Answer
a) $(-\infty, -1)\cup(-1, 1)\cup(1,\infty)$
b) $(-\infty, \infty)$
Work Step by Step
To find the domain, we look for which values of $x$ the function is undefined. For both a) and b), it is when the denominator equals $0$.
a)
$1-e^{1-x^2}=0$
$1=e^{1-x^2}$
$\ln(1)=\ln(e^{1-x^2})$
$0=\ln(e^{1-x^2})$
Using a log property (where $\ln{x^y}=y\ln{x}$):
$0=(1-x^2)\ln(e)$
Remembering that $\ln(e)=1$
$1=x^2$
$x=-1,\thinspace 1$
So for a), the domain is $(-\infty, -1)\cup(-1, 1)\cup(1,\infty)$
b)
$e^{cos{x}}=0$
There is no $x$ value where this equation holds true, so the domain is all real numbers.
