1. State the problem: Given $f(x)=\frac{1}{x}$ and $g(x)=x^2-1$, find the domain of $f\circ g$. 2. Formula and rules: The composite is defined by $(f\circ g)(x)=f(g(x))$ and it is defined exactly when $g(x)$ lies in the domain of $f$. The domain of $f$ is all real numbers except 0, i.e., $\{x\in\mathbb{R}:x\neq 0\}$. The domain of $g$ is all real numbers $\mathbb{R}$ since a polynomial is defined for every real $x$. 3. Compute the composite and restrictions: Substitute to get $$ (f\circ g)(x)=\frac{1}{x^2-1} $$ This expression is defined exactly when the denominator $x^2-1$ is not zero. Factor the denominator: $x^2-1=(x-1)(x+1)$. Set the denominator equal to zero to find excluded points: $x^2-1=0$ gives $x=1$ and $x=-1$. 4. Conclusion: Therefore the domain of $f\circ g$ is all real numbers except $1$ and $-1$. In set notation the domain is $\mathbb{R}\setminus\{1,-1\}$. 5. Match to the choices: This matches option (c) $\mathbb{R} - \{1, -1\}$. Final answer: (c).