MHB 6.2.15 Find the domain of each function.

[karush]

$\textsf{6.2.15 Find the domain of each function.}$
(a) $f(x)=\dfrac{1-e^{x^2}}{1-e^{1-x^2}}$
set the denominator to zero and solve
$1-e^{1-x^2}=0$
then
$x=1,-1$
from testing the domain is
$(-1,1)$(b) $f(x)=\dfrac{1+x}{e^{ \cos x}}$
set $e^{\cos x}=0$ which is $x\in \mathbb{R}$
so domain is
$(-\infty,\infty)$Ok, I think these are correct don't know the book answer
I did this mostly via obervation with the denominator but presume limit should be used otherwise

 

[skeeter]

(a) $1-e^{1-x^2} \ne 0 \implies \text{ domain is } x \in (-\infty,-1) \cup (-1,1) \cup (1,\infty) \text{ or } x \in \mathbb{R} \, ; \, x \ne \pm 1$

(b) $e^{\cos{x}} \ne 0 \text{ for all } x \in \mathbb{R} \implies \text{ domain is } x \in \mathbb{R}$

 

[karush]

oh...

 

[HallsofIvy]

Do you understand why you set the denominator equal to 0?