Question 178169
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The values of <i><b>x</b></i> that are excluded are any values that would cause the denominator to be zero.


So take the denominator of *[tex \Large f(x) = {(x-1) \over (x^2+1)}]


and set it equal to zero:


*[tex \Large x^2 + 1 = 0] which has no solutions over the real numbers.  Hence, the domain of <i><b>f</b></i> is *[tex \Large \{x | x \in \R\}]


Compare with *[tex \Large g(x) = {(x-1) \over (x^2-1)}]


Here setting the denominator equal to zero yields:


*[tex \Large x^2 - 1 = 0]


which has two real solutions, namely 1 and -1.  Therefore the the domain of <i><b>g</b></i> is *[tex \Large \{x | x \in \R\, x \neq 1, x \neq -1}]


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