The problem asks to find the values of $x$ where $f'(x) = 0$, given that

$f(x) = \frac{x^2 + 4}{x}$

Step 1: Differentiate $f(x)$

We start by simplifying the function:

$f(x) = \frac{x^2}{x} + \frac{4}{x} = x + \frac{4}{x}$

Now, we differentiate $f(x)$:

$f'(x) = \frac{d}{dx}\left( x + \frac{4}{x} \right)$

Using the power rule and the chain rule:

$f'(x) = 1 - \frac{4}{x^2}$

Step 2: Set $f'(x) = 0$

To find the values of $x$ where $f'(x) = 0$, we solve:

$1 - \frac{4}{x^2} = 0$

Step 3: Solve for $x$

Rearrange the equation:

$\frac{4}{x^2} = 1$

Multiply both sides by $x^2$:

$4 = x^2$

Solve for $x$:

$x = \pm 2$

Thus, the values of $x$ where $f'(x) = 0$ are $x = 2$ and $x = -2$.

Would you like further explanation on any step or more details?

Related Questions:

1. What is the second derivative of $f(x)$ and its significance?
2. How does the behavior of $f(x)$ change at $x = 2$ and $x = -2$?
3. What are the critical points of $f(x)$ besides where $f'(x) = 0$?
4. Can you explain the physical or geometrical interpretation of setting $f'(x) = 0$?
5. What is the domain and range of $f(x)$?

Tip: Always check the domain of the function before solving for critical points, especially if there are restrictions on $x$.