Answer
$$
f(x)=e^{0.001x}-\ln{x}
$$
(a) $ f^{\prime}(x)\gt 0 $ on about $( 567, \infty)$ so $f(x)$ is
increasing on about $( 567, \infty)$
(b) $ f^{\prime}(x) \lt 0 $ on about $(0,567)$ so $f(x)$ is
decreasing on about$(0,567)$.
Work Step by Step
$$
f(x)=e^{0.001x}-\ln{x}
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=0.001e^{0.001x}-\frac{1}{x}\\
\end{aligned}
$$
Note that $f(x)$ is only defined for $x\gt 0 $
look for an $x$-value that makes $f^{\prime}(x)=0$, and solve for $x$.
$$
\begin{aligned}
f^{\prime}(x) =0.001e^{0.001x}-\frac{1}{x}=0\\
\end{aligned}
$$
Use a graphing calculator to plot $f^{\prime}(x)$ for $x\gt 0 $. we find that:
$$
\begin{aligned}
f^{\prime}(x) =0 \quad \text {when } \quad x\approx 567
\end{aligned}
$$
Now we check the sign of $f^{\prime}(x)$ in the two intervals $(0, 567) , (567,\infty)$.
* Test a number in the interval $(0, 567)$ say $ 1$
$$
\begin{aligned}
f^{\prime}(1) &=0.001e^{0.001(1)}-\frac{1}{(1)} \\
&\approx -0.99
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is negative in that interval $(1, 567)$, so $f(x)$ is decreasing on $(1, 567)$.
** Test a number in the interval $(567,\infty)$ say $ 600$
$$
\begin{aligned}
f^{\prime}(600) &=0.001e^{0.001(600)}-\frac{1}{(600)} \\
&\approx 0.00015
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is positive in that interval $( 567, \infty)$, so $f(x)$ is increasing on $( 567, \infty)$.
So, we conclude that :
(a) $ f^{\prime}(x)\gt 0 $ on about $( 567, \infty)$ so $f(x)$ is
increasing on about $( 567, \infty)$
(b) $ f^{\prime}(x) \lt 0 $ on about $(0,567)$ so $f(x)$ is
decreasing on about$(0,567)$.
