Problem 104 - Elementary and Intermediate Algebra

Question

Complete each table. See Example 11. $$ \begin{array}{|c|c|} \hline x & {\frac{1}{x+8}} \\ \hline-7 & {} \\ \hline-9 & {} \\ \hline-8 & {} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

For \( x = -7 \): 1, for \( x = -9 \): -1, for \( x = -8 \): undefined.

Step 1: Understand the Function

The function given in the table is \( f(x) = \frac{1}{x+8} \). Our task is to evaluate this function for different values of \( x \).

Step 2: Calculate for \( x = -7 \)

Substitute \( x = -7 \) into the function: \[ f(-7) = \frac{1}{-7+8} = \frac{1}{1} = 1 \] So, when \( x = -7 \), \( f(x) = 1 \).

Step 3: Calculate for \( x = -9 \)

Substitute \( x = -9 \) into the function: \[ f(-9) = \frac{1}{-9+8} = \frac{1}{-1} = -1 \] So, when \( x = -9 \), \( f(x) = -1 \).

Step 4: Calculate for \( x = -8 \)

Substitute \( x = -8 \) into the function: \[ f(-8) = \frac{1}{-8+8} = \frac{1}{0} \] The expression is undefined because division by zero is not possible. Hence, for \( x = -8 \), \( f(x) \) is undefined.

Key Concepts

Function Evaluation

In algebra, function evaluation is a fundamental concept that helps us find the output of a function for a given input. In simple terms, it means putting a specific value into a function to see what it produces. Here, we have the function \( f(x) = \frac{1}{x+8} \).
To evaluate this function:

• Take the value of \( x \) given in each case.
• Substitute \( x \) into the function.
• Simplify the expression to find \( f(x) \).

For instance, when \( x = -7 \), we plug it into the function, giving us \( f(-7) = \frac{1}{-7 + 8} = 1 \). This process helps us understand how the function behaves with different inputs.

Division by Zero

One of the essential rules in mathematics is that you cannot divide by zero. Division by zero is undefined because it does not produce a meaningful number. Let's take a closer look at why this is the case:

• If you have \( \frac{1}{0} \), there's no number that you can multiply by 0 to get 1.
• Zero times anything is zero, which makes it impossible to reverse the operation.
• In terms of limits, as a number approaches zero in a division expression, the result grows very large or very small, but never truly resolves.

In our function, when \( x = -8 \), we encounter division by zero: \( f(-8) = \frac{1}{-8+8} = \frac{1}{0} \). Thus, for \( x = -8 \), the function value is undefined.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. The function \( f(x) = \frac{1}{x+8} \) is an example.
Here is what makes a function rational:

• The numerator can be any polynomial (In our example, it's just 1).
• The denominator must not equal zero.
• It can take many forms, but the basic principle is having a variable in the denominator.

In any rational function, it's vital to identify the values for \( x \) that make the denominator zero, as these are the points where the function is undefined. By analyzing the denominator, we see that \( x = -8 \) would make \( x+8 = 0 \), resulting in division by zero. Recognizing these restrictions is crucial in working accurately with rational functions.