Hello,

calculate first the 1rst and 2nd derivative:

1)

2)

3)

The conditions of critical points are: f'(x) = 0andf''(x) ≠ 0

f(0) = 0 and f''(0) = 0 You've got a point of inflection with a horizontal gradient. Such a point is called in German (literally translated) a "terrace point".

f(3/2) = -27/16 and f''(3/2) = 9 > 0 that means the function has a minimum. From this point the function is continuously increasing.

The condition for the existence of points of inflection is

From 3) you get x = 0 or x = 1

The 2 points are:

The slope in I_1 is f'(0) = 0 therefore the tangent is the x-axis.

The slope in I_2 is f'(1) = -2. Use the point-slope-formula to get the equation of the tangent:

The graph is concave upward if f''(x) > 0 and the graph is concave downward if f''(x) < 0:

I've marked the concave downward part of the graph in red.

EB