Minimum of x^3-3x for x>0

Take the derivative and set this to 0......and we have

3x^2 - 3   = 0

3(x^2 - 1) = 0    divide by 3

x^2 - 1   = 0        factor

(x + 1) (  x - 1)  = 0    setting each factor to 0 and we have that

x = - 1     and  x = 1    so  these are the critical points

Taking the second derivative, we have

6x   .......   and putting x = 1   into this  we get   6(1)  = 6

This indicates a minimum when x = 1   and y = (1)^3 - 3(1)   =  -2

So....the minimum point is -2 when x > 0