Problem 26 - A Graphical Approach to College Algebra

Question

For each pair of functions, (a) find \((f+g)(x),(f-g)(x),\) and \((f g)(x)\) (b) give the domains of the functions in part (a); (c) find \(\frac{f}{8}\) and give its domain; (d) find \(f \circ g\) and give its domain; and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=|2 x-4|, g(x)=x+1$$

Step-by-Step Solution

Answer

All functions and compositions have the domain \(\{x \in \mathbb{R}\}\).

Step 1: Find Sums and Differences of Functions

Calculate the sum and the difference of the functions: 1. For \((f+g)(x)\), we have: \[(f+g)(x) = f(x) + g(x) = |2x - 4| + (x + 1) = |2x - 4| + x + 1\]2. For \((f-g)(x)\), we have: \[(f-g)(x) = f(x) - g(x) = |2x - 4| - (x + 1) = |2x - 4| - x - 1\]

Step 2: Find Product of Functions

Calculate the product of the functions: For \((fg)(x)\), we have: \[(fg)(x) = f(x) \cdot g(x) = |2x - 4| \cdot (x + 1)\] This expression represents the product and cannot be simplified further without specific values of \(x\).

Step 3: Determine Domains of the Functions

Since both \(|2x-4|\) and \(x+1\) are defined for all \(x \in \mathbb{R}\), the domains of \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\) are all \(\{x \in \mathbb{R}\}\).

Step 4: Find f Divided by 8 and its Domain

Calculate \(\frac{f}{8}\): \[\frac{f(x)}{8} = \frac{|2x - 4|}{8}\]Since absolute value and division by 8 do not impose any additional restrictions, the domain remains \(\{x \in \mathbb{R}\}\).

Step 5: Find Composition f ∘ g and its Domain

Evaluate \(f \circ g\): \[f(g(x)) = f(x + 1) = |2(x + 1) - 4| = |2x + 2 - 4| = |2x - 2|\]The domain of \(f \circ g\) is \(\{x \in \mathbb{R}\}\) since \(f(x) = |2x-2|\) can accept any real number.

Step 6: Find Composition g ∘ f and its Domain

Evaluate \(g \circ f\): \[g(f(x)) = g(|2x - 4|) = |2x - 4| + 1\] Since \(g(x) = x+1\) is defined for any real number, the domain of \(g \circ f\) is also \(\{x \in \mathbb{R}\}\).

Key Concepts

Function Composition

Function composition involves creating a new function by applying one function to the results of another. If you have two functions, say \(f(x)\) and \(g(x)\), composing \(f\) with \(g\) is notated as \(f \circ g\), and is evaluated as \(f(g(x))\). Similarly, \(g \circ f\) is evaluated as \(g(f(x))\).
Let's take a closer look: In our given problem, \(f(x) = |2x - 4|\) and \(g(x) = x + 1\). To find \(f \circ g\), replace \(x\) in \(f(x)\) with \(g(x)\):

• \(f(g(x)) = f(x + 1) = |2(x + 1) - 4| = |2x + 2 - 4| = |2x - 2|\)

Similarly, for \(g \circ f\), replace \(x\) in \(g(x)\) with \(f(x)\):

• \(g(f(x)) = g(|2x - 4|) = |2x - 4| + 1\)

Function composition creates new relationships between the input and output variables, allowing us to explore complex transformations of data.

Function Domains

Understanding the domain of a function is crucial as it tells us the set of input values (\(x\)-values) for which the function is defined. In simpler terms, it's what \(x\)-values we can plug into the function without causing any mathematical errors such as division by zero or taking the square root of a negative number.
In the exercise example, the functions involved are \(f(x) = |2x - 4|\) and \(g(x) = x + 1\). Both functions are defined for all real numbers, which means their domains are \(\{x \in \mathbb{R}\}\). This is true for:

• Sum and difference: \((f+g)(x) = |2x - 4| + x + 1\) and \((f-g)(x) = |2x - 4| - x - 1\)
• Product: \((fg)(x) = |2x - 4| \cdot (x + 1)\)
• Division by 8: \(\frac{|2x - 4|}{8}\)
• Compositions \(f \circ g\) and \(g \circ f\)

This is because absolute value and linear functions like \(x + 1\) do not impose additional constraints beyond \(x \in \mathbb{R}\). Always pay careful attention to the operations involved to correctly determine the domain of a function.

Absolute Value Function

The absolute value function is one of the most fundamental functions in mathematics. Given as \(f(x) = |2x - 4|\) in our exercise, it describes the distance of a number from zero on the number line regardless of direction. This means for any expression inside the absolute value, like \(2x - 4\), the output is always a non-negative number.
In the context of operations such as addition, subtraction, and multiplication with other functions, the absolute value retains its property of non-negativity, influencing the shape of the graph and the range of outputs it can produce.
For example, whether you replace \(x\) with values that make \(2x - 4\) positive or negative, \(|2x - 4|\) always results in a positive outcome. This is a key feature that ensures the domain stays expansive as \(x\) traverses all real numbers. To solve more complex exercises involving absolute value functions, ensure an understanding of how it manipulates post-expression inputs and outputs.

Algebraic Expressions

Algebraic expressions form the building blocks of algebra. They are combinations of numbers, variables, and operations representing a value or set of values. In our example, it's crucial to understand the manipulation of expressions like \(|2x - 4|\), \(x+1\), and their combinations.
When you perform operations such as addition, subtraction, multiplication, or composition on algebraic expressions, they transform but retain an underlying structure. For instance:

• Sum: \((f+g)(x) = |2x - 4| + x + 1\)
• Difference: \((f-g)(x) = |2x - 4| - x - 1\)
• Product: \((fg)(x) = |2x - 4| \cdot (x + 1)\)

Each operation involves carefully combining the parts of the expression by following algebraic rules. Simplification may not always be possible without specific values for variables, but understanding each part's role in the expression is key to manipulating them correctly.
As you work with algebraic expressions, focus on understanding how operations modify their structure and what that means for their mathematical behavior and applications.