Problem 81 - Elementary Algebra

Question

Calculate \((f+g)(x)\) and \((f-g)(x)\) and state the restrictions to the domain. \(f(x)=13 x\) and \(g(x)=1 x-2\)

Step-by-Step Solution

Answer

\((f+g)(x) = 14x - 2\), \((f-g)(x) = 12x + 2\), domain: all real numbers.

Step 1: Understanding Functions f(x) and g(x)

We have two functions given: \(f(x) = 13x\) and \(g(x) = 1x - 2\). The goal is to derive \((f+g)(x)\) and \((f-g)(x)\), which are the sum and the difference of the two functions, respectively.

Step 2: Finding (f+g)(x)

To find \((f+g)(x)\), add the functions \(f(x)\) and \(g(x)\). This gives us: \[(f+g)(x) = f(x) + g(x) = 13x + (1x - 2) = (13x + 1x) - 2 = 14x - 2.\]

Step 3: Finding (f-g)(x)

To find \((f-g)(x)\), subtract \(g(x)\) from \(f(x)\). This calculation gives: \[(f-g)(x) = f(x) - g(x) = 13x - (1x - 2) = (13x - 1x) + 2 = 12x + 2.\]

Step 4: Determining Domain Restrictions

The domain of each function \(f(x)\) and \(g(x)\) is all real numbers because they are defined by linear equations with no denominators or square roots. Thus, the domain of \((f+g)(x)\) and \((f-g)(x)\) is also all real numbers since they are combinations of \(f(x)\) and \(g(x)\).

Step 5: Conclusion

Therefore, \((f+g)(x) = 14x - 2\), \((f-g)(x) = 12x + 2\) and there are no restrictions to the domain; it is all real numbers.

Key Concepts

Domain Restrictions

When working with functions, understanding domain restrictions is crucial. A function's domain includes all the values of "x" that can be input into the function without causing any undefined behavior, such as division by zero or negative square roots.

In this exercise, we have two linear functions, which are simple in terms of domain issues. Linear functions like \( f(x) = 13x \) and \( g(x) = 1x - 2 \) don't have any restrictions since they don't involve division or square roots. This means:

• The domain for \( f(x) \), \( g(x) \), \( (f+g)(x) \), and \( (f-g)(x) \) is all real numbers.

Recognizing this simplicity early on can help you know when and where to look for potential problems with the domain.

Linear Functions

Linear functions are fundamental in algebra and take the form \( f(x) = mx + b \), where "m" is the slope and "b" is the y-intercept.

In our case:

• \( f(x) = 13x \) is a linear function with a slope of 13 and an intercept of 0.
• \( g(x) = 1x - 2 \) is another linear function with a slope of 1 and an intercept of -2.

These functions produce straight lines when graphed on a coordinate plane. Linear functions are easy to manipulate, making operations like addition and subtraction straightforward.

Sum of Functions

Adding two functions together involves performing the operation on their terms. To find \((f+g)(x)\), you simply add \( f(x) \) and \( g(x) \):

• \( (f+g)(x) = f(x) + g(x) = 13x + (1x - 2) = 14x - 2 \)

This operation combines the like terms, with each term added individually. Adding functions is useful in combining relationships represented by these equations, resulting in another linear function that's simply a sum of the original ones.

In real-world contexts, this can mean combining rates (like speed) or other additive quantities.

Difference of Functions

Subtracting functions means taking one function and subtracting its terms from another. For \((f-g)(x)\), we subtract \( g(x) \) from \( f(x) \):

• \( (f-g)(x) = f(x) - g(x) = 13x - (1x - 2) = 12x + 2 \)

Subtraction impacts each term directly, often changing signs. It's a powerful tool for identifying changes or differences between two relationships modeled by functions.

This can be applied to scenarios like calculating net balances or comparing differing rates.