extreme f(x)=(2+x)^2(1-x)^2

Lösung

extrempunkte

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

Minimum (- 2, 0), Maximum (- \frac{1}{2}, \frac{81}{16}), Minimum (1, 0)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Ansteigend

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

Absteigend

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

Ansteigend

: 1 < x < \infty

Stecke

x = - 2

(2 + x)^{2} (1 - x)^{2} : 0

Minimum (- 2, 0)

Stecke

x = - \frac{1}{2}

(2 + x)^{2} (1 - x)^{2} : \frac{81}{16}

Maximum (- \frac{1}{2}, \frac{81}{16})

Stecke

x = 1

(2 + x)^{2} (1 - x)^{2} : 0

Minimum (1, 0)

Minimum (- 2, 0), Maximum (- \frac{1}{2}, \frac{81}{16}), Minimum (1, 0)

e x t re m e y = 6 x - ln (6 x)

e x t re m e f (x) = 7 + 81 x - 3 x^{3}

e x t re m e f (x, y) = (x - y) (9 - x y)

e x t re m e f (x) = x^{2} + \frac{200}{x}

e x t re m e - 0.1 t^{2} + 0.8 t + 98.8, 0 \leq t \geq 8