Problem 54 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine the domains of (a) \(f,\) (b) \(g\) and (c) \(f \circ g .\) Use a graphing utility to verify your results. $$f(x)=\frac{3}{x^{2}-1}, \quad g(x)=x+1$$

Step-by-Step Solution

Answer

The domain of function f is all real numbers except -1 and 1. The domain of function g is all real numbers. The domain of the composite function f(g(x)) is all real numbers except -2 and 0.

Step 1: Find the domain of function f

The function is \( f(x) = \frac{3}{x^{2}-1} \). We must find the values of x for which the denominator is not zero. Solve \( x^{2}-1 \neq 0 \) for x. The solutions are \( x \neq -1,1 \). Therefore, the domain of \( f \) is all real numbers except -1 and 1.

Step 2: Find the domain of function g

The function \( g(x)=x+1 \) is a linear function. Linear functions are defined for all real numbers. Therefore, the domain of \( g \) is all real numbers.

Step 3: Find the domain for the composite function f(g(x))

The composite function is \( f(g(x)) \). We already determined the domains of the individual functions. However, the composite function also requires that \( g(x) \), the inside function, does not produce a result that is outside the domain of the outer function \( f \). We find \( f(g(x)) = f(x+1) = \frac{3}{(x+1)^2 -1} \). As we established for \( f \), it's necessary to solve for \( (x+1)^2 -1 \neq 0 \). This gives us \( x \neq -2, 0 \). Therefore, the domain of \( f \circ g \) is all real numbers excluding -2 and 0.

Key Concepts

Composite Functions

Understanding composite functions is like learning how to assemble a puzzle. A composite function, denoted as \( f \circ g \), combines two functions where the output of one function becomes the input for another. To illustrate, if you have functions \( f \) and \( g \) with \( g \) acting first, you create \( f(g(x)) \) by taking \( g(x) \) and then applying \( f \).
In our exercise, we have \( f(x) = \frac{3}{x^{2}-1} \) and \( g(x) = x + 1 \). When we combine them, we get \( f(g(x)) = f(x+1) \) which requires careful examination to ensure that the outputs of \( g \) do not fall into the values excluded from \( f's \) domain. This process shows the intricate dance between the two functions, where the limitations of each affect the overall outcome of the composite function.

Domain of a Function

Imagine the domain of a function as a party guest list. It determines who is allowed to 'enter' a function. Formally, the domain is the set of all possible input values (usually \( x \) values) for which the function is defined and produces real numbers as outputs. To identify the domain, we look at restrictions such as division by zero and square roots of negative numbers, which are not permitted as they are undefined or result in imaginary numbers.
In the given exercise, we found the domain of \( f(x) \) by ensuring that the denominator does not equal zero (\( x^2 - 1 eq 0 \)), excluding \( x = -1 \) and \( x = 1 \). Such an analysis is crucial for understanding the behavior of functions and preventing mathematical 'faux pas'.

Rational Functions

A rational function is like a fraction in which both the numerator and the denominator are polynomials. Picture a fraction where, instead of simple numbers in the numerator and denominator, there are entire expressions. For example, \( f(x) = \frac{3}{x^{2}-1} \) in our exercise is a rational function.

Non-Zero Denominators

The most important rule of thumb with rational functions is avoiding zero in the denominator, as division by zero is undefined. Therefore, finding the domain of a rational function involves identifying and excluding the \( x \) values that would make the denominator zero. It's as important as making sure you don't divide your pie into zero pieces—you can't have a pie if it doesn't exist!

Linear Functions

Linear functions are the straight-forward, straight-line relations you encounter in algebra. They have the form \( y = mx + b \) where \( m \) is the slope and \( b \) is the y-intercept. In our exercise, \( g(x) = x + 1 \) is a linear function.

No Restrictions on Domain

The beauty of linear functions is their simplicity: they are defined for all real numbers, meaning their domain is \( (-\infty, +\infty) \). There are no 'x-marks' on the number line that are forbidden. This makes linear functions highly predictable and easy to work with, especially when combining them with other types of functions in composite functions.