To solve the problem of finding the minimum value of the function (f(x) = \frac{x^2 + 3x + 12}{x - 1}) for (x > 1), follow these steps:

Step 1: Substitute (t = x - 1)

Since (x > 1), (t > 0), and (x = t + 1). Substitute into the function:
[ f(x) = \frac{(t + 1)^2 + 3(t + 1) + 12}{t} ]

Step 2: Simplify the numerator

Expand and combine like terms:
[ (t^2 + 2t + 1) + 3t + 3 + 12 = t^2 + 5t + 16 ]
Thus:
[ f(x) = \frac{t^2 + 5t + 16}{t} = t + \frac{16}{t} + 5 ]

Step 3: Apply AM-GM inequality

For positive (t), (t + \frac{16}{t} \geq 2\sqrt{t \cdot \frac{16}{t}} = 8) (equality when (t = \frac{16}{t}), i.e., (t = 4)).

Step 4: Calculate the minimum value

[ f(x)_{\text{min}} = 8 + 5 = 13 ]

Answer: (\boxed{13})

作者声明:本文包含人工智能生成内容。