extreme x^2+4x-5

Lösung

extrempunkte

x^{2} + 4 x - 5

Minimum (- 2, - 9)

Schritte zur Lösung

f^{'} (x) = 2 x + 4

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < \infty

Stecke

x = - 2

x^{2} + 4 x - 5 : - 9

Minimum (- 2, - 9)

e x t re m e f (x) = 2 sin (x) - x, 0 \leq x \leq 2 π

e x t re m e f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

e x t re m e f (x) = \frac{20}{3 ( 2 + x ) ^{3}}

e x t re m e f (x) = x^{3} + 3 x^{2} + 5, - 3 \leq x \leq 2

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