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

Lösung

extrempunkte

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

Maximum (- \frac{1 + 13}{3}, \frac{38 + 26 13}{27} - 3), Minimum (\frac{- 1 + 13}{3}, \frac{38 - 26 13}{27} - 3)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 13 - 1}{3},

Absteigend

: \frac{- 13 - 1}{3} < x < \frac{13 - 1}{3},

Ansteigend

: \frac{13 - 1}{3} < x < \infty

Stecke

x = - \frac{1 + 13}{3}

x^{3} + x^{2} - 4 x - 3 : \frac{38 + 26 13}{27} - 3

Maximum (- \frac{1 + 13}{3}, \frac{38 + 26 13}{27} - 3)

Stecke

x = \frac{- 1 + 13}{3}

x^{3} + x^{2} - 4 x - 3 : \frac{38 - 26 13}{27} - 3

Minimum (\frac{- 1 + 13}{3}, \frac{38 - 26 13}{27} - 3)

Maximum (- \frac{1 + 13}{3}, \frac{38 + 26 13}{27} - 3), Minimum (\frac{- 1 + 13}{3}, \frac{38 - 26 13}{27} - 3)

cr i t i c a l 2 x ln (x) + x

in t erce pt s f (x) = - (x + 3)^{2}

d o main - \frac{7}{( 2 + x ) ^{2}}

d o main 2 cos (x) - 2 x

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