Answer
$$
f(x) =3-(8+3 x)^{2 / 3}
$$
A relative maximum occurs at $x=\frac{-8}{3},$ and we find that:
$$
f(\frac{-8}{3}) =3.
$$
Work Step by Step
$$
f(x) =3-(8+3 x)^{2 / 3}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=-\frac{2}{3}(8+3 x)^{-1 / 3}(3) \\
&=-\frac{2}{(8+3 x)^{1 / 3}}
\end{aligned}
$$
To find the critical numbers, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $(8+3 x)^{1 / 3}$ equal to 0, when $x=\frac{-8}{3}$. So,
the only critical number is $x=\frac{-8}{3} $ .
We can still apply the first derivative test, however, to find where $f$ is increasing or decreasing. Now, check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, \frac{-8}{3} ), \quad (\frac{-8}{3} , \infty ) .
$$
(1)
Test a number in the interval $(-\infty, \frac{-8}{3} ) $ say $-3 $:
$$
\begin{aligned}
f^{\prime}(-3) &=-\frac{2}{(8+3 (-3))^{1 / 3}} \\
&=2\\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty , \frac{-8}{3} ) $.
(2)
Test a number in the interval $( \frac{-8}{3} , \infty) $ say $ 0$:
$$
\begin{aligned}
f^{\prime}(0) &=-\frac{2}{(8+3 (0))^{1 / 3}} \\
&=-1\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( \frac{-8}{3} , \infty )$.
From (1), (2) we find that:
The function $f$ is increasing on intervals $ (-\infty ,\frac{-8}{3} ), $ and is decreasing on interval $(\frac{-8}{3} , \infty). $ So, a relative maximum occurs at $x=\frac{-8}{3},$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(\frac{-8}{3}) &=3-(8+3 (\frac{-8}{3}))^{2 / 3} \\
&=3 \\
\end{aligned}
$$
