To find points of discontinuity for the given function

$f(x) = x + \sqrt{x^2 - 1},$

we analyze its domain and behavior.

Step 1: Analyze the domain

The square root function $\sqrt{x^2 - 1}$ is defined only when $x^2 - 1 \geq 0$. This inequality implies: $x^2 \geq 1.$

Solving this: $x \leq -1 \quad \text{or} \quad x \geq 1.$

Thus, the domain of $f(x)$ is: $(-\infty, -1] \cup [1, \infty).$

Step 2: Check continuity at the endpoints of the domain

At $x = -1$ and $x = 1$, the square root becomes $\sqrt{(-1)^2 - 1} = \sqrt{1 - 1} = 0$. Therefore, $f(-1)$ and $f(1)$ are well-defined: $f(-1) = -1 + 0 = -1, \quad f(1) = 1 + 0 = 1.$

Step 3: Identify discontinuities

The function $f(x)$ is discontinuous where the domain is broken, i.e., in the intervals $(-1, 1)$, because the square root $\sqrt{x^2 - 1}$ is not defined for $x \in (-1, 1)$.

Conclusion

The points of discontinuity for $f(x)$ occur in the interval $(-1, 1)$, where the function is not defined. There are no discontinuities at $x = -1$ or $x = 1$.

If you'd like, I can explain this in more depth or help with similar problems.

Related Questions:

1. What is the range of the function $f(x)$?
2. How does the function behave as $x \to \pm\infty$?
3. Is $f(x)$ continuous on each interval in its domain?
4. Can we plot the function to visualize its behavior?
5. How do we compute the derivative of $f(x)$?

Tip:

When analyzing continuity, always check both the domain and the limits at boundary points!