Answer
$$
f(x) =\frac{(5-9 x)^{2 / 3}}{7}+1
$$
A relative minimum occurs at $x=\frac{5}{9},$ and we find that:
$$
\begin{aligned}
f(\frac{5}{9}) &=1. \\
\end{aligned}
$$
Work Step by Step
$$
f(x) =\frac{(5-9 x)^{2 / 3}}{7}+1
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=\left(\frac{2}{3}\right) \frac{(5-9 x)^{-1 / 3}}{7}(-9) \\
&=-\frac{6}{7(5-9 x)^{1 / 3}}
\end{aligned}
$$
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 $7(5-9 x)^{1 / 3}$ equal to 0, when $x=\frac{5}{9}$. So,
the only critical number is $x=\frac{5}{9} $ .
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
$$
(-\infty, \frac{5}{9} ), \quad (\frac{5}{9} , \infty ) .
$$
(1)
Test a number in the interval $(-\infty, \frac{5}{9}) $ say $0 $:
$$
\begin{aligned}
f^{\prime}(0) &=-\frac{6}{7(5-9 (0))^{1 / 3}} \\
&=-\frac{6\cdot \:5^{\frac{2}{3}}}{35}\\
&\approx -0.50126
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty , \frac{5}{9}) $.
(2)
Test a number in the interval $( \frac{5}{9} , \infty) $ say $ 1$:
$$
\begin{aligned}
f^{\prime}(1) &=-\frac{6}{7(5-9 (1))^{1 / 3}} \\
&=\frac{3\cdot \:2^{\frac{1}{3}}}{7}\\
&\approx 0.53996
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $( \frac{5}{9} , \infty )$.
From (1), (2) we find that:
The function $f$ is decreasing on interval $ ( -\infty , \frac{5}{9} ) , $ and is increasing on interval $ ( \frac{5}{9} , \infty ) . $ So, a relative minimum occurs at $x=\frac{5}{9},$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(\frac{5}{9}) &=\frac{(5-9 (\frac{5}{9}))^{2 / 3}}{7}+1 \\
&=1. \\
\end{aligned}
$$
