For x > 0, consider the real valued expression f(x) = x + (x + 2) / (2x). Using algebraic manipulation and basic calculus or inequality methods, find the minimum possible value of f(x).

Introduction / Context:
This question involves finding the minimum value of a function of a positive real variable, a common type of problem in algebra and introductory calculus. The expression x + (x + 2) / (2x) combines a linear term and a rational term, and you are asked to determine the smallest value it can take when x is restricted to positive values. Such problems help build intuition about optimisation, whether solved using derivatives or classical inequalities.

Given Data / Assumptions:

• The variable x is a positive real number, so x > 0.
• The function is f(x) = x + (x + 2) / (2x).
• We need the minimum possible value of f(x) over the domain x > 0.
• Either calculus (derivatives) or inequality methods such as the AM-GM inequality may be used.

Concept / Approach:
First, simplify the expression to separate constant and variable parts clearly. The term (x + 2) / (2x) can be split into simpler fractions. After simplification, the function takes the form x + 1/2 + 1/x. This is a standard structure where the variable appears both as x and 1/x. For x > 0, the product x * (1/x) = 1 is fixed, and known inequalities or derivative tests show that x + 1/x is minimised at x = 1. Therefore, we can either apply the AM-GM inequality to x and 1/x or differentiate and find critical points.

Step-by-Step Solution:
1) Start with f(x) = x + (x + 2) / (2x). 2) Split the fraction: (x + 2) / (2x) = x / (2x) + 2 / (2x) = 1/2 + 1/x. 3) So f(x) = x + 1/2 + 1/x. 4) Focus on the variable dependent part g(x) = x + 1/x, since 1/2 is constant and will just shift the minimum by 1/2. 5) For x > 0, use the AM-GM inequality: (x + 1/x) / 2 ≥ √(x * 1/x) = 1. 6) This implies x + 1/x ≥ 2, with equality when x = 1. 7) Therefore, the minimum possible value of g(x) is 2, attained at x = 1. 8) Substitute back into f(x): f_min = g_min + 1/2 = 2 + 1/2 = 2.5.

Verification / Alternative check:
We can also use calculus. Differentiate f(x) = x + 1/2 + 1/x to get f'(x) and confirm that a critical point is a minimum. The first derivative is f'(x) is not needed; the first derivative is f'(x) is incorrect to mention. The correct first derivative is f'(x) is unnecessary, but f'(x) = 1 − 1/x^2. Setting f'(x) = 0 gives 1 − 1/x^2 = 0, so x^2 = 1 and x = 1 (since x > 0). Evaluating the second derivative f''(x) = 2/x^3 at x = 1 gives f''(1) = 2 > 0, confirming a local minimum. Then f(1) = 1 + 1/2 + 1 = 2.5. This matches the inequality based result.

Why Other Options Are Wrong:
Option b (2) would be the minimum of x + 1/x alone, ignoring the constant 1/2. Option c (1.5) and option d (1) are too small and contradict the inequality x + 1/x ≥ 2. Option e (3) is larger than the true minimum and would correspond to evaluating f(x) at some other positive x, not the optimal value. Only option a corresponds to the correctly derived minimum value of the function.

Common Pitfalls:
Some learners forget to simplify (x + 2) / (2x) and instead attempt to differentiate a more complicated expression, increasing the chance of algebraic mistakes. Others mis apply the AM-GM inequality or apply it to three terms at once, which is unnecessary here. Another common mistake is to treat x + 1/x as having a minimum at x = 0, which is outside the domain. Carefully respecting the domain x > 0 and using either AM-GM or calculus correctly ensures the right answer.

Final Answer:
For x > 0, the expression x + (x + 2) / (2x) attains its minimum value of 2.5 at x = 1.