extreme f(x)=x^3-4x^2+5

Lösung

extrempunkte

f (x) = x^{3} - 4 x^{2} + 5

Maximum (0, 5), Minimum (\frac{8}{3}, - \frac{121}{27})

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 8 x

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{8}{3},

Ansteigend

: \frac{8}{3} < x < \infty

Stecke

x = 0

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

Maximum (0, 5)

Stecke

x = \frac{8}{3}

x^{3} - 4 x^{2} + 5 : - \frac{121}{27}

Minimum (\frac{8}{3}, - \frac{121}{27})

Maximum (0, 5), Minimum (\frac{8}{3}, - \frac{121}{27})

a sy m pt o t es f (x) = \frac{3 x ^{2} - 6 x}{x ^{2} - 25}

p er p e n d i c u l a r 3 = 3 (2) - 12

d o main f (x) = - (x - 2)^{2} + 5

in t erce pt s f (x) = 5^{x}

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