To determine where the given function $f(x) = \frac{\ln(x) + \tan^{-1}(x)}{x^2 - 1}$ is continuous, we need to analyze both the numerator and the denominator of the function.

Step-by-step analysis:

1. Numerator: The numerator is $\ln(x) + \tan^{-1}(x)$.

   • The natural logarithm function $\ln(x)$ is only defined for $x > 0$, which means that for $\ln(x)$ to be valid, $x$ must be positive.
   • The inverse tangent function $\tan^{-1}(x)$ is defined for all real numbers, so it does not impose any additional restrictions.

   Therefore, the numerator is defined for $x > 0$.

2. Denominator: The denominator is $x^2 - 1$, which can be factored as: $x^2 - 1 = (x - 1)(x + 1)$ The denominator equals zero when $x = 1$ or $x = -1$. The function is not defined where the denominator is zero, meaning $x = 1$ and $x = -1$ are points of discontinuity.

Conclusion:

• The function $f(x)$ is continuous where both the numerator and denominator are defined. This implies:
  • $x > 0$ (from the logarithm function)
  • $x \neq 1$ and $x \neq -1$ (from the denominator)

Thus, $f(x)$ is continuous for $0 < x < 1$ and $x > 1$, i.e., in the intervals $(0, 1) \cup (1, \infty)$.

Would you like further details or clarifications?

Here are some related questions you might find helpful:

1. How can we analyze the differentiability of $f(x)$?
2. What happens to the behavior of $f(x)$ as $x$ approaches 1 or -1?
3. How do the inverse trigonometric functions impact the continuity of other functions?
4. Can you describe the domain of the function $f(x)$ if the numerator were $e^x$ instead of $\ln(x)$?
5. What is the significance of the function being undefined at $x = 1$ and $x = -1$?

Tip: When checking continuity, always ensure that both the numerator and denominator are defined and examine points where the denominator could become zero.