extreme f(x)=e^x(15-x^2)

Lösung

extrempunkte

f (x) = e^{x} (15 - x^{2})

Minimum (- 5, - \frac{10}{e ^{5}}), Maximum (3, 6 e^{3})

Schritte zur Lösung

f^{'} (x) = - e^{x} (x - 3) (x + 5)

Intervalle finden:

Absteigend

: - \infty < x < - 5,

Ansteigend

: - 5 < x < 3,

Absteigend

: 3 < x < \infty

Stecke

x = - 5

e^{x} (15 - x^{2}) : - \frac{10}{e ^{5}}

Minimum (- 5, - \frac{10}{e ^{5}})

Stecke

x = 3

e^{x} (15 - x^{2}) : 6 e^{3}

Maximum (3, 6 e^{3})

Minimum (- 5, - \frac{10}{e ^{5}}), Maximum (3, 6 e^{3})

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

e x t re m e f (x) = 8 x^{3} - 17 x^{2} + 12 x + 14

e x t re m e f (x, y) = 3 y^{3} + 7 x y - 9 x^{2} - 5

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

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