extreme f(x)=(x^2+3x-1)^{1/3}

Lösung

extrempunkte

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

Minimum (- \frac{3}{2}, - (\frac{13}{4})^{\frac{1}{3}})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 3 - 1 3 ^{\frac{1}{2}}}{2},

Absteigend

: \frac{- 3 - 1 3 ^{\frac{1}{2}}}{2} < x < - \frac{3}{2},

Ansteigend

: - \frac{3}{2} < x < \frac{- 3 + 1 3 ^{\frac{1}{2}}}{2},

Ansteigend

: \frac{- 3 + 1 3 ^{\frac{1}{2}}}{2} < x < \infty

Stecke

x = - \frac{3}{2}

(x^{2} + 3 x - 1)^{\frac{1}{3}} : - (\frac{13}{4})^{\frac{1}{3}}

Minimum (- \frac{3}{2}, - (\frac{13}{4})^{\frac{1}{3}})

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

e x t re m e f (x, y) = \frac{20000}{3 + x ^{2} + y ^{2}}

e x t re m e y = \frac{x}{x ^{2} + 64}

e x t re m e f (x) = - x^{2} + 1, - 1 \leq x \leq 0

e x t re m e f (x) = x^{5} - x^{3} + 8, - 1 \leq x \leq 1