Problem 18 - 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\()\) \(f(x)=\frac{4}{x-2}\) and \(g(x)=\frac{3}{4 x}\)

Step-by-Step Solution

Answer

(f ∘ g)(x) = \(\frac{16x}{3 - 8x}\), (g ∘ f)(x) = \(\frac{3x - 6}{16}\); Domains: (f ∘ g) is x≠0, x≠3/8; (g ∘ f) is x≠2.

Step 1: Identify f(g(x))

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). We have \(g(x) = \frac{3}{4x}\). Substitute this into \(f(x) = \frac{4}{x-2}\) to get \[f(g(x)) = \frac{4}{\frac{3}{4x} - 2}.\] Simplifying, we find \[f(g(x)) = \frac{4 \cdot 4x}{3 - 8x} = \frac{16x}{3 - 8x}.\]

Step 2: Determine the domain of f(g(x))

The domain of \(f(g(x))\) is given by the set of \(x\) such that \(g(x)\) is defined and \(g(x) - 2 eq 0\). \(g(x)\) is undefined for \(x = 0\) because this causes the denominator in \(g(x)\) to be zero. Additionally, \(g(x) = 2\) when \(\frac{3}{4x} = 2\), which simplifies to \(x = \frac{3}{8}\). Therefore, the domain is \(x eq 0\) and \(x eq \frac{3}{8}\).

Step 3: Identify g(f(x))

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). We have \(f(x) = \frac{4}{x-2}\). Substitute this into \(g(x) = \frac{3}{4x}\) to get \[g(f(x)) = \frac{3}{4 \cdot \frac{4}{x-2}} = \frac{3 \cdot (x-2)}{16}.\] Simplifying, this becomes \[g(f(x)) = \frac{3x - 6}{16}.\]

Step 4: Determine the domain of g(f(x))

The domain of \(g(f(x))\) is given by the set of \(x\) such that \(f(x)\) is defined and does not make the denominator in \(g(x)\) equal to zero. \(f(x)\) is undefined for \(x = 2\) because it would result in division by zero. Therefore, the domain is \(x eq 2\).

Key Concepts

Understanding the Domain of a Function

The domain of a function is the set of all possible input values (usually represented by \(x\)) for which the function is defined. Think of it as all the numbers you can substitute into the function without facing issues like division by zero, which makes the function undefined.

When working with composed functions like \((f \circ g)(x)\) or \((g \circ f)(x)\), determining the domain can be tricky because you have to account for the domains of both the individual functions and the composition.

• The domain of \(f(g(x))\) is made up of all \(x\) such that \(g(x)\) is defined, and \(f\) can handle the output from \(g(x)\).
• If \(f(x)\) or \(g(x)\) has a restriction (e.g., a denominator that yields zero), those restrictions also apply to their compositions.

By examining what makes the functions undefined, like dividing by zero or taking square roots of negative numbers, you can outline the domain effectively.

Exploring Rational Functions

Rational functions come about when you have a ratio of two polynomials. They look like \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials.

The primary concern with rational functions is their domain—specifically, finding where they are undefined.

• These functions are undefined where the denominator \(Q(x)\) becomes zero because division by zero is not possible.
• To find the domain, solve \(Q(x) = 0\) and exclude these solutions from real numbers.

For example, with \(f(x) = \frac{4}{x - 2}\) and \(g(x) = \frac{3}{4x}\), you must exclude \(x = 2\) for \(f(x)\) and \(x = 0\) for \(g(x)\) to determine where these functions are undefined. Thus, understanding the behaviors at these key points helps in one quickly finding where the function can be used without errors.

Performing Function Operations

Function operations involve combining two or more functions using basic arithmetic operations—addition, subtraction, multiplication, or division—or by composing them.

Composition, which you're currently working with, involves "plugging" one function into another.

• The notation \((f \circ g)(x)\) means substitute \(g(x)\) into \(f(x)\).
• Similarly, \((g \circ f)(x)\) means substitute \(f(x)\) into \(g(x)\).

This process sometimes appears complex but can reveal a lot about how functions work together. When performing composition:

• Always check if the resulting expression is defined across the domain.
• Factor and simplify expressions where possible to find a cleaner understanding of the composed function.

Through these operations, you can see how functions change and affect each other, broadening your understanding of mathematics.