Firstly, we need to find the second derivative of the function f(x). It is given as f''(x) = 3x^2 - 6x. We set this equal to zero to find the critical points.
So, we solve
3x^2 - 6x = 0
The solutions to this equation are the critical points x = 0 and x = 2.
Next, we test these values and values around them to understand where the function is concave up and where it is concave down. We take four test points - one to the left of the smallest critical point, one between the two critical points, one to the right of the largest critical point, and one exactly in the middle. The test points we use are -3, -1, 1 and 3.
Inserting these four points into the second derivative, we find:
- f''(-3) = 3(-3)^2 - 6(-3) = 27 > 0, meaning the graph is concave up in this interval.
- f''(-1) = 3(-1)^2 - 6(-1) = 9 > 0, meaning the graph is concave up in this interval too.
- f''(1) = 3*1^2 - 6*1 = -3 < 0, indicating that the graph is concave down in this interval.
- f''(3) = 3*3^2 - 6*3 = 9 > 0, indicating that the graph is concave up in this interval.
Therefore, we can see that the function is concave up on the intervals (-∞,0) and (2,∞), and it's concave down on the interval (0,2). So, the final answer is:
- Concave Up: (-∞,0) ∪ (2,∞)
- Concave Down: (0,2)