f(x) = x^3 + 3x^2 + x + 2
f'(x) = 3x^2 + 6x + 1
f''(x) = 6x + 6 = 0 → x = -1
Gesucht ist also die Tangente an der Stelle a = -1
f(a) = f(-1) = 3
f'(a) = f'(-1) = -2
Tangentengleichung
t(x) = f'(a)·(x - a) + f(a)
t(x) = -2·(x - (-1)) + 3 = -2·(x + 1) + 3 = -2·x + 1