Find the minimum value of
\[2 \sqrt{x} + \frac{1}{x}\]for $x > 0.$
I notice that this is a function of the form $f(x) = a \sqrt{x} + \frac{b}{x}$ where $a = 2$ and $b = 1.$ I also notice that this function is always positive for $x > 0,$ since both terms are positive. This means that the minimum value of the function must occur at a critical point, where the derivative is zero or undefined, or at an endpoint of the domain. The domain of the function is $(0, \infty),$ so there is no endpoint to consider. Therefore, I only need to find the critical points and compare their function values. To find the derivative of the function, I use the power rule and the quotient rule. I get $f'(x) = a \cdot \frac{1}{2} \cdot x^{-\frac{1}{2}} - b \cdot \frac{1}{x^2}.$ I simplify this to $f'(x) = \frac{a}{2 \sqrt{x}} - \frac{b}{x^2}.$ I set this equal to zero and solve for x. I get $\frac{a}{2 \sqrt{x}} - \frac{b}{x^2} = 0.$ I multiply both sides by $2 \sqrt{x}$ and get $a - \frac{2b \sqrt{x}}{x^2} = 0.$ I isolate $\sqrt{x}$ and get $\sqrt{x} = \frac{a x^2}{2b}.$ I square both sides and get $x = \frac{a^2 x^4}{4b^2}.$ I divide both sides by $x$ and get $1 = \frac{a^2 x^3}{4b^2}.$ I multiply both sides by $\frac{4b^2}{a^2}$ and get $\frac{4b^2}{a^2} = x^3.$ I take the cube root of both sides and get $\sqrt[3]{\frac{4b^2}{a^2}} = x.$ In this case, $a = 2$ and $b = 1,$ so I plug those values in and get $\sqrt[3]{\frac{4}{4}} = x.$ I simplify and get $x = 1.$ This is the only critical point of the function, so it must be the minimum value. To verify this, I plug it back into the original function and get $f(1) = 2 \sqrt{1} + \frac{1}{1} = 2 + 1 = 3.$ 