#1**+2 **

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

CPhill
Mar 21, 2017