extreme f(x)=(x^2-5x+7)/(x-3)

Lösung

extrempunkte

f (x) = \frac{x ^{2} - 5 x + 7}{x - 3}

Maximum (2, - 1), Minimum (4, 3)

Schritte zur Lösung

f^{'} (x) = \frac{x ^{2} - 6 x + 8}{( x - 3 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < 2,

Absteigend

: 2 < x < 3,

Absteigend

: 3 < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = 2

\frac{x ^{2} - 5 x + 7}{x - 3} : - 1

Maximum (2, - 1)

Stecke

x = 4

\frac{x ^{2} - 5 x + 7}{x - 3} : 3

Minimum (4, 3)

Maximum (2, - 1), Minimum (4, 3)

e x t re m e f (x) = (3 x^{3} - 9)

e x t re m e f (x) = \frac{x ^{2} - 4}{e ^{x^{4} - 1}}

e x t re m e f (x) = 2 x - 5 x^{2} + 4

e x t re m e f (x, y) = x^{4} + y^{4} - 4 x y + x

e x t re m e - 2 x^{3} + 14 x^{2} + 2 y^{2} + 4 x y