The function f(x) = 4x^3-7x^2 has point(s) of local minima at

The correct answer is Option (3) → \(x = \displaystyle\frac{7}{6}\)

A local minimum occurs at a critical point where \(f''(x) \gt 0\) (second derivative test).

    \(f'(x) = 12x^2 - 14x = 2x(6x - 7)\)

Putting \(f'(x) = 0\):

    \( \displaystyle x = 0 \quad\text{or}\quad x = \frac{7}{6} \)

    \(f''(x) = 24x - 14\)

At \(x = 0\):

    \(f''(0) = -14 \lt 0\), a local maximum.

At \(x = \frac{7}{6}\):

    \( \displaystyle f''\left(\frac{7}{6}\right) = 28 - 14 = 14 \gt 0 \), a local minimum.

So the only point of local minimum is \(x = \frac{7}{6}\); \(x = 0\) is a point of local maximum.