Problem 70 - Precalculus Essentials

Question

Find; a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=\frac{x}{x+5}, g(x)=\frac{6}{x}$$

Step-by-Step Solution

Answer

a. The function \((f \circ g)(x)\) is given by \(\frac{6}{6 + 5x}\). b. The domain of \(f \circ g(x)\) is all real numbers except \(0\) and \(-\frac{6}{5}\).

Step 1: Find the composite function (f o g)(x)

To find \((f \circ g)(x)\), replace every \(x\) in \(f(x)\) with \(g(x)\). This gives: \(f(g(x)) = \frac{g(x)}{g(x) + 5} = \frac{\frac{6}{x}}{\frac{6}{x} + 5}\.

Step 2: Simplify the composite function

Simplify the fraction within the fraction by multiplying the numerator and the denominator by \(x\). This gives \(f(g(x)) = \frac{6}{6 + 5x}\.

Step 3: Find the domain of the composite function

The domain of \(f \circ g\) is all \(x\) for which \(g(x)\) and \(f(g(x))\) are defined. Since \(g(x) = \frac{6}{x}\), we know that \(g(x)\) is defined for all \(x ≠ 0\). Similarly, since \(f(g(x)) = \frac{6}{6 + 5x}\), \(f(g(x))\) is defined for all \(x ≠ -\frac{6}{5}\). Therefore, the domain of \(f \circ g\) is all real numbers except \(0\) and \(-\frac{6}{5}\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values, often represented by the variable \( x \), which will produce a valid output when substituted into the function. Understanding the domain of a function is crucial, as it tells us the set of values for which the function is defined.

For example, consider the function \( g(x) = \frac{6}{x} \). Here, the expression involves division by \( x \), meaning \( x \) cannot be \( 0 \), because division by zero is undefined. Thus, the domain of \( g \) excludes \( x = 0 \).

• Keywords: Domain, function, valid input values, exclusion.
• To determine the domain, find all values for \( x \) that will not cause division by zero or result in an undefined expression.

Similarly, when examining the composite function \( f(g(x)) = \frac{6}{6 + 5x} \), it's important to identify any constraints that \( x \) must satisfy. Here, the function implies \( 6 + 5x eq 0 \). Solving for \( x \) in this equation gives \( x eq -\frac{6}{5} \). Thus, the domain of the composite function \( f \circ g \) is all real numbers, except \( x = 0 \) and \( x = -\frac{6}{5} \).

When composing functions, always check for restrictions in the domain from both the inner and outer functions.

Function Composition

Function composition involves combining two functions in order to create a new function by applying one function to the result of another. It is denoted as \((f \circ g)(x)\), which means applying \(g(x)\) first and then \(f(x)\).

The process is straightforward: substitute the entire function \(g(x)\) into every occurrence of \(x\) in \(f(x)\). For instance, given \( f(x) = \frac{x}{x+5} \) and \( g(x) = \frac{6}{x} \), the composite function \( (f \circ g)(x) \) is found by replacing \( x \) in \( f(x) \) with \( g(x) \), resulting in \( \frac{g(x)}{g(x) + 5} \).

• Process of composition: Replace \( x \) in \( f(x) \) with \( g(x) \).
• Further simplification might be needed to write the composite function in a simpler form.

In our example, the fraction requires simplification: \( f(g(x)) = \frac{\frac{6}{x}}{\frac{6}{x} + 5} \). This can be simplified further by multiplying both the numerator and the denominator by \( x \), leading to \( \frac{6}{6 + 5x} \).

This results in a clean expression without fractions inside of fractions, making the function easier to handle and understand. So, simplify where possible for clarity and ease of interpretation.

Rational Expressions

Rational expressions are fractions in which both the numerator and the denominator are polynomials. They can sometimes be tricky because they may introduce restrictions on the domain if the denominator includes variables that can lead to a division by zero.

Take for example the function \( f(x) = \frac{x}{x+5} \). Here, the denominator \( x + 5 \) cannot be zero, because it would make the fraction undefined. This indicates that \( x eq -5 \) is a necessary condition for the function to exist.

• Key features: Numerator and denominator are polynomials.
• Division by zero is forbidden, leading to domain restrictions.

When dealing with rational expressions as composite functions, like \( f(g(x)) = \frac{6}{6 + 5x} \), similar rules apply. The denominator must not equal zero to avoid undefined expressions. Here, the domain restriction comes from \( 6 + 5x eq 0 \), translating to \( x eq -\frac{6}{5} \).

When simplifying or working with rational expressions, always remember to check and derive the permissible values for the variable involved, as ignoring these could lead to incorrect conclusions about the function's behavior.