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Chapter 2: Problem 11

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

Short Answer

Expert verified
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 by step solution

01

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 \(x^{2}-1\).
02

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.
03

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 \(x \in (-\infty,-1) \cup (-1,1) \cup (1,\infty)\).

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