The function f(x)=(-3)/4x^4-8x^3-45/2x^2 + 163 has a local maxima at

The correct answer is Option (3) → \(x=0\) and \(x = -5\)

The critical points come from \(f'(x) = 0\):

    \(f'(x) = -3x^3 - 24x^2 - 45x\)
    \(f'(x) = -3x(x^2 + 8x + 15)\)
    \(f'(x) = -3x(x + 3)(x + 5)\)

    \(f'(x) = 0 \Rightarrow x = 0,\ -3,\ -5\)

By the second derivative test, with \(f''(x) = -9x^2 - 48x - 45\):

    \(f''(0) = -45 \lt 0\): local maximum
    \(f''(-3) = -81 + 144 - 45 = 18 \gt 0\): local minimum
    \(f''(-5) = -225 + 240 - 45 = -30 \lt 0\): local maximum

So f has local maxima at x = 0 and x = −5.

Why the other options are incorrect

• Option (1) \(x=0\) and \(x = 3\) — Incorrect. x = 3 is not a critical point: \(f'(3) = -9 \times 6 \times 8 \ne 0\).

• Option (2) \(x=-3\) and \(x = -5\) — Incorrect. x = −3 is a local minimum, since \(f''(-3) \gt 0\).

• Option (4) \(x=-3\) — Incorrect. x = −3 is the local minimum.

NCERT Exemplar:
Application of Derivatives