The function \( f(x)=81-3x^3 \) has
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A point of inflexion at \( x=0 \)
\( f'(x)=-9x^2 \)
\( f'(x)=0\ \Rightarrow\ x=0 \)
For x < 0 and for x > 0, \( f'(x)=-9x^2\lt 0 \). The derivative does not change sign at x = 0, so f is decreasing on both sides and there is no local maximum or minimum there.
\( f''(x)=-18x \)
f''(x) > 0 for x < 0 and f''(x) < 0 for x > 0, so the concavity changes at x = 0. Hence x = 0 is a point of inflexion.
Why the other options are incorrect
Option (1) A local maxima at \( x=0 \) — Incorrect. f is decreasing on both sides of x = 0, so it has no local maximum there; f'(0) = 0 alone does not decide.
Option (2) A local minima at \( x=0 \) — Incorrect. f is decreasing on both sides of x = 0, so it has no local minimum there either.
Option (4) A local maxima at \( x=3 \) and a local minima at \( x=0 \) — Incorrect. \( f'(3)=-81\ne 0 \), so x = 3 is not even a critical point, and x = 0 is not a minimum.