To solve the problem (inferred from the example solution), we use the Arithmetic Mean-Geometric Mean (AM-GM) Inequality for positive numbers.

Step 1: Rewrite the function

Given (f(x) = x + \frac{1}{x-1}) (with (x > 1)), rewrite it to group terms for AM-GM:
[f(x) = (x-1) + \frac{1}{x-1} + 1]

Step 2: Apply AM-GM Inequality

For positive numbers (a) and (b), the AM-GM inequality states:
[a + b \geq 2\sqrt{ab}]

Here, (a = x-1) and (b = \frac{1}{x-1}) (both positive since (x > 1)). Thus:
[ (x-1) + \frac{1}{x-1} \geq 2\sqrt{(x-1) \cdot \frac{1}{x-1}} = 2 \cdot 1 = 2]

Step 3: Find the minimum value

Adding 1 to both sides:
[f(x) = (x-1) + \frac{1}{x-1} + 1 \geq 2 + 1 = 3]

Step 4: Equality condition

Equality holds when (a = b):
[x-1 = \frac{1}{x-1} \implies (x-1)^2 = 1 \implies x-1 = 1 \implies x = 2]

At (x=2), (f(2) = 2 + \frac{1}{2-1} = 3), confirming the value is achievable.

Answer: (\boxed{3}) (Note: The example solution likely had a typo; this is the minimum value.)

(\boxed{3})

作者声明:本文包含人工智能生成内容。