y'=3x^2-3=3(x+ 1)(x- 1)

Let y' & gt=0, x∈(-∞,-1]∩[ 1, +∞).

be your

Therefore, the function monotonically decreases at (-1, 1) and monotonically increases at (-∞,-1] and [1, +∞).

(2)

Find the extreme value so that y'=3(x+ 1)(x- 1)=0.

Two stagnation points are obtained: x 1= 1, x2=- 1.

Because the function decreases first and then increases on both sides of x 1= 1, x 1 is a minimal point.

On both sides of x2=- 1, the function increases first and then decreases, so x2=- 1 is the maximum point.

Therefore, the maximum value f(- 1)=2 and the minimum value f( 1)=-2.