extreme f(x)=(x^2+2)/(2x-1)

Lösung

extrempunkte

f (x) = \frac{x ^{2} + 2}{2 x - 1}

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

Schritte zur Lösung

f^{'} (x) = \frac{2 x ^{2} - 2 x - 4}{( 2 x - 1 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Absteigend

: - 1 < x < \frac{1}{2},

Absteigend

: \frac{1}{2} < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 1

\frac{x ^{2} + 2}{2 x - 1} : - 1

Maximum (- 1, - 1)

Stecke

x = 2

\frac{x ^{2} + 2}{2 x - 1} : 2

Minimum (2, 2)

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

e x t re m e f (x) = - 2 cos (x) - 3 x

e x t re m e \frac{6 x}{x ^{2} + 1}

e x t re m e f (x) = - 1 - x^{2}, 0 \leq x \leq 1

e x t re m e f (x) = x^{2} + x y + \frac{1}{2} y^{2} - 3 x + y

e x t re m e x^{2} + \frac{400}{x}