For which of the following real functions, x=0 is a point of local ma…

The correct answer is Option (1) → (A), (B) and (C) only

Each function is a polynomial with no x term, so f'(0) = 0 in every case. By the second derivative test, x = 0 is a local maximum when f''(0) < 0. At x = 0 only the \( x^2 \) term contributes to f''(0), so f''(0) = 2 × (coefficient of \( x^2 \)).

For (A):

    \( \displaystyle f''(x)=48x^2-18x-4 \)
    \( \displaystyle f''(0)=-4 \)

For (B):

    \( \displaystyle f''(x)=60x^3+12\sqrt{2}x^2-14 \)
    \( \displaystyle f''(0)=-14 \)

For (C), where e is a constant:

    \( \displaystyle f''(x)=60x^2-6 \)
    \( \displaystyle f''(0)=-6 \)

For (D):

    \( \displaystyle f''(x)=-18x+4 \)
    \( \displaystyle f''(0)=4 \)

So f''(0) is −4, −14, −6 and 4 for (A), (B), (C) and (D).

Checking the statements

(A) \( f(x)=4x^{4}-3x^{3}-2x^{2}+1 \) — Correct. f''(0) = −4 < 0, a local maximum.

(B) \( f(x)=3x^{5}+\sqrt{2}x^{4}-7x^{2}+4 \) — Correct. f''(0) = −14 < 0, a local maximum.

(C) \( f(x)=5x^{4}-3x^{2}+e \) — Correct. f''(0) = −6 < 0, a local maximum.

(D) \( f(x)=-3x^{3}+2x^{2}-9 \) — Incorrect. f''(0) = 4 > 0, so x = 0 is a local minimum.