Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 16

Find the domain of

a.) $f(x) = \sin (e^{-x})$

b.) $g(x) = \sqrt{1 - 2^x}$

a.) Recall that $f(x)$ is a trigonometric function that is defined for all values of $x$. Thus, the domain is $(- \infty, \infty)$.

b.) Since $g(x)$ contains square root,


$
\begin{equation}
\begin{aligned}

& 1 - 2^x \geq 0
\\
\\
& 1 \geq 2^x
\end{aligned}
\end{equation}
$


If we take the natural logarithm of both sides..

$\ln 1 \geq \ln 2^x$

By the property of logarithm


$
\begin{equation}
\begin{aligned}

& \ln 1 \geq x (\ln 2)
\\
\\
& x \leq \frac{\ln 1}{\ln 2}

\end{aligned}
\end{equation}
$


Thus, the domain of $g(x)$ is

$\displaystyle \left(- \infty, \frac{\ln 1}{\ln 2}\right]$ or $\displaystyle \left( -\infty, 0 \right]$