Final answer:
The equation y = x^3 - x^2 + 2x + 1 is always increasing because its first derivative, which is 3x^2 - 2x + 2, is always positive, indicating that there are no real critical points where the function changes its behavior from increasing to decreasing or vice versa.
Step-by-step explanation:
To determine the intervals where the equation y = x^3 - x^2 + 2x + 1 is increasing or decreasing, we need to find its first derivative which will give us the slope of the tangent to the curve at any point x. The derivative is denoted as y' or f'(x). For the function given, let's calculate the first derivative:
y' = 3x^2 - 2x + 2.
Now, to find where the function is increasing or decreasing, we set the derivative equal to zero and solve for x to find critical points:
0 = 3x^2 - 2x + 2.
This quadratic equation does not have any real solutions because the discriminant (b^2 - 4ac) is negative. Since the derivative is always positive (as the leading coefficient of the squared term is positive), the original function is always increasing. Therefore, option A is incorrect because it assumes there is a point a where the function changes from increasing to decreasing or vice versa, which is not true in this case.
Options B, C, and D all imply there is some changing behavior at points a or b, which is also incorrect. Since the function has no intervals of decreasing, the correct description is that the function is increasing for all x.