The function \( f(x)=\ 2x^3-15x^2+36x+4 \) has local maxima at \( x= \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
The critical points come from f′(x) = 0:
\( f'(x)=6x^2-30x+36 \)
\( =6(x-2)(x-3) \)
\( f''(x)=12x-30 \)
\( f''(2)=-6\lt 0 \), so x = 2 is a point of local maximum.
\( f''(3)=6\gt 0 \), so x = 3 is a point of local minimum.
Why the other options are incorrect
Option (1) 4 — Incorrect. \( f'(4)=12\ne 0 \), so x = 4 is not a critical point.
Option (2) 3 — Incorrect. x = 3 is a critical point, but \( f''(3)=6\gt 0 \), so it gives a local minimum.
Option (3) 0 — Incorrect. \( f'(0)=36\ne 0 \), so x = 0 is not a critical point.