extreme f(x)=32e^{4x}x-28e^{4x}

Lösung

extrempunkte

f (x) = 32 e^{4 x} x - 28 e^{4 x}

Minimum (\frac{5}{8}, - 8 e^{\frac{5}{2}})

Schritte zur Lösung

f^{'} (x) = 16 e^{4 x} (8 x - 5)

Intervalle finden:

Absteigend

: - \infty < x < \frac{5}{8},

Ansteigend

: \frac{5}{8} < x < \infty

Stecke

x = \frac{5}{8}

32 e^{4 x} x - 28 e^{4 x} : - 8 e^{\frac{5}{2}}

Minimum (\frac{5}{8}, - 8 e^{\frac{5}{2}})

e x t re m e f (x, y) = 5 x^{2} - 4 x y + 4 y^{2} + 12 x + 25

e x t re m e f (x) = 2 x^{3} - \frac{12 x ^{2}}{2} - 2 [- 1.3]

e x t re m e \frac{27 x ^{2}}{( 2 - x ) ^{3}}

e x t re m e \frac{8 x ^{2}}{x - 6}, - 3 \leq x \leq 2

e x t re m e f (x, y) = x + 3 y