The minimum value of $x^2+\frac{1}{x}$ is:

$\left(\frac{1}{2}\right)^{\frac{2}{3}}+(2)^{\frac{1}{3}}$

The correct answer is Option (1) → $\left(\frac{1}{2}\right)^{\frac{2}{3}}+(2)^{\frac{1}{3}}$

Differentiate $f(x)$ with respect to $x$,

$f'(x)=\frac{d}{dx}(x^2+\frac{1}{x})$

$=2x-\frac{1}{x^2}$

To find critical points, $f'(x)=0$

$2x-\frac{1}{x^2}=0$

$2x^3=1$

$x^3=\frac{1}{2}$

$x=(\frac{1}{2})^{\frac{1}{3}}$

Taking second derivative test

$f''(x)=\frac{d}{dx}(2x-\frac{1}{x^2})$

$=2+\frac{2}{x^3}>0$

Substituting $x=\frac{1}{\sqrt[3]{2}}$

$f\left(\frac{1}{\sqrt[3]{2}}\right)=\left(\frac{1}{\sqrt[3]{2}}\right)^2+\frac{1}{\frac{1}{\sqrt[3]{2}}}$

$=\frac{1}{\sqrt[3]{4}}+\sqrt[3]{2}$

$=\left(\frac{1}{2}\right)^{\frac{2}{3}}+(2)^{\frac{1}{3}}$