Question

The minimum value of the function \[ f(x) = x + \frac{4}{x} \quad {for} \quad x>0 { is:} \]

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When finding the minimum of a function, use the first derivative to find critical points, and confirm the nature of the critical points with the second derivative test.

Answer 1

The Correct Option is D

We are tasked with finding the minimum value of the function \( f(x) = x + \frac{4}{x} \) for \( x > 0 \).
Step 1: Take the Derivative of \( f(x) \)
To find the critical points, we first take the derivative of \( f(x) \) with respect to \( x \):
\[ f'(x) = 1 - \frac{4}{x^2}. \] Step 2: Solve for Critical Points
Set the derivative equal to zero to find the critical points:
\[ 1 - \frac{4}{x^2} = 0 \quad \Rightarrow \quad \frac{4}{x^2} = 1 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = 2. \] (Note that \( x > 0 \), so we only consider the positive solution \( x = 2 \)).
Step 3: Verify the Minimum
To verify that \( x = 2 \) is a minimum, we compute the second derivative of \( f(x) \):
\[ f''(x) = \frac{8}{x^3}. \] Since \( f''(x) > 0 \) for \( x > 0 \), the function is concave up at \( x = 2 \), confirming that it is a minimum.
Step 4: Calculate the Minimum Value
Now, substitute \( x = 2 \) into the original function:
\[ f(2) = 2 + \frac{4}{2} = 2 + 2 = 4. \] Thus, the minimum value of the function is \( \boxed{4} \).