Problem 37 - College Algebra with Modeling and Visualization

Question

Exercises 37 and 38: Use the table to evaluate each expression, if possible. (a) \((f+g)(2)\) (b) \((f-g)(4)\) (c) \((f g)(-2)\) (d) \((f / g)(0)\) $$ \begin{array}{rrrrr} x & -2 & 0 & 2 & 4 \\ f(x) & 0 & 5 & 7 & 10 \\ g(x) & 6 & 0 & -2 & 5 \end{array} $$

Step-by-Step Solution

Answer

(a) 5, (b) 5, (c) 0, (d) undefined

Step 1: Understand the Problem

We are given a table with values of two functions, \( f(x) \) and \( g(x) \), for specific values of \( x \). We need to evaluate expressions that combine these functions at specified \( x \) values.

Step 1: Evaluate \((f+g)(2)\)

To find \((f+g)(2)\), we add the values of \( f(x) \) and \( g(x) \) at \( x = 2 \). From the table, \( f(2) = 7 \) and \( g(2) = -2 \). Thus, \((f+g)(2) = f(2) + g(2) = 7 + (-2) = 5\).

Step 2: Evaluate \((f-g)(4)\)

For \((f-g)(4)\), we subtract \( g(x) \) from \( f(x) \) at \( x = 4 \). The table gives \( f(4) = 10 \) and \( g(4) = 5 \). Therefore, \((f-g)(4) = f(4) - g(4) = 10 - 5 = 5\).

Step 3: Evaluate \((f \cdot g)(-2)\)

In order to find \((f \cdot g)(-2)\), we multiply \( f(x) \) and \( g(x) \) at \( x = -2 \). According to the table, \( f(-2) = 0 \) and \( g(-2) = 6 \). Hence, \((f \cdot g)(-2) = f(-2) \cdot g(-2) = 0 \cdot 6 = 0\).

Step 4: Evaluate \((f / g)(0)\)

To determine \((f / g)(0)\), divide \( f(x) \) by \( g(x) \) at \( x = 0 \). The table provides \( f(0) = 5 \) and \( g(0) = 0 \). Division by zero is undefined, so \((f / g)(0)\) is not possible.

Key Concepts

Addition of Functions

The addition of functions is a straightforward operation that involves combining the outputs of two functions. When we want to find the sum of functions \( f(x) \) and \( g(x) \), it involves adding the respective outputs of these functions:
\[(f+g)(x) = f(x) + g(x)\]
It's similar to adding numbers, but here we add the outputs of the functions at given values of \( x \).
For instance, if you are asked to find \((f+g)(2)\), you simply look at the values of \( f(2) \) and \( g(2) \) from the table.

• For \( f(x) \), when \( x = 2 \), \( f(2) = 7 \).
• For \( g(x) \), when \( x = 2 \), \( g(2) = -2 \).

So, the result of \((f+g)(2) = 7 + (-2) = 5\).
This kind of operation helps in understanding how functions behave together at given points.

Subtraction of Functions

Subtracting functions involves finding the difference between the outputs of two functions at the same \( x \) value. This can be useful when comparing how two functions behave in terms of their outputs.
Mathematically, this is expressed as:
\[(f-g)(x) = f(x) - g(x)\]
The subtraction process can be visualized as taking the value of the first function and removing the value of the second function from it. For the problem of finding \((f-g)(4)\):

• Look at the table to determine \( f(4) = 10 \).
• Also, you find that \( g(4) = 5 \).

Thus, \((f-g)(4) = 10 - 5 = 5\).
This shows how the functions compare at \( x = 4 \), allowing for direct comparison of their behavior.

Multiplication of Functions

When multiplying functions, you are finding a product that comes from multiplying their outputs at the same point. This gives us another way to explore how functions compare.
The process is represented by:
\[(f \cdot g)(x) = f(x) \cdot g(x)\]
In our case for \((f \cdot g)(-2)\):

• We use the table to find \( f(-2) = 0 \).
• Next, observe that \( g(-2) = 6 \).

Multiplying them results in \((f \cdot g)(-2) = 0 \cdot 6 = 0\).
The zero result highlights the interesting property of multiplication where any factor being zero makes the product zero, a useful concept in functions intersections.

Division of Functions

Dividing functions involves taking the output of one function and dividing it by the other. This operation can be tricky because division by zero is undefined and must be avoided.
We express it mathematically as:
\[(f / g)(x) = \frac{f(x)}{g(x)}\]
When dealing with \((f / g)(0)\):

• The table tells us that \( f(0) = 5 \).
• However, \( g(0) = 0 \).

Since division by zero is undefined, \((f / g)(0)\) is not a valid operation.
This reminds us to always check the denominator when dividing so that we maintain valid mathematical operations.