To solve the problem, we can use the method of completing the square, which is efficient for analyzing quadratic functions.

Step 1: Prove (f(x) = x^2 - 2x + 3) is always positive

Complete the square for the quadratic term:
[ \begin{align} f(x) &= x^2 - 2x + 3 \ &= (x^2 - 2x + 1) + 2 \quad \text{(add and subtract 1 to complete the square)} \ &= (x - 1)^2 + 2. \end{align} ]
Since ((x - 1)^2 \geq 0) for all real (x), adding 2 gives:
[ f(x) = (x - 1)^2 + 2 \geq 2 > 0. ]
Thus, (f(x)) is always positive.

Step 2: Find the minimum value of (f(x))

From the vertex form ((x - 1)^2 + 2):

• The term ((x - 1)^2) reaches its minimum value of 0 when (x = 1).
• Substituting (x = 1) into (f(x)):
  [ f(1) = (1 - 1)^2 + 2 = 2. ]

Final Answers:

• (f(x)) is always positive because it equals ((x-1)^2 + 2 \geq 2 > 0).
• The minimum value of (f(x)) is (\boxed{2}).

(\boxed{2})

作者声明:本文包含人工智能生成内容。