Problem 11 - College Algebra Essentials

Question

Find the domain of each function. $$g(x)=\frac{1}{x^{2}+1}-\frac{1}{x^{2}-1}$$

Step-by-Step Solution

Answer

The domain of the function \(g(x)=\frac{1}{x^{2}+1}-\frac{1}{x^{2}-1}\) is all real numbers except \(x=1\) and \(x=-1\). In interval notation, the domain is \(x \in (-\infty,-1) \cup (-1,1) \cup (1,\infty)\).

Step 1: Identify the potential problematic terms

The function \(g(x)\) is composed of two fractions. Both denominators cannot be equal to zero as division by zero is undefined. These are the terms \(x^{2}+1\) and \(x^{2}-1\). It's easy to see that the first doesn't lead to a restriction, because \(x^{2}+1\geq 1\), for all \(x\) which means it cannot be zero. So we only need to consider the second denominator \(x^{2}-1\).

Step 2: Find the values of x for which the denominator equals zero

Set the denominator \(x^{2}-1\) equal to zero and solve for \(x\). Thus we obtain:\(x^{2}-1=0\) \This leads to \(x^{2}=1\) which gives \(x=1\) or \(x=-1\). These are the problematic values, as if \(x\) takes any of these values, the function \(g(x)\) becomes undefined.

Step 3: Exclude the problematic values from the domain

The domain of a function is the set of all real numbers excluding any values that make the function undefined. So the domain of the function \(g(x)\) is all real numbers except \(x=1\) and \(x=-1\). In interval notation, this is written as \(x \in (-\infty,-1) \cup (-1,1) \cup (1,\infty)\).