The largest open interval in which the function f(x) = 4x^3 - 5x^2 -…

The correct answer is Option (2) → \(\left(-\infty, -\displaystyle\frac{1}{2}\right) \cup \left(\displaystyle\frac{4}{3}, \infty\right)\)

A function is increasing where \(f'(x) \gt 0\).

    \(f'(x) = 12x^2 - 10x - 8\)
    \(= 2(6x^2 - 5x - 4)\)
    \(= 2(2x + 1)(3x - 4)\)

The roots are \(x = -\frac{1}{2}\) and \(x = \frac{4}{3}\). The product is positive when both factors have the same sign, that is, outside the roots:

    \( \displaystyle f'(x) \gt 0 \text{ for } x \lt -\frac{1}{2} \text{ or } x \gt \frac{4}{3} \)

So f is increasing on each of the open intervals \(\left(-\infty, -\frac{1}{2}\right)\) and \(\left(\frac{4}{3}, \infty\right)\), and the set where it increases is their union.

Why the other options are incorrect

• Option (1) \(\left(-\infty, -\frac{4}{3}\right) \cup \left(\frac{4}{3}, \infty\right)\) — Incorrect. The smaller root is \(-\frac{1}{2}\), not \(-\frac{4}{3}\): \(f'(-1) = 12 + 10 - 8 = 14 \gt 0\), yet x = −1 is left out.

• Option (3) \(\left(-\infty, -\frac{1}{2}\right) \cap \left(\frac{4}{3}, \infty\right)\) — Incorrect. The two intervals have no common point, so their intersection is empty.

• Option (4) \(\left(-\infty, \frac{1}{2}\right) \cup \left(\frac{4}{3}, \infty\right)\) — Incorrect. \(f'(0) = -8 \lt 0\), so f decreases at x = 0, which this set includes.