For the function \(f(x) = x^2 - 6x + 10\), consider the following statements:
(A) f has a local maximum at x = 3.
(B) The minimum value of f is 1.
(C) f is strictly increasing in \((3, \infty)\).
(D) \(f'(0) = 6\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (C) only
Differentiating:
\(f'(x) = 2x - 6\)
\(f''(x) = 2 \gt 0\)
f′(x) = 0 at x = 3, and since f″(3) > 0 this is a minimum:
\(f(3) = 9 - 18 + 10 = 1\)
So f has its minimum value 1 at x = 3, f′(x) = 2x − 6 is positive for x > 3, and f′(0) = −6.
Checking the statements
(A) f has a local maximum at x = 3. — Incorrect. \(f''(3) = 2 \gt 0\), so x = 3 is a local minimum.
(B) The minimum value of f is 1. — Correct. f(3) = 9 − 18 + 10 = 1.
(C) f is strictly increasing in \((3, \infty)\). — Correct. For x > 3, f′(x) = 2x − 6 > 0.
(D) \(f'(0) = 6\). — Incorrect. f′(0) = 0 − 6 = −6.