extreme f(x)=x^3+x^2-x+3

Lösung

extrempunkte

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

Maximum (- 1, 4), Minimum (\frac{1}{3}, \frac{76}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Absteigend

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

Ansteigend

: \frac{1}{3} < x < \infty

Stecke

x = - 1

x^{3} + x^{2} - x + 3 : 4

Maximum (- 1, 4)

Stecke

x = \frac{1}{3}

x^{3} + x^{2} - x + 3 : \frac{76}{27}

Minimum (\frac{1}{3}, \frac{76}{27})

Maximum (- 1, 4), Minimum (\frac{1}{3}, \frac{76}{27})

e x t re m e sin (4 x), 0 \leq x \leq \frac{π}{2}

e x t re m e f (x) = \frac{x ^{2} - 4 x + 4}{x - 6}

e x t re m e f (x) = x^{3} - 588 x

e x t re m e f (x) = 23 n - 166 n^{2}

e x t re m e f (x) = \frac{- 2}{3} x^{3} + 9 x^{2} - 28 x + 1