Question

The minimum value of \( f(x) = x + \frac{4}{x + 2} \) is

• A. -1
• B. -2
• C. 1
• D. 2

Hint:

To find minimum of functions with rational terms, apply derivative test and evaluate at critical points.

Answer 1

The Correct Option is D

Let \( f(x) = x + \frac{4}{x + 2} \). 

To find the minima, take the derivative:
\[ f'(x) = 1 - \frac{4}{(x + 2)^2} \] Set \( f'(x) = 0 \Rightarrow 1 = \frac{4}{(x + 2)^2} \Rightarrow (x + 2)^2 = 4 \Rightarrow x = 0 \text{ or } -4 \)

Check values:
At \( x = 0 \), \( f(0) = 0 + \frac{4}{2} = 2 \)
At \( x = -4 \), \( f(-4) = -4 + \frac{4}{-2} = -4 - 2 = -6 \) (Not in the valid domain for minimum in this context).

Hence, the minimum value is \( \boxed{2} \).