extreme f(x)=((-8x^3+3x^2))/4

Lösung

extrempunkte

f (x) = \frac{( - 8 x ^{3} + 3 x ^{2} )}{4}

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{1}{64})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

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

Absteigend

: \frac{1}{4} < x < \infty

Stecke

x = 0

\frac{- 8 x ^{3} + 3 x ^{2}}{4} : 0

Minimum (0, 0)

Stecke

x = \frac{1}{4}

\frac{- 8 x ^{3} + 3 x ^{2}}{4} : \frac{1}{64}

Maximum (\frac{1}{4}, \frac{1}{64})

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{1}{64})

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

e x t re m e f (x, y) = y (e^{x} - 1)

e x t re m e f (x) = \frac{x}{x ^{2} + 9}, - 4 \leq x \leq 5

e x t re m e x^{3} + 3 x^{2} - 24 x

e x t re m e - x^{3} + 4 x y - 2 y^{2} + 1