Kurvendiskussion: f(x) = (x^2 - 3·x + 5)/(2·x^2 - 4·x + 6)

Funktion und Ableitungen

f(x) = (x^2 - 3·x + 5)/(2·x^2 - 4·x + 6) = 1/2 + (2 - x)/(2·x^2 - 4·x + 6)

f'(x) = (2·x^2 - 8·x + 2)/(2·x^2 - 4·x + 6)^2

f''(x) = (- 4·x^3 + 24·x^2 - 12·x - 16)/(2·x^2 - 4·x + 6)^3

Symmetrie

Keine erkennbare Symmetrie

Polstellen Nenner = 0

2·x^2 - 4·x + 6 = 0
Keine Polstellen

Asymptote

Horizontale Asymptote: y = 0.5

Y-Achsenabschnitt f(0)

f(0) = 5/6 = 0.833

Nullstellen f(x) = 0

(x^2 - 3·x + 5)/(2·x^2 - 4·x + 6) = 0

x^2 - 3·x + 5 = 0

Keine reellen Nullstellen

Extremstellen f'(x) = 0

(2·x^2 - 8·x + 2)/((2·x^2 - 4·x + 6)^2) = 0

2·x^2 - 8·x + 2 = 0

x = 2 - √3 ∨ x = x = 2 + √3

x = 0.268 ∨ x = 3.732

f(0.268) = √3/8 + 5/8 = 0.842 | Hochpunkt

f(3.732) = 5/8 - √3/8 = 0.408 | Tiefpunkt

Wendestellen f''(x) = 0

(- 4·x^3 + 24·x^2 - 12·x - 16)/(2·x^2 - 4·x + 6)^3 = 0

- 4·x^3 + 24·x^2 - 12·x - 16 = 0 | Hier hilft nur ein Näherungsferfahren

x = -0.584 ∨ x = 1.294 ∨ x = 5.290

f(-0.584) = 0.787

f(1.294) = 0.669

f(5.290) = 0.419

Skizze