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

Lösung

extrempunkte

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

Maximum (\frac{4 - 10}{3}, \frac{20 10 - 56}{27}), Minimum (\frac{4 + 10}{3}, \frac{- 56 - 20 10}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 10 + 4}{3},

Absteigend

: \frac{- 10 + 4}{3} < x < \frac{10 + 4}{3},

Ansteigend

: \frac{10 + 4}{3} < x < \infty

Stecke

x = \frac{4 - 10}{3}

x^{3} - 4 x^{2} + 2 x : \frac{20 10 - 56}{27}

Maximum (\frac{4 - 10}{3}, \frac{20 10 - 56}{27})

Stecke

x = \frac{4 + 10}{3}

x^{3} - 4 x^{2} + 2 x : \frac{- 56 - 20 10}{27}

Minimum (\frac{4 + 10}{3}, \frac{- 56 - 20 10}{27})

Maximum (\frac{4 - 10}{3}, \frac{20 10 - 56}{27}), Minimum (\frac{4 + 10}{3}, \frac{- 56 - 20 10}{27})

in v erse f (x) = x^{(3)} + 1

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

a sy m pt o t es f (x) = \frac{7}{( x - 8 ) ^{3}}

d o main 6 - x

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