a)
fk(x) = - k^2·x^3 - 6·k·x^2 - 9·x = - x·(k·x + 3)^2
F(x) = - 0.25·k^2·x^4 - 2·k·x^3 - 4.5·x^2
Nullstellen f(x) = 0
- x·(k·x + 3)^2 = 0 --> x = - 3/k ∨ x = 0
A = ∫ (- 3/k bis 0) f(x) dx = 27/(4·k^2)
b)
27/(4·k^2) = 3 --> k = - 3/2 ∨ k = 3/2
c)
Wendestelle f''(x) = 0
- 6·k^2·x - 12·k = 0 --> x = - 2/k
t(x) = f'(- 2/k) * (x - (- 2/k)) + f(- 2/k) = 3·x + 8/k