The derivative of the polynomial is:
$$y' = 18x^2 + 3$$
To find the critical points, we set the derivative equal to zero and solve for x:
$$18x^2 + 3 = 0$$
$$18x^2 = -3$$
$$x^2 = -\frac{1}{6}$$
Since the square of a real number cannot be negative, there are no
Show more…