Problem 81 - Elementary and Intermediate Algebra

Question

Use the tables of values for functions \(f\) and \(g\) to find each of the following. a. \((f+g)(1)\) b. \((f-g)(5)\) c. \((f \cdot g)(1)\) d. \((g / f)(5)\) $$ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 1 & 3 \\ 5 & 8 \\ \hline \end{array} $$ $$ \begin{array}{|c|c|} \hline x & g(x) \\ \hline 1 & 4 \\ 5 & 0 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

(f+g)(1) = 7, (f-g)(5) = 8, (f⋅g)(1) = 12, (g/f)(5) = 0.

Step 1: Identify Given Values

From the table for function \(f\), the given values are \(f(1) = 3\) and \(f(5) = 8\). From the table for function \(g\), the given values are \(g(1) = 4\) and \(g(5) = 0\).

Step for (f+g)(1)

To find \((f+g)(1)\), calculate \(f(1) + g(1)\). Using the given values, it becomes \(3 + 4\).

Step 3: Calculating (f+g)(1)

\(f(1) + g(1) = 3 + 4 = 7\). So, \((f+g)(1) = 7\).

Step for (f-g)(5)

To find \((f-g)(5)\), calculate \(f(5) - g(5)\). Use the given values, so it becomes \(8 - 0\).

Step 5: Calculating (f-g)(5)

\(f(5) - g(5) = 8 - 0 = 8\). So, \((f-g)(5) = 8\).

Step for (f⋅g)(1)

To find \((f \cdot g)(1)\), calculate \(f(1) \cdot g(1)\). Use the given values, so it becomes \(3 \cdot 4\).

Step 7: Calculating (f⋅g)(1)

\(f(1) \cdot g(1) = 3 \cdot 4 = 12\). So, \((f \cdot g)(1) = 12\).

Step for (g/f)(5)

To find \((g/f)(5)\), calculate \(\frac{g(5)}{f(5)}\). Use the given values, so it becomes \(\frac{0}{8}\).

Step 9: Calculating (g/f)(5)

\(\frac{g(5)}{f(5)} = \frac{0}{8} = 0\). So, \((g/f)(5) = 0\).

Key Concepts

Function Evaluation

Function evaluation is a core concept in mathematics where we find the output of a function based on a given input. Think of it as simply plugging a number into a formula or rule. For instance, consider a function \( f(x) \). To evaluate \( f(x) \) for a specific \( x \), you replace \( x \) in the function with your desired value. This gives you the result or the output of the function: \( y = f(x) \). In this exercise, function evaluation is performed using tables of values. For example:

• If you need to find \( f(1) \), you look at the table and find that \( f(1) = 3 \).
• Similarly, to evaluate \( g(5) \), you locate 5 in the table for \( g(x) \) and see that \( g(5) = 0 \).

This technique is foundational to understanding how functions work, as it allows you to determine specific outputs from given inputs without necessarily knowing the algebraic form of the function.

Arithmetic Operations on Functions

Arithmetic operations on functions enable you to combine two functions using basic math operations such as addition, subtraction, multiplication, and division. This builds a new function from existing ones. Let's break down each operation:

• Addition (\(f+g\)): To find \( (f+g)(x) \), you add the corresponding outputs from both functions: \( f(x) + g(x) \).


• Subtraction (\(f-g\)): To determine \( (f-g)(x) \), you subtract one function's output from the other's: \( f(x) - g(x) \).


• Multiplication (\(f \cdot g\)): For \( (f \cdot g)(x) \), multiply the outputs of the functions: \( f(x) \cdot g(x) \).


• Division (\(g/f\)): To find \( (g/f)(x) \), divide the output of one function by the other: \( \frac{g(x)}{f(x)} \), ensuring \( f(x) eq 0 \) to avoid division by zero.

In our problem, these operations were applied to find results like \( (f+g)(1) = 7 \) and \( (g/f)(5) = 0 \), using the values from given tables.

Tables of Values in Functions

Tables of values are a handy tool for representing functions and evaluating them. When you have a table of values, it shows specific inputs \( x \) and their corresponding outputs \( f(x) \). This gives you a visual guide and makes it easier to perform function evaluation and operations without complex calculations.In our example, we have:

• The table for \( f(x) \), where an input of \( x = 1 \) gives an output of \( f(1) = 3 \), and \( x = 5 \) yields \( f(5) = 8 \).


• The table for \( g(x) \), where \( g(1) = 4 \) and \( g(5) = 0 \).

Using these values, you can easily understand and compute operations like \( (f \cdot g)(1) \), which was calculated by multiplying \( f(1) = 3 \) with \( g(1) = 4 \). Tables make the process straightforward by eliminating the need for deriving the functional form, thus providing a concrete way to handle operations on the provided data.