Problem 30 - College Algebra

Question

Find: a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=\frac{5}{x+4}, g(x)=\frac{1}{x}$$

Step-by-Step Solution

Answer

The composite function \((f \circ g)(x) = \frac{5x}{1+4x}\) and its domain are all real numbers except \(x = -\frac{1}{4}\).

Step 1: Compute (f \circ g)(x)

Compute the composite function \(f(g(x))\). This involves substituting \(g(x)\) into the function \(f(x)\). So we replace \(x\) in \(f(x)\) with \(g(x)\). This gives us \(f(g(x)) = f(\frac{1}{x}) = \frac{5}{(\frac{1}{x}) + 4}\). Whenever the opportunity arises, it's recommended to simplify this expression.

Step 2: Simplify (f \circ g)(x)

We continue on from the previous step to simplify \(f(g(x))\). Given \(f(\frac{1}{x}) = \frac{5}{(\frac{1}{x}) + 4}\), we can simplify further by eliminating the complex fraction: \(\frac{5}{\frac{1}{x} + 4}= \frac{5x}{1+4x}\). Thus, \(f \circ g(x) = \frac{5x}{1+4x}\).

Step 3: Find Domain of (f \circ g)

The domain of a function is the set of all real numbers for which the function is defined. For \(f(g(x)) = \frac{5x}{1+4x}\), we need to find all \(x\) for which the denominator is not equal to zero. Therefore, we solve \(1 + 4x ≠ 0\), which yields \(x ≠ -\frac{1}{4}\).

Key Concepts

Function Composition

Understanding function composition can seem tricky at first, but it's quite straightforward once you break it down. Function composition, denoted as \(f \circ g\), involves combining two functions to create a new function. This is done by inserting the output of one function into the input of another. In our example, we are given two functions: \(f(x) = \frac{5}{x+4}\) and \(g(x) = \frac{1}{x}\). To find the composite function \(f(g(x))\), you replace the \(x\) in \(f(x)\) with \(g(x)\).
This gives us \(f(g(x)) = f\left(\frac{1}{x}\right) \). Once you substitute this in, you simplify the math expression which results in \(f(g(x)) = \frac{5x}{1 + 4x}\).
This final expression represents the combination of \(f\) and \(g\). Function composition is particularly useful because it enables us to combine functions seamlessly and operate them within a single framework.

• Replace input value of \(f\) with \(g(x)\)
• Simplify the resulting function

Rational Functions

Rational functions are an important concept in algebra and calculus. These functions are expressed as the ratio of two polynomials. In simpler terms, they are fractions where the numerator and the denominator are both polynomials. In our composite function example, \(f(g(x)) = \frac{5}{x+4}\) becomes \(\frac{5x}{1+4x}\). This is a rational function because both the numerator and the denominator are polynomials.
Rational functions have unique characteristics:

• The most critical aspect is determining where these functions are undefined. This usually occurs where the denominator equals zero.
• Understanding this is essential, as it directly ties into finding the domain of a rational function.

Rational functions often include asymptotes and intercepts, which are vital for graphing and further analysis. Handling these functions requires careful manipulation and simplification techniques to make them easier to work with in different scenarios.

Domain of a Function

The domain of a function is a set of all possible input values (usually denoted by \(x\)) for which the function is defined. When dealing with rational functions, like our composite \(f(g(x)) = \frac{5x}{1+4x}\), finding the domain can be slightly more involved.
The main factor to bear in mind here is that the function cannot be defined where the denominator is zero. To find where the function is undefined, you solve the equation \(1 + 4x = 0\). Solving this gives \(x eq -\frac{1}{4}\), meaning that at \(x = -\frac{1}{4}\), the function doesn't exist.
Additionally, because \(g(x) = \frac{1}{x}\) is part of the composite function, \(x eq 0\) must also be considered, as \(g(x)\) is undefined at \(x=0\).

• The overall domain of \(f \circ g\) is thus all real numbers, except \(x = -\frac{1}{4}\) and \(x = 0\).

Determining the domain is crucial, as it informs us where our function operates correctly and helps to identify any potential limitations when graphing or analyzing the function.