Question

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

eq 0 \iff e^{1 - x^2} \
eq 1$

Explanation:

Step1: Set denominator not equal to 0

$1 - e^{1-x^2}
eq 0$

Step2: Rearrange the inequality

$e^{1-x^2}
eq 1$

Step3: Use $e^0=1$ to simplify

$1-x^2
eq 0$

Step4: Solve for $x$

$x^2
eq 1 \implies x
eq 1 \text{ and } x
eq -1$

Answer:

All real numbers except $x=1$ and $x=-1$, or in interval notation: $(-\infty, -1) \cup (-1, 1) \cup (1, +\infty)$