To find the value of $x$ that minimizes the function $f(x) = 2x^2 + \frac{64}{x}$, we will use differentiation. Here are the steps:

1. Differentiate the function:

First, let's find the first derivative of $f(x)$ with respect to $x$.

$f(x) = 2x^2 + \frac{64}{x}$

The first derivative, $f'(x)$, is calculated as:

$f'(x) = \frac{d}{dx} \left( 2x^2 \right) + \frac{d}{dx} \left( \frac{64}{x} \right)$

$f'(x) = 4x - \frac{64}{x^2}$

2. Set the first derivative equal to zero:

To find the critical points, set $f'(x) = 0$ and solve for $x$.

$4x - \frac{64}{x^2} = 0$

3. Solve for $x$:

Multiply both sides of the equation by $x^2$ to eliminate the fraction:

$4x^3 - 64 = 0$

$4x^3 = 64$

$x^3 = 16$

$x = \sqrt[3]{16}$

The value of $x$ that minimizes the function is $x = \sqrt[3]{16}$.

4. Confirm that this is a minimum:

To confirm that this critical point is a minimum, we can check the second derivative, $f''(x)$.

$f''(x) = \frac{d}{dx} \left( 4x - \frac{64}{x^2} \right)$

$f''(x) = 4 + \frac{128}{x^3}$

Since $f''(x) > 0$ for all $x > 0$, the function is concave up at $x = \sqrt[3]{16}$, confirming that this is indeed a minimum.

Thus, the value of $x$ that minimizes the function is $x = \sqrt[3]{16}$.

Would you like details or have any questions about the process?

---

Here are 5 related questions for further exploration:

1. How would the minimum value change if the function were $3x^2 + \frac{64}{x}$?
2. What happens to the minimum value if $f(x) = 2x^2 + \frac{64}{x} + 10$?
3. How do you find the maximum value of a function using differentiation?
4. What is the significance of the second derivative in determining the nature of a critical point?
5. Can you apply this method to find the minimum of a more complex function, such as $2x^2 + \frac{64}{x} + x^4$?

Tip: Always verify if a critical point is a minimum or maximum by checking the second derivative or using the first derivative test.