero must be excluded from the domain.

Set the denominator equal to zero

Set the denominator equal to zero and solve for \(x\). The denominator is \(x\), so: \[ x = 0 \]

Solve for excluded value

Solving \(x = 0\), we find that \(x = 0\) must be excluded from the domain.

Compare with provided values

Compare the excluded value \(x = 0\) with the provided values (a) through (d):(a) \(x = 0\)(b) \(x = 1\)(c) \(x = 0\)(d) \(x = -1\)

Identify excluded values

From the comparison, the values \(x = 0\) in items (a) and (c) must be excluded from the domain.

Key concepts

rational functions

A rational function is a fraction where both the numerator and the denominator are polynomials. Understanding rational functions is vital because they help us model a lot of real-world situations. The general form of a rational function is: \( R(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials. The key thing to remember is that the denominator, \( Q(x) \), cannot be zero because division by zero is undefined.

For example, in the expression \( \frac{x^2 + 1}{x} \), \( x^2 + 1 \) (the numerator) is a polynomial, and \( x \) (the denominator) is also a polynomial. To avoid the expression becoming undefined, we need to make sure the denominator does not equal zero.

domain restrictions

Domain restrictions tell us the values that a variable can take for which a given mathematical expression is defined. For rational functions, the primary restriction comes from the denominator.

The domain of a rational function is all real numbers except for the values that make the denominator zero. To find these restricted values, set the denominator equal to zero and solve for the variable.

Let's consider the expression from the exercise: \( \frac{x^2 + 1}{x} \). Here, the denominator is \( x \). Setting \( x \) equal to zero, we get:

\[ x = 0 \]

So, \( x = 0 \) must be excluded from the domain. This means the domain of \( \frac{x^2 + 1}{x} \) is all real numbers except zero.

undefined values

Undefined values in a rational function are those that make the denominator zero. Any time the variable takes on an undefined value, the overall expression becomes undefined.

In the exercise, we are asked to find whether certain values need to be excluded from the domain of \( \frac{x^2 + 1}{x} \). After setting the denominator equal to zero, we found that \( x = 0 \) makes the denominator zero, and thus, it is an undefined value.

To summarize, for the given expression:

• (a) \( x = 0 \) is an undefined value and must be excluded.
• (b) \( x = 1 \) does not make the denominator zero, so it's in the domain.
• (c) again, \( x = 0 \) must be excluded.
• (d) \( x = -1 \) is not an undefined value, so it's in the domain.

Understanding how to find and exclude these values ensures the function remains defined and usable.