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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 \( x \).

Find the values of x, where f'(x) = 0, given that f(x) = (x^2 + 4) / x.

This problem explores finding the critical points of the function f(x) = (x^2 + 4) / x by setting its derivative f'(x) = 0. It uses basic differentiation rules and solves for x. Suitable for students in Grades 11-12 or early university calculus.

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