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

To find the local minima of $f(x) = 4x^3 - 7x^2$, find where the derivative equals zero.

The first derivative is:

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

Setting the derivative equal to zero:

$12x^2 - 14x = 0$

$2x(6x - 7) = 0$

This gives two critical points:

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

---

To determine which critical points are local minima, use the second derivative test.

The second derivative is:

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

At a critical point:

• If $f''(x) > 0$, the point is a local minimum
• If $f''(x) < 0$, the point is a local maximum

---

For $x = 0$:

$f''(0) = 24(0) - 14$

$f''(0) = -14 < 0$

This is a local maximum.

---

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

$f''\left(\frac{7}{6}\right) = 24 \times \frac{7}{6} - 14$

$f''\left(\frac{7}{6}\right) = 28 - 14$

$f''\left(\frac{7}{6}\right) = 14 > 0$

This is a local minimum.

---

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