The largest open interval in which the function $f(x) = 4x^3-5x^2-8x + 12$ increases, is:

$\left(-∞,-\frac{1}{2}\right)∪\left(\frac{4}{3},∞\right)$

The correct answer is Option (2) → $\left(-∞,-\frac{1}{2}\right)∪\left(\frac{4}{3},∞\right)$

Given: $f(x) = 4x^3 - 5x^2 - 8x + 12$

Derivative: $f'(x) = \frac{d}{dx}(4x^3 - 5x^2 - 8x + 12) = 12x^2 - 10x - 8$

Set $f'(x) > 0$ for increasing intervals:

12x^2 - 10x - 8 > 0

Divide by 2: 6x^2 - 5x - 4 > 0

Solve quadratic equation 6x^2 - 5x - 4 = 0:

Discriminant: $D = (-5)^2 - 4*6*(-4) = 25 + 96 = 121$

Roots: $x = \frac{5 \pm \sqrt{121}}{12} = \frac{5 \pm 11}{12}$

$x_1 = \frac{16}{12} = \frac{4}{3}$, $x_2 = \frac{-6}{12} = -\frac{1}{2}$

Quadratic opens upwards (coefficient of x^2 positive), so $f'(x) > 0$ for $x < -1/2$ or $x > 4/3$

Increasing at: $\left(-∞,-\frac{1}{2}\right)∪\left(\frac{4}{3},∞\right)$