extreme f(x)=4+x+x^2-x^3

Lösung

extrempunkte

f (x) = 4 + x + x^{2} - x^{3}

Minimum (- \frac{1}{3}, \frac{103}{27}), Maximum (1, 5)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Absteigend

: 1 < x < \infty

Stecke

x = - \frac{1}{3}

4 + x + x^{2} - x^{3} : \frac{103}{27}

Minimum (- \frac{1}{3}, \frac{103}{27})

Stecke

x = 1

4 + x + x^{2} - x^{3} : 5

Maximum (1, 5)

Minimum (- \frac{1}{3}, \frac{103}{27}), Maximum (1, 5)

r an g e 3 x - 4

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

in v erse f (x) = - \frac{2}{3} lo g_{10} (x - 1) + 2

in v erse f (x) = \frac{2 x + 3}{5}

p er i o d i c i t y f (x) = - \frac{1}{5} cos (\frac{1}{5} x)