Verify if the function y=x^3+3x^2-3x+6 has local extrema?

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.