f(x) = (a·x^2 + b·x + c)/(2·x + d) = 0.5·a·x - 0.25·a·d + 0.5·b + R
0.5·a = 0.5 --> a = 1
- 0.25·1·d + 0.5·b = 1 --> d = 2·b - 4
f(1) = (a·0^2 + b·0 + c)/(2·0 + d) = 0 --> c = 0
f'(x) = (2·a·x^2 + 2·a·d·x + b·d - 2·c)/(2·x + d)^2
f'(0) = (2·1·0^2 + 2·1·d·0 + b·d - 2·0)/(2·0 + d)^2 = b/d = b/(2·b - 4) = -1.5 --> b = 1.5
Damit haben wir die Funktion
f(x) = (1·x^2 + 1.5·x)/(2·x - 1)