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

A function is increasing when its derivative is positive, where $f'(x) > 0$.

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

$f'(x) = 12x^2 - 10x - 8$

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To find where the function is increasing:

$12x^2 - 10x - 8 > 0$

$6x^2 - 5x - 4 > 0$

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Setting the derivative equal to zero:

$6x^2 - 5x - 4 = 0$

Using the quadratic formula where $a = 6$, $b = -5$, $c = -4$:

$x = \frac{5 \pm \sqrt{(-5)^2 - 4(6)(-4)}}{2(6)}$

$x = \frac{5 \pm \sqrt{25 + 96}}{12}$

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

$x = \frac{5 \pm 11}{12}$

Critical points:

$x = \frac{5 + 11}{12} = \frac{4}{3}$

$x = \frac{5 - 11}{12} = -\frac{1}{2}$

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Testing the sign of $f'(x) = 6x^2 - 5x - 4$ in each interval:

For $x < -\frac{1}{2}$, test $x = -1$:

$f'(-1) = 6(1) - 5(-1) - 4 = 7 > 0$ (positive)

For $-\frac{1}{2} < x < \frac{4}{3}$, test $x = 0$:

$f'(0) = 6(0) - 5(0) - 4 = -4 < 0$ (negative)

For $x > \frac{4}{3}$, test $x = 2$:

$f'(2) = 6(4) - 5(2) - 4 = 10 > 0$ (positive)

Since $6x^2 - 5x - 4$ is a parabola opening upward, it is positive outside the roots and negative between them.

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The function is increasing where $f'(x) > 0$:

$\left(-\infty, -\frac{1}{2}\right) \cup \left(\frac{4}{3}, \infty\right)$