Answer
$$
f(x)=\ln \left(x^{2}+1\right)-x^{0.3}
$$
(a) $ f^{\prime}(x)\gt 0 $ on about $(0, 558)$ so $f(x)$ is
increasing on about $(0, 558)$,
(b) $ f^{\prime}(x) \lt 0 $ on about $( 558, \infty) $ so $f(x)$ is
decreasing on about $( 558, \infty)$.
Work Step by Step
$$
f(x)=\ln \left(x^{2}+1\right)-x^{0.3}
$$
The first derivative is
$$
f^{\prime}(x)=\frac{2 x}{x^{2}+1}-0.3 x^{-0.7}
$$
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 $x$.
$$
\begin{aligned}
f^{\prime}(x) =\frac{2 x}{x^{2}+1}-0.3 x^{-0.7}=0\\
\end{aligned}
$$
Use a graphing calculator to plot $f^{\prime}(x)$ for $x\gt 0 $. we find that:
$$
\begin{aligned}
f^{\prime}(x) =0 \quad \text {when } \quad x=558
\end{aligned}
$$
Now we check the sign of $f^{\prime}(x)$ in the two intervals $(0, 558) , (558,\infty)$.
* Test a number in the interval $(0, 558)$ say $ 1$
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2 (1)}{(1)^{2}+1}-0.3 (1)^{-0.7} \\
&= 0.7
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is positive in that interval $(0, 558)$, so $f(x)$ is increasing on $(0, 558)$.
** Test a number in the interval $(558,\infty)$ say $ 600$
$$
\begin{aligned}
f^{\prime}(600) &=\frac{2 (600)}{(600)^{2}+1}-0.3 (600)^{-0.7} \\
& =-0.00007
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is negative in that interval $( 558, \infty)$, so $f(x)$ is decreasing on $( 558, \infty)$.
So, we conclude that :
(a) $ f^{\prime}(x)\gt 0 $ on about $(0, 558)$ so $f(x)$ is
increasing on about $(0, 558)$,
(b) $ f^{\prime}(x) \lt 0 $ on about $( 558, \infty) $ so $f(x)$ is
decreasing on about $( 558, \infty)$.
