f(x) = x^2·e^(5·x + 2)
f'(x) = e^(5·x + 2)·(5·x^2 + 2·x)
f''(x) = e^(5·x + 2)·(25·x^2 + 20·x + 2)
a) f''(-0.39) = -2.100 → konkav
b) f'(-0.31) = -0.2188 --> größer als -0.22
c) f(0) = 0 ; f'(0) = 0 ; f''(0) = 14.78 → Lokales Minimum
d) f'(-0.29) = -0.2765 → fallend
e) f''(-0.76) = 0.2050 → positiv