Let \(f(x) = 4x^3- 18x^2 +27x - 5,\ x \in \mathrm{R}\). Then which of the following statements are TRUE?
(A) \(f''(x) = 24x - 36\)
(B) \(f\) has local maxima at \(x =\displaystyle\frac{3}{2}\) but no minima
(C) \(f\) has neither maxima nor minima
(D) \(f\) has both maxima and minima
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
\(f'(x) = 12x^2 - 36x + 27\)
\(= 3(4x^2 - 12x + 9)\)
\(= 3(2x - 3)^2\)
\(f''(x) = 24x - 36\)
\(f'(x) = 0\) only at \(x = \frac{3}{2}\), and \(f'(x) \gt 0\) on both sides of it (the second derivative test fails there, since \(f''\left(\frac{3}{2}\right) = 0\)). As f′ does not change sign, the first derivative test gives no local maximum or minimum at \(x = \frac{3}{2}\); f is increasing on R.
So \(f''(x) = 24x - 36\), and f has no local maximum and no local minimum.
Checking the statements
(A) \(f''(x) = 24x - 36\) — Correct. The derivative of \(12x^2 - 36x + 27\) is \(24x - 36\).
(B) \(f\) has local maxima at \(x = \frac{3}{2}\) but no minima — Incorrect. \(f' \gt 0\) on both sides of \(\frac{3}{2}\), so f keeps increasing there.
(C) \(f\) has neither maxima nor minima — Correct. \(f'(x) = 3(2x - 3)^2\) never changes sign.
(D) \(f\) has both maxima and minima — Incorrect. The only critical point, \(x = \frac{3}{2}\), is not an extremum.