extreme f(x)=(-x^{2/3})(x-3)

Lösung

extrempunkte

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

Minimum (0, 0), Maximum (\frac{6}{5}, \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Schritte zur Lösung

f^{'} (x) = \frac{- 5 x + 6}{3 x ^{\frac{1}{3}}}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{6}{5},

Absteigend

: \frac{6}{5} < x < \infty

Stecke

x = 0

(- x^{\frac{2}{3}}) (x - 3) : 0

Minimum (0, 0)

Stecke

x = \frac{6}{5}

(- x^{\frac{2}{3}}) (x - 3) : \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}

Maximum (\frac{6}{5}, \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Minimum (0, 0), Maximum (\frac{6}{5}, \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

e x t re m e f (x) = f (x, y) = 3 x^{3} - 9 x + 9 x y^{2}

e x t re m e f (x) = 5 x + 5 x^{- 1}

e x t re m e x + 2 sin (x)

e x t re m e - 0.2 x^{3} - 0.7 x^{2} + 3 x - 6

e x t re m e h (t) = \frac{t}{t - 2} [3.5]