Let f be a function defined by \(f(x) = 2x^3-3x^2-36x+2\), then which of the following are correct?
(A) The critical points of f(x) are -2 and 3.
(B) The function f(x) increases in the interval (3, ∞)
(C) The function f(x) decreases in the interval (-2, 3)
(D) The function f(x) increases in the interval (-2, 3)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
\( f'(x) = 6x^2 - 6x - 36 \)
\( = 6(x^2 - x - 6) \)
\( = 6(x - 3)(x + 2) \)
The critical points are where \(f'(x) = 0\), that is x = −2 and x = 3. The sign of f′(x):
\( x \lt -2:\ f'(x) \gt 0 \)
\( -2 \lt x \lt 3:\ f'(x) \lt 0 \)
\( x \gt 3:\ f'(x) \gt 0 \)
So the critical points are −2 and 3, f increases on \((3, \infty)\) and f decreases on \((-2, 3)\).
Checking the statements
(A) The critical points of f(x) are -2 and 3. — Correct. \(6(x - 3)(x + 2) = 0\) at these points.
(B) The function f(x) increases in the interval (3, ∞) — Correct. Both factors are positive there, so \(f'(x) \gt 0\).
(C) The function f(x) decreases in the interval (-2, 3) — Correct. \(f'(x) \lt 0\) there, e.g. \(f'(0) = -36\).
(D) The function f(x) increases in the interval (-2, 3) — Incorrect. f decreases there, since \(f'(x) \lt 0\).