f(x) = 3·k·x^3 - 3·(k + 1)·x^2 + 4·x
f'(x) = 9·k·x^2 - 6·x·(k + 1) + 4
f''(x) = 18·k·x - 6·(k + 1)
a)
Maximum f'(x) = 0
9·k·x^2 - 6·x·(k + 1) + 4 = 0 --> x = 2/(3·k) ∨ x = 2/3
f''(2/(3·k)) = 6·(1 - k) < 0 für k > 1
f(2/(3·k)) = 4·(3·k - 1)/(9·k^2) > 1 --> nie erfüllt
f''(2/3) = 6·(k - 1) < 0 --> k < 1
f(2/3) = 4·(3 - k)/9 > 1 --> k < 3/4
b)
t(x) = f'(1) * (x - 1) + f(1) = 3·k·x - 2·x - 3·k + 3
Nullstellen t(x) = 0
3·k·x - 2·x - 3·k + 3 = 0 --> x = 3·(k - 1)/(3·k - 2) < 0 --> 2/3 < k < 1