Problem 58 - College Algebra

Question

Find the domain of each function. $$f(x)=\frac{15}{(x+8)(x-3)}$$

Step-by-Step Solution

Answer

The domain of the function \(f(x) = \frac{15}{(x+8)(x-3)}\) is \(-\infty

Step 1: Set Denominator Equal to Zero

Start by setting the denominator \((x+8)(x-3)\) equal to zero. That gives us two equations: \(x + 8 = 0\) and \(x - 3 = 0\).

Step 2: Solve for x

Solve each equation separately. The solutions from the first equation, \(x + 8 = 0\), will give us \(x = -8\). From the second one, \(x - 3 = 0\), we will find \(x = 3\).

Step 3: Determine the Domain

The domain of the function is all real numbers, denoted as \(-\infty\leq x \leq \infty\) except the two values -8 and 3 because these two numbers will make the denominator zero. Therefore the domain of the given function will be \(-\infty

Key Concepts

Rational Functions

Rational functions are an essential concept in mathematics and they appear in various problem-solving scenarios. A rational function is defined as a ratio of two polynomials. In other words, it takes the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.

In our example, the function is \( f(x) = \frac{15}{(x+8)(x-3)} \). Here, our numerator is the constant polynomial \( 15 \) and the denominator is the product of two binomials \( (x+8)(x-3) \). Understanding the structure of rational functions helps in determining their behavior especially when approaching specific values and intervals.

Rational functions can represent complex relationships and often involve variables in both numerator and denominator. The values that the variables take can significantly impact the function's validity and continuity.

Denominator Zero

One significant aspect of working with rational functions is dealing with the denominator. Specifically, we need to identify when the denominator becomes zero. This is crucial as the function becomes undefined whenever the denominator equals zero:

• Write down the denominator in its expanded form, as we see with \( (x+8)(x-3) \).
• Set the denominator equal to zero, \( (x+8)(x-3) = 0 \).
• This gives us multiple equations based on factors, such as \( x+8 = 0 \) and \( x-3 = 0 \).

Solving these equations gives us the individual x-values where the denominator is zero, therefore, where the function becomes undefined.

For our example, solving \( x+8=0 \) yields \( x=-8 \), and \( x-3=0 \) yields \( x=3 \). These are the critical points where the function doesn't exist.

Exclusion from Domain

Having found the values where the denominator of a rational function is zero helps us determine which x-values must be excluded from the domain.

The domain of a function is basically all the possible input values (x-values) that you can substitute into the function without causing any undefined situations such as division by zero. With rational functions such as \( f(x) = \frac{15}{(x+8)(x-3)} \), you need to consider:

• Removing any x-values that make the denominator zero as these are undefined.
• Include all other real numbers in the domain.

This means, for the given function, we exclude \( x = -8 \) and \( x = 3 \) from the domain. Thus, the domain can be expressed as the union of intervals: \(-\infty < x < -8\) and \(-8 < x < 3\) and \(3 < x < \infty\). Ensuring exclusions are correctly identified ensures proper function definition.

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Other exercises in this chapter

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Key Concepts

Rational Functions

Rational functions are an essential concept in mathematics and they appear in various problem-solving scenarios. A rational function is defined as a ratio of two polynomials. In other words, it takes the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.

In our example, the function is \( f(x) = \frac{15}{(x+8)(x-3)} \). Here, our numerator is the constant polynomial \( 15 \) and the denominator is the product of two binomials \( (x+8)(x-3) \). Understanding the structure of rational functions helps in determining their behavior especially when approaching specific values and intervals.

Rational functions can represent complex relationships and often involve variables in both numerator and denominator. The values that the variables take can significantly impact the function's validity and continuity.

Denominator Zero

One significant aspect of working with rational functions is dealing with the denominator. Specifically, we need to identify when the denominator becomes zero. This is crucial as the function becomes undefined whenever the denominator equals zero:

• Write down the denominator in its expanded form, as we see with \( (x+8)(x-3) \).
• Set the denominator equal to zero, \( (x+8)(x-3) = 0 \).
• This gives us multiple equations based on factors, such as \( x+8 = 0 \) and \( x-3 = 0 \).

Solving these equations gives us the individual x-values where the denominator is zero, therefore, where the function becomes undefined.

For our example, solving \( x+8=0 \) yields \( x=-8 \), and \( x-3=0 \) yields \( x=3 \). These are the critical points where the function doesn't exist.

Exclusion from Domain

Having found the values where the denominator of a rational function is zero helps us determine which x-values must be excluded from the domain.

The domain of a function is basically all the possible input values (x-values) that you can substitute into the function without causing any undefined situations such as division by zero. With rational functions such as \( f(x) = \frac{15}{(x+8)(x-3)} \), you need to consider:

• Removing any x-values that make the denominator zero as these are undefined.
• Include all other real numbers in the domain.

This means, for the given function, we exclude \( x = -8 \) and \( x = 3 \) from the domain. Thus, the domain can be expressed as the union of intervals: \(-\infty < x < -8\) and \(-8 < x < 3\) and \(3 < x < \infty\). Ensuring exclusions are correctly identified ensures proper function definition.