Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
As we see below, the inductor voltages lead the current by 90$^\circ$, while the capacitor voltage lags the current by 90$^\circ$. The resistor voltage and current are in phase.
The capacitor, the inductor, and the resistor are in parallel, so
$$V_L=V_C=V_{R} \tag 1$$
We know that a resistor's voltage and current are in phase while the current leads the voltage in a capacitor by 90$^\circ$, and the voltage leads the current in the inductor by 90$^\circ$.
This means that the angle between $I_C$ and $I_L$ is 180$^\circ$, and the angle between $I_R$, and $I_C$ is 90$^\circ$ while the angle between $I_R$ and $I_L$ is -90$^\circ$.
Hence,
$$I^2 =(I_L-I_C)^2+I_R^2$$
So,
$$\left[ \dfrac{\varepsilon_0}{Z} \right]^2 =\left[ \dfrac{\varepsilon_0}{X_L}-\dfrac{\varepsilon_0}{X_C} \right]^2+\left[ \dfrac{\varepsilon_0}{R} \right]^2$$
where $\varepsilon_0$ is a common factor,
$$\left[ \dfrac{1}{Z} \right]^2 =\left[ \dfrac{1}{X_L}-\dfrac{1}{X_C} \right]^2+\left[ \dfrac{1}{R} \right]^2$$
$$ Z =\left(\left[ \dfrac{1}{\omega L}-\omega C \right]^2+ \dfrac{1}{R^2}\right)^{-1/2} \tag 2$$
Recalling that
$$I=\dfrac{\varepsilon_0}{Z}$$
Plug from (2),
$$I=\dfrac{\varepsilon_0}{\left(\left[ \dfrac{1}{\omega L}-\omega C \right]^2+ \dfrac{1}{R^2}\right)^{-1/2}}$$
$$I=\varepsilon_0\left(\left[ \dfrac{1}{\omega L}-\omega C \right]^2+ \dfrac{1}{R^2}\right)^{1/2}$$
$$\boxed{I=\varepsilon_0\sqrt{\left[ \dfrac{1}{\omega L}-\omega C \right]^2+ \dfrac{1}{R^2}}}$$
$$\color{blue}{\bf [b]}$$
From the boxed formula above,
When $\omega \rightarrow 0$, then $I\rightarrow\infty$.
And when $\omega \rightarrow \infty$, then $I\rightarrow\infty$.
$$\color{blue}{\bf [c]}$$
To find the minimum current, we need to find the point at which $dI/d\omega=0$.
$$\dfrac{dI}{d\omega}=0$$
$$\dfrac{d }{d\omega}\left[ \varepsilon_0^2 \left[ \omega^{-1} L^{-1}-\omega C \right]^2+ \dfrac{1}{R^2}\right]^{1/2}=0$$
$$\dfrac{1}{2}\left[ \varepsilon_0^2 \left[ \omega^{-1} L^{-1} -\omega C \right]^2+ \dfrac{1}{R^2}\right]^{-1/2}\left[\varepsilon_0^2(2)(-\omega^{-2}L^{-1}-C)\right][\omega^{-1} L^{-1}-\omega C]=0$$
Then,
$$\dfrac{\dfrac{1}{2}\left[\varepsilon_0^2(2)(-\omega^{-2}L^{-1}-C)\right][\omega^{-1} L^{-1}-\omega C]}{\left[ \varepsilon_0^2 \left[ \omega^{-1} L^{-1} -\omega C \right]^2+ \dfrac{1}{R^2}\right]^{1/2} }=0$$
$$ (-\omega^{-2}L^{-1}-C) (\omega^{-1} L^{-1}-\omega C ) =0$$
Thus,
$$ -\omega^{-2}L^{-1}-C =0\;\;\;\;\Rightarrow \omega^2=\dfrac{1}{\sqrt{-CL}}\tag{not real}$$
and,
$$ (\omega^{-1} L^{-1}-\omega C ) =0\;\;\;\;\Rightarrow \omega^2=\dfrac{1}{\sqrt{ CL}}=\omega_0^2$$
The minimum value of $I$ occurs at
$$\boxed{\omega=\omega_0=\dfrac{1}{\sqrt{ CL}}}$$
$$\color{blue}{\bf [d]}$$
From above, we can see that the minimum value of the current is at $\omega=\omega_0$, while it reaches high values when $\omega\rightarrow 0$ or $\omega\rightarrow \infty$
See the graph below.
