Find the intervals of decreasing function f (x) = x ^ 3 – 6x ^ 2 + 5

f ’(X) = (X ^ 3 – 6 * X ^ 2 + 5)’ = 3 * X ^ 2 – 12 * X.

Let us equate the derivative to zero and determine the critical points.

3 * X ^ 2 – 12 * X = 0;

3 * X * (X – 4) = 0.

The product is equal to zero if at least one of the factors is equal to zero.

3 * X = 0;

X1 = 0;

X – 4 = 0;

X2 = 4.

Let us determine the signs of the derivative on the intervals (-∞; 0), (0; 4), (4; + ∞).

When X = -1, f ‘(- 1) = 15. The function increases.

When X = 1, f ‘(- 1) = -9. The function is decreasing.

When X = 5, f ‘(- 1) = 15. The function increases.

Answer: The function increases as X Є (-∞; 0) U (4; + ∞).