Find the intervals of increase of the function y = x ^ 3 + x ^ 2-8x.

Consider the function y = x ^ 3 + x ^ 2 – 8x.

Find the extremum points of the function, i.e. points at which y ‘= 0:

y ‘= (x ^ 3 + x ^ 2 – 8x)’ = 3x ^ 2 + 2x – 8,

3x ^ 2 + 2x – 8 = 0;

D = 4 + 4 * 8 * 3 = 100,

x1 = (-2 + 10) / 6 = 8/6 = 4/3,

x2 = (-2 – 10) / 6 = -12/6 = -2.

Extremum points: -2 and 4/3.

Consider the intervals of decreasing / increasing function.

When x <-2, y ’> 0, the function increases.

At -2 <x <4/3, y ’<0, the function decreases.

For x> 4/3, y ‘> 0, the function increases.

Thus, the function increases on the interval: (-∞; -2] and [4/3; + ∞).

Answer: (-∞; -2] and [4/3; + ∞).