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

$x = \frac{7}{6}$

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

Given function:

$f(x) = 4x^3 - 7x^2$

First derivative:

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

Critical points from $f'(x)=0$:

$x = 0$ or $x = \frac{7}{6}$

Second derivative:

$f''(x) = 24x - 14$

At $x = 0$: $f''(0) = -14 < 0$ ⇒ local maximum.

At $x = \frac{7}{6}$: $f''\left(\frac{7}{6}\right) = 24\left(\frac{7}{6}\right) - 14 = 28 - 14 = 14 > 0$ ⇒ local minimum.

Therefore, the function has a local minimum at $x = \frac{7}{6}$.