Answer
$$
f(x) =x^{2}+\frac{1}{x}
$$
A relative minimum occurs at $x=\frac{1}{\sqrt[3] {2}}=\left(\frac{3 \sqrt{4}}{2}\right) ,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f\left(\frac{3 \sqrt{4}}{2}\right) &=\left(\frac{\sqrt[3]{4}}{2}\right)^{2}+\left(\frac{\sqrt[3]{4}}{2}\right)^{-1} \\
&=\left(\frac{\sqrt[3]{4}}{2}\right)^{-1}\left[\left(\frac{\sqrt[3]{4}}{2}\right)^{3}+1\right] \\
&=\frac{2}{\sqrt[3]{4}}\left(\frac{4}{8}+1\right)=\frac{2}{\sqrt[3]{4}}\left(\frac{3}{2}\right) \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} \\
&=\frac{3 \sqrt[3]{2}}{2} \approx 1.890
\end{aligned}
$$
Work Step by Step
$$
f(x) =x^{2}+\frac{1}{x}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=2x-\frac{1}{x^{2}}\\
&=\frac{2x^{3}-1}{x^{2}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{2x^{3}-1}{x^{2}}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2x^{3}-1&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2x^{3}&=1 \\
x^{3}&=\frac{1}{2} \\
x &=\frac{1}{\sqrt[3] {2}} \\
\end{aligned}
$$
Next, 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 $x^{2}$ equal to 0, when $x=0$ But , $f(x) $ is not defined at $x=0 $. therefore,
The only critical number is : $x =\frac{1}{\sqrt[3] {2}} $ .
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty, \frac{1}{\sqrt[3] {2}} ), \quad (\frac{1}{\sqrt[3] {2}} ,\infty ) .
$$
(1)
Test a number in the interval $(-\infty, \frac{1}{\sqrt[3] {2}} ) $ say $-1 $:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{2(-1)^{3}-1}{(-1)^{2}}\\
&=-3\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, \frac{1}{\sqrt[3] {2}} ) $.
(2)
Test a number in the interval $ (\frac{1}{\sqrt[3] {2}} ,\infty ) $ say $ 1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2(1)^{3}-1}{(1)^{2}}\\
&=1\\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(\frac{1}{\sqrt[3] {2}} ,\infty )$.
From (1), (2) we find that:
(*)
The function $f$ is decreasing on interval $ (-\infty, \frac{1}{\sqrt[3] {2}} ) , $ and is increasing on interval $ (\frac{1}{\sqrt[3] {2}} ,\infty ) . $ So, a relative minimum occurs at $x=\frac{1}{\sqrt[3] {2}}=\left(\frac{3 \sqrt{4}}{2}\right) ,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f\left(\frac{3 \sqrt{4}}{2}\right) &=\left(\frac{\sqrt[3]{4}}{2}\right)^{2}+\left(\frac{\sqrt[3]{4}}{2}\right)^{-1} \\
&=\left(\frac{\sqrt[3]{4}}{2}\right)^{-1}\left[\left(\frac{\sqrt[3]{4}}{2}\right)^{3}+1\right] \\
&=\frac{2}{\sqrt[3]{4}}\left(\frac{4}{8}+1\right)=\frac{2}{\sqrt[3]{4}}\left(\frac{3}{2}\right) \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} \\
&=\frac{3 \sqrt[3]{2}}{2} \approx 1.890
\end{aligned}
$$
