A function has extreme values at the points where the first derivative is equal to 0.

y=x^3+3x^2-3x+6

y'= 3x^2 + 6x - 3

3x^2 + 6x - 3 = 0

=> x^2 + 2x - 1 = 0

x1 = -2/2 + sqrt (4 + 4)/2

=> -1 + 2*sqrt 2/2

=> -1 + sqrt 2

x2 = -1 - sqrt 2

**The function has extremes at the points where x = -1 + sqrt 2 and x = -1 - sqrt 2**

To determine the local extrema of a function, we must calculate the first derivative zeroes.

We'll diiferentiate with respect to x:

dy/dx = (x^3+3x^2-3x+6)'

dy/dx = 3x^2 + 6x - 3

We'll cancel dy/dx = 0:

3x^2 + 6x - 3 = 0

We'll divide by 3:

x^2 + 2x - 1 = 0

We'll determine the zeroes of the quadratic:

x1 = [-2+sqrt(4 + 4)]/2

x1 = (-2+2sqrt2)/2

x1 = -1 + sqrt2

x2 = -1 - sqrt2

Since the function has critical points x1 and x2 and the local extrema f(x1) and f(x2).

**Since the function is decreasing between x1 and x2 and it is increasing over the ranges (-infinite, -1-sqrt2) and (-1+sqrt2 , +infinite), then f(-1-sqrt2) is a maximum point and f(-1+sqrt2) is a minimum point.**