extreme f(x)=(e^{3x})/(3x-3)

Lösung

extrempunkte

f (x) = \frac{e ^{3 x}}{3 x - 3}

Minimum (\frac{4}{3}, e^{4})

Schritte zur Lösung

f^{'} (x) = \frac{e ^{3 x} ( 3 x - 4 )}{3 ( x - 1 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < 1,

Absteigend

: 1 < x < \frac{4}{3},

Ansteigend

: \frac{4}{3} < x < \infty

Stecke

x = \frac{4}{3}

\frac{e ^{3 x}}{3 x - 3} : e^{4}

Minimum (\frac{4}{3}, e^{4})

e x t re m e f (x) = - 2 sin^{2} (x)

e x t re m e f (x) = \frac{1}{4} x^{4} + \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 4

e x t re m e f (x) = x y^{2} = 9

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

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