Problem 12 - Intermediate Algebra

Question

Determine \((f \circ g)(x)\) and \((g \circ f)(x)\) for each pair of functions. Also specify the domain of \((f \circ g)(x)\) and \((g \circ f)(x)\). (Objective 1\()\) \(f(x)=7 x-2\) and \(g(x)=\sqrt{2 x-1}\)

Step-by-Step Solution

Answer

(f ∘ g)(x) = 7(\sqrt{2x-1}) - 2; domain: [1/2, ∞). (g ∘ f)(x) = \sqrt{14x-5}; domain: [5/14, ∞).

Step 1: Understand Function Composition

Function composition involves substituting one function into another. For two functions \(f(x)\) and \(g(x)\), the notation \((f \circ g)(x)\) means replacing \(x\) in \(f(x)\) with \(g(x)\), giving us \(f(g(x))\). Similarly, \((g \circ f)(x)\) means substituting \(f(x)\) into \(g(x)\), giving us \(g(f(x))\).

Step 2: Determine (f ∘ g)(x)

Substitute \(g(x)\) into \(f(x)\). The function \(g(x)\) is \(\sqrt{2x-1}\). Substitute this into \(f\):\[f(g(x)) = f(\sqrt{2x-1}) = 7(\sqrt{2x-1}) - 2\]This is the composition \((f \circ g)(x)\).

Step 3: Determine the Domain of (f ∘ g)(x)

The domain of \((f \circ g)(x)\) depends on \(g(x)\) being defined and \(f(g(x))\) being defined. Since \(g(x) = \sqrt{2x-1}\), it is defined when the expression under the square root is non-negative: \\[2x - 1 \geq 0 \Rightarrow x \geq \frac{1}{2}.\] So, the domain of \((f \circ g)(x)\) is \([\frac{1}{2}, \infty)\), with consideration that \(f\) handles real numbers without additional restrictions.

Step 4: Determine (g ∘ f)(x)

Substitute \(f(x)\) into \(g(x)\). The function \(f(x)\) is \(7x - 2\). Substitute this into \(g\):\[g(f(x)) = g(7x - 2) = \sqrt{2(7x - 2) - 1} = \sqrt{14x - 4 - 1} = \sqrt{14x - 5}.\]This is the composition \((g \circ f)(x)\).

Step 5: Determine the Domain of (g ∘ f)(x)

The domain of \((g \circ f)(x)\) requires \(14x - 5 \geq 0\). Solve for \(x\) to determine the domain:\[14x - 5 \geq 0 \Rightarrow 14x \geq 5 \Rightarrow x \geq \frac{5}{14}.\]The domain of \((g \circ f)(x)\) is \([\frac{5}{14}, \infty)\).

Key Concepts

Domain of a Function

The domain of a function tells us all the input values for which the function is defined. In simpler terms, it's a collection of all possible values of \(x\) we can use in a function without encountering any mathematical issues, like dividing by zero or taking the square root of a negative number.

• For basic algebraic expressions like \(7x - 2\), the domain is all real numbers because there are no restrictions.
• However, if the function includes a square root, such as \(\sqrt{2x-1}\), the domain is only values of \(x\) that make the inside of the square root non-negative. This is because you cannot take the square root of a negative number within the set of real numbers.

In the given exercise, when finding the domain of \((f \circ g)(x)\), you consider where \(g(x)\), or \(\sqrt{2x-1}\), is defined, requiring \(2x-1 \geq 0\). Hence, the domain is \([\frac{1}{2}, \infty)\).
For \((g \circ f)(x)\), the expression becomes \(\sqrt{14x - 5}\), so the domain requires \(14x - 5 \geq 0\), giving us \([\frac{5}{14}, \infty)\).
Understanding the domain ensures we work with valid inputs and avoid undefined operations.

Function Notation

Function notation is a way to represent functions in a concise manner, usually expressed as \(f(x)\), where \(f\) names the function and \(x\) represents the input variable.

• The expression \(f(x) = 7x - 2\) tells us that if we substitute a value for \(x\), the output will be determined by multiplying \(x\) by 7 and subtracting 2.
• In \(g(x) = \sqrt{2x - 1}\), the expression implies that after substituting \(x\), you first multiply \(x\) by 2, subtract 1, and then take the square root of the result.

Function notation is not only a shorthand but also acts as a precise mapping of inputs to outputs.
The composition of functions, indicated by \((f \circ g)(x)\) and \((g \circ f)(x)\), uses this notation to further illustrate how functions can be combined. In \(f \circ g\), you apply \(g(x)\) first and then \(f\) to that result. While in \(g \circ f\), the order is reversed. This notation aids in effectively managing complex operations and compositions.

Algebraic Functions

Algebraic functions include operations that involve addition, subtraction, multiplication, division, and powers of variables. These functions are at the core of understanding more complex mathematical concepts.

• The function \(f(x) = 7x - 2\) is linear as it forms a straight line when graphed and involves basic operations like multiplication and subtraction.
• The function \(g(x) = \sqrt{2x - 1}\) is more complex, as it includes a square root, making it a radical function.

Understanding algebraic functions is important for function composition:
- When composing \((f \circ g)(x)\), we substitute the algebraic function \(g(x)\) into \(f(x)\).
- Similarly, \((g \circ f)(x)\) involves substituting \(f(x)\) into the algebraic function \(g(x)\).
Each step can change the nature of the function and affect its domain. By understanding algebraic functions, you become adept at predicting and manipulating these transformations during function compositions.