f(x) = e-x - e- 2·x
f'(x) = 2·e- 2·x - e-x
f''(x) = e-x - 4·e- 2·x
F(x) = 0.5·e- 2·x - e-x
Nullstellen f(x) = 0
e-x - e- 2·x = 0 | z = e-x
z - z2 = 0
z(1 - z) = 0
z = 0 --> x = -ln(0) = keine Lösung
z = 1 --> x = -ln(1) = 0
Horizontale Tangenten f'(x) = 0
2·e- 2·x - e-x = 0 | Lösen wie vorher über Substitution
x = ln(2)
f(ln(2)) = 0.25
t(x) = 0.25
Wendepunkte f''(x) = 0
e-x - 4·e- 2·x = 0 | Auch hier wieder Lösen über Substitution
x = 2·ln(2)
f(2·ln(2)) = 3/16
WP(2·ln(2) | 3/16) = WP(1.386 | 0.1875)
lim x → -∞ f(x) = -∞
lim x → ∞ f(x) = 0
Fläche
F(∞) - F(0) = 0 - (-0.5) = 0.5
Skizze:
