Let's analyze the domain of each function provided:

(a) $f(x) = \frac{1 - e^{x^2}}{1 - e^{1-x^2}}$

Domain Analysis:

The function is defined for all $x$ except where the denominator is zero. To find these values: $1 - e^{1-x^2} = 0$ $e^{1-x^2} = 1$ $1 - x^2 = 0$ $x^2 = 1$ $x = \pm 1$

So, $x \neq \pm 1$. Additionally, $e^{x^2}$ is defined for all $x$, so it doesn't impose any further restrictions.

Conclusion:

The domain of $f(x)$ is $x \in \mathbb{R} \setminus \{ -1, 1 \}$.

(b) $f(x) = \frac{1 + x}{e \cdot \cos x}$

Domain Analysis:

The function is undefined when $\cos x = 0$ because this would make the denominator zero: $\cos x = 0$ The cosine function is zero at $\frac{\pi}{2} + k\pi$ for $k$ integer.

Conclusion:

The domain of $f(x)$ is $x \in \mathbb{R} \setminus \left\{ \frac{\pi}{2} + k\pi : k \in \mathbb{Z} \right\}$.

Would you like more detailed steps on any of these calculations or have other questions about the functions?

Here are some additional questions to consider:

1. What would the graph of each function look like?
2. How do the points where the functions are undefined affect their continuity?
3. What is the range of these functions?
4. How does the behavior of $f(x)$ in part (b) change near the points where the denominator is zero?
5. What would happen if you were to integrate each of these functions over a suitable interval?

Tip: Always check the behavior of functions not only where they are undefined but also near those points to understand limits and potential asymptotes.