extreme f(x)=4x^3-x^2-x+8

Lösung

extrempunkte

f (x) = 4 x^{3} - x^{2} - x + 8

Maximum (\frac{1 - 13}{12}, \frac{13 13 - 19}{216} + 8), Minimum (\frac{1 + 13}{12}, \frac{- 19 - 13 13}{216} + 8)

Schritte zur Lösung

f^{'} (x) = 12 x^{2} - 2 x - 1

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 13 + 1}{12},

Absteigend

: \frac{- 13 + 1}{12} < x < \frac{13 + 1}{12},

Ansteigend

: \frac{13 + 1}{12} < x < \infty

Stecke

x = \frac{1 - 13}{12}

4 x^{3} - x^{2} - x + 8 : \frac{13 13 - 19}{216} + 8

Maximum (\frac{1 - 13}{12}, \frac{13 13 - 19}{216} + 8)

Stecke

x = \frac{1 + 13}{12}

4 x^{3} - x^{2} - x + 8 : \frac{- 19 - 13 13}{216} + 8

Minimum (\frac{1 + 13}{12}, \frac{- 19 - 13 13}{216} + 8)

Maximum (\frac{1 - 13}{12}, \frac{13 13 - 19}{216} + 8), Minimum (\frac{1 + 13}{12}, \frac{- 19 - 13 13}{216} + 8)