extreme f(x)=4x^3-3x^2-6x+17

Lösung

extrempunkte

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 17

Maximum (- \frac{1}{2}, \frac{75}{4}), Minimum (1, 12)

Schritte zur Lösung

f^{'} (x) = 12 x^{2} - 6 x - 6

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{1}{2}

4 x^{3} - 3 x^{2} - 6 x + 17 : \frac{75}{4}

Maximum (- \frac{1}{2}, \frac{75}{4})

Stecke

x = 1

4 x^{3} - 3 x^{2} - 6 x + 17 : 12

Minimum (1, 12)

Maximum (- \frac{1}{2}, \frac{75}{4}), Minimum (1, 12)

d o main \frac{x ^{2} + 25}{x}

m o n o t o n e f (x) = x^{4} - 2 x^{3}

p a r i t y f (x) = \frac{x}{x + 1}

in v erse \frac{4}{x + 7}

s hi f t f (t) = sin (2 t - \frac{π}{3}) + 1