fk(x) = 1/(6·k)·x^3 - x^2 + 3/2·k·x
fk'(x) = 1/(2·k)·x^2 - 2·x + 3/2·k
fk'(x) = 1/k·x - 2
Symmetrie
Keine untersuchte Symmetrie
Nullstellen f(x) = 0
1/(6·k)·x^3 - x^2 + 3/2·k·x = 0 | ·(6·k)
x^3 - 6·k·x^2 + 9·k^2·x = x·(x^2 - 6·k·x + 9·k^2) = x·(x - 3·k)^2 = 0
x1 = 0 einfache Nullstelle
x2 = 3·k doppelte Nullstelle
Extrempunkte fk'(x) = 0
1/(2·k)·x^2 - 2·x + 3/2·k = 0 | ·(2·k)
x^2 - 4·k·x + 3·k^2 = (x - k)·(x - 3·k) = 0
x1 = k
x2 = 3·k
f(k) = 2/3·k^2 --> HP(k | 2/3·k^2)
f(3·k) = 0 --> TP(3·k | 0)
Wendepunkte fk''(x) = 0
1/k·x - 2 = 0 | *k
x - 2·k = 0
x = 2·k
f(2·k) = 1/3·k^2 --> WP(2·k | 1/3·k^2)