1. **Problem statement:** Determine where the function $$f(x) = \frac{x - 1}{x^2 - 1}$$ is continuous and if possible, extend it to a larger domain continuously. 2. **Analyze the denominator:** The denominator is $$x^2 - 1$$, which factors as $$x^2 - 1 = (x-1)(x+1)$$. 3. **Domain restrictions:** The function is undefined where the denominator is zero, i.e., at $$x=1$$ and $$x=-1$$. 4. **Simplify the function:** \[ f(x) = \frac{x - 1}{(x - 1)(x + 1)} \] For $$x \neq 1$$, we can cancel $$x - 1$$: \[ f(x) = \frac{1}{x + 1} \] 5. **Check the behavior at $$x=1$$:** The original function has a removable discontinuity at $$x=1$$ since the factor cancels out. 6. **Extend the function:** Define a new function $$g(x)$$ by \[ g(x) = \begin{cases} f(x) = \frac{x-1}{x^2 -1}, & x \neq 1 \\ \lim_{x \to 1} f(x), & x=1 \end{cases} \] Calculate the limit: \[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{1}{x + 1} = \frac{1}{2} \] So, \[ g(1) = \frac{1}{2} \] 7. **Domain and continuity:** - Original function $$f$$ is continuous on $$\mathbb{R} \setminus \{ -1,1 \}$$. - Extended function $$g$$ is continuous on $$\mathbb{R} \setminus \{ -1 \}$$ since the discontinuity at $$1$$ is removed. 8. **Conclusion:** - $$f$$ is discontinuous at $$x=\pm 1$$. - By defining $$g(1) = \frac{1}{2}$$, $$g$$ is continuous at $$x=1$$ and everywhere except at $$x=-1$$. **Final answer:** $$\boxed{\text{Extend } f \text{ by defining } g(1) = \frac{1}{2} \text{, then } g \text{ is continuous on } \mathbb{R} \setminus \{-1\} }$$