extreme y=x^{4/5}(x-1)

Lösung

extrempunkte

y = x^{\frac{4}{5}} (x - 1)

Maximum (0, 0), Minimum (\frac{4}{9}, - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

Schritte zur Lösung

f^{'} (x) = \frac{9 x - 4}{5 x ^{\frac{1}{5}}}

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{4}{9},

Ansteigend

: \frac{4}{9} < x < \infty

Stecke

x = 0

x^{\frac{4}{5}} (x - 1) : 0

Maximum (0, 0)

Stecke

x = \frac{4}{9}

x^{\frac{4}{5}} (x - 1) : - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}}

Minimum (\frac{4}{9}, - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

Maximum (0, 0), Minimum (\frac{4}{9}, - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

e x t re m e f (x) = \frac{( x ^{2} - 11 )}{x + 6}

e x t re m e \frac{64}{x ^{2} + 6 x - 7}

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

e x t re m e y = x^{3} - 12 x + 3

e x t re m e y = - x - ln ∣ x ∣