For what value of x does f(x) = (x - 3)² + 1 reach its minimum?

For what value of x does f(x) = (x - 3)² + 1 reach its minimum?

• A. 3
• B. 2
• C. -3
• D. 0

Answer: (A)

The squared part can never be negative, so the smallest f(x) occurs when that square is zero. (x − 3)² = 0 exactly when x = 3, giving f(x) = 0 + 1 = 1. Since the square term is at least 0 for all x, you can’t get a value smaller than 1, so the minimum happens at x = 3. Substituting other options would yield larger results: for example, x = 2 gives f = (−1)² + 1 = 2, x = −3 gives f = (−6)² + 1 = 37, and x = 0 gives f = (−3)² + 1 = 10.

The squared part can never be negative, so the smallest f(x) occurs when that square is zero. (x − 3)² = 0 exactly when x = 3, giving f(x) = 0 + 1 = 1. Since the square term is at least 0 for all x, you can’t get a value smaller than 1, so the minimum happens at x = 3. Substituting other options would yield larger results: for example, x = 2 gives f = (−1)² + 1 = 2, x = −3 gives f = (−6)² + 1 = 37, and x = 0 gives f = (−3)² + 1 = 10.