For \(x \in \mathrm{R}\), if \(f(x) = -(x-1)^2 + 2\), then
(A) \(f\) is an increasing function on \((-\infty, 1]\)
(B) \(f\) has no critical points
(C) \(f\) has a maximum value at \(x = 1\)
(D) \(f\) has a minimum value at \(x = 1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (C) only
Differentiating:
\(f'(x) = -2(x - 1)\)
\(f'(x) = 0 \text{ at } x = 1\)
\(f'(x) \gt 0 \text{ for } x \lt 1,\quad f'(x) \lt 0 \text{ for } x \gt 1\)
\(f''(x) = -2 \lt 0\)
Also \(-(x - 1)^2 \le 0\), so \(f(x) \le 2 = f(1)\) for every x.
So x = 1 is the only critical point, f is increasing up to x = 1, and f has its maximum value 2 at x = 1.
Checking the statements
(A) \(f\) is an increasing function on \((-\infty, 1]\) — Correct. \(f'(x) \gt 0\) for \(x \lt 1\), and f’(1) = 0 only at the end point.
(B) \(f\) has no critical points — Incorrect. \(f'(1) = 0\), so x = 1 is a critical point.
(C) \(f\) has a maximum value at \(x = 1\) — Correct. \(f''(1) = -2 \lt 0\); the maximum value is f(1) = 2.
(D) \(f\) has a minimum value at \(x = 1\) — Incorrect. f(1) = 2 is the largest value, not the smallest.