Find the minimum value of
\[2 \sqrt{x} + \frac{1}{x}\]for $x > 0.$
By AM-GM,
\[2 \sqrt{x} + \frac{1}{x} = \sqrt{x} + \sqrt{x} + \frac{1}{x} \ge 3 \sqrt[3]{\sqrt{x} \cdot \sqrt{x} \cdot \frac{1}{x}} = 3.\]Equality occurs when $x = 1,$ so the minimum value is $\boxed{3}.$
3
Level 2
Intermediate Algebra
Problem: Find the minimum value of
\[2 \sqrt{x} + \frac{1}{x}\]for $x > 0.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the function \( f(x) = 2 \sqrt{x} + \frac{1}{x} \) for \( x > 0 \), we first need to determine the critical points of the function. This involves taking the derivative of \( f(x) \) and setting it equal to zero.

1. Calculate the derivative \( f'(x) \):
   - The derivative of \( 2 \sqrt{x} \) is \( \frac{1}{\sqrt{x}} \).
   - The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \).
   - Therefore, \( f'(x) = \frac{1}{\sqrt{x}} - \frac{1}{x^2} \).

2. Set the derivative equal to zero to find critical points:
   - \( \frac{1}{\sqrt{x}} - \frac{1}{x^2} = 0 \).
   - Rearranging gives \( \frac{1}{\sqrt{x}} = \frac{1}{x^2} \).
   - Cross-multiplying leads to \( x^2 = \sqrt{x} \).
   - Squaring both sides results in \( x^4 = x \).
   - Factoring gives \( x(x^3 - 1) = 0 \), leading to \( x = 1 \) as the only positive solution.

3. To confirm that this critical point is a minimum, we can use the second derivative test:
   - Calculate the second derivative \( f''(x) \):
     - The derivative of \( f'(x) = \frac{1}{\sqrt{x}} - \frac{1}{x^2} \) is \( f''(x) = -\frac{1}{2} x^{-3/2} + \frac{2}{x^3} \).
   - Evaluate \( f''(1) \):
     - \( f''(1) = -\frac{1}{2} + 2 = \frac{3}{2} \), which is positive.

4. Since \( f''(1) > 0 \), the function has a local minimum at \( x = 1 \).

5. Finally, evaluate \( f(1) \):
   - \( f(1) = 2 \sqrt{1} + \frac{1}{1} = 2 + 1 = 3 \).

Thus, the minimum value of \( 2 \sqrt{x} + \frac{1}{x} \) for \( x > 0 \) is \( 3 \).<|END_OF_LATENT|> The final answer is $3$.