extreme f(x)=x^3-6x^2+5

Lösung

extrempunkte

f (x) = x^{3} - 6 x^{2} + 5

Maximum (0, 5), Minimum (4, - 27)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 12 x

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = 0

x^{3} - 6 x^{2} + 5 : 5

Maximum (0, 5)

Stecke

x = 4

x^{3} - 6 x^{2} + 5 : - 27

Minimum (4, - 27)

Maximum (0, 5), Minimum (4, - 27)

a sy m pt o t es \frac{2 x ^{2} - 7 x + 3}{x ^{2} - 3 x - 2}

d o main f (x) = \frac{4 y ^{2} + 16 y + 64}{y ^{3} - 6 y ^{2}}

d o main f (x) = \frac{1}{\frac{20}{x - 5} - 2}

in f l ec t i o n f (x) = x^{3} - 3 x^{2}

r an g e - (x - 7)^{2} + 4