Problem 16 - 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\()\) \(f(x)=\frac{2}{x}\) and \(g(x)=|x|\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = \frac{2}{|x|}\) with domain \(x \neq 0\). \((g \circ f)(x) = \left|\frac{2}{x}\right|\) with domain \(x \neq 0\).

Step 1: Understand the Composition

To find the composition of functions \((f \circ g)(x)\) or \((g \circ f)(x)\), we substitute one function into the other.**Definition:** \((f \circ g)(x) = f(g(x))\) and \((g \circ f)(x) = g(f(x))\).First, compute \(f(g(x))\) and then compute \(g(f(x))\).

Step 2: Compute (f o g)(x)

To find \((f \circ g)(x)\), substitute \(g(x) = |x|\) into \(f(x)\):1. **Start with**: \(f(x) = \frac{2}{x}\)2. **Substitute**: \(f(g(x)) = f(|x|) = \frac{2}{|x|}\).So, \((f \circ g)(x) = \frac{2}{|x|}\).

Step 3: Determine Domain of (f o g)(x)

The domain of \((f \circ g)(x)\) = \(\frac{2}{|x|}\) is determined by the restriction of the denominator not being zero.- Since \(|x| = 0\) only when \(x = 0\), \(x\) cannot be 0. - Therefore, the domain is all real numbers except \(x = 0\): \(x \in \mathbb{R}, x eq 0\).

Step 4: Compute (g o f)(x)

To find \((g \circ f)(x)\), substitute \(f(x) = \frac{2}{x}\) into \(g(x)\):1. **Start with**: \(g(x) = |x|\)2. **Substitute**: \(g(f(x)) = g\left(\frac{2}{x}\right) = \left|\frac{2}{x}\right|\).So, \((g \circ f)(x) = \left|\frac{2}{x}\right|\).

Step 5: Determine Domain of (g o f)(x)

The domain of \((g \circ f)(x)\) = \(\left|\frac{2}{x}\right|\) is determined similarly to the domain for \((f \circ g)(x)\).- The function requires \(x\) to not be 0 as it results in division by zero. - Thus, the domain is all real numbers except \(x = 0\): \(x \in \mathbb{R}, x eq 0\).

Key Concepts

Understanding the Function Domain

A function's domain is the set of all possible input values (x-values) that the function can accept without causing any mathematical errors, like division by zero or taking the square root of a negative number. Considering this, finding the domain is crucial in understanding how a function behaves.

In our exercise, the composition

• For \((f \circ g)(x) = \frac{2}{|x|}\), the denominator, \(|x|\), must not be zero, as this would make the fraction undefined. Therefore, the domain excludes \(x=0\).
• For \((g \circ f)(x) = \left|\frac{2}{x}\right|\), a similar restriction applies. The term \(x=0\) would again create a division by zero, rendering the function undefined.

Overall, both functions share the domain of all real numbers except zero, denoted as \(x \in \mathbb{R}, \ x eq 0\). Remember, always check inside the functions to identify and exclude any problematic values from the domain.

Exploring the Absolute Value Function

The absolute value function, denoted by \(|x|\), is quite straightforward: it measures the distance of a number from zero on the number line. Therefore, \(|x|\) is always non-negative.

For any number \(x\), whether positive or negative,

• \(|x| = x\) if \(x\) is positive or zero
• \(|x| = -x\) if \(x\) is negative

Absolute value functions come in handy when addressing questions with no concern for signs, just magnitudes.
This property was useful in our exercise because the composition \((f \circ g)(x)\) involves replacing \(x\) in \(\frac{2}{x}\) with \(|x|\). Doing so ensures that the divisor is always positive, apart from \(x=0\), which must be avoided.

Diving into Rational Functions

Rational functions are functions that are defined by the ratio of two polynomials. In a simple term, they are fractions \(\frac{P(x)}{Q(x)}\) where both \(P(x)\) and \(Q(x)\) are polynomials.

The critical part of rational functions is to ensure their denominators do not equal zero, as this creates undefined outputs.

• For instance, the function \(f(x) = \frac{2}{x}\) is rational because it is a fraction with polynomial numerator \(2\) and denominator \(x\).
• When dealing with rational functions, we must address any restrictions originating from the denominator. In this case, \(x eq 0\) to avoid dividing by zero.

This is key in both composed functions \((f \circ g)(x)\) and \((g \circ f)(x)\), where the nature of rational functions and their restrictions play a defining role in establishing the allowable domain.