f(0) = 0
f'(0) = 0
f(6) = 1
f'(6) = 0
Ich hätte es aber wie folgt gemacht
f(x) = a·x^2·(x - b) = a·x^3 - a·b·x^2
f'(6) = 0 --> 12·a·(9 - b) = 0 --> b = 9
f(6) = 1 --> 36·a·(6 - b) = 1 --> 36·a·(6 - 9) = 1 --> a = -1/108
f(x) = -1/108·x^2·(x - 9) = 1/12·x^2 - 1/108·x^3