extreme-x^3+8x^2-15x

Lösung

extrempunkte

- x^{3} + 8 x^{2} - 15 x

Minimum (\frac{8 - 19}{3}, - 5 (8 - 19) + \frac{1024 - 173 19}{27}), Maximum (\frac{8 + 19}{3}, - 5 (8 + 19) + \frac{1024 + 173 19}{27})

Schritte zur Lösung

f^{'} (x) = - 3 x^{2} + 16 x - 15

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 19 + 8}{3},

Ansteigend

: \frac{- 19 + 8}{3} < x < \frac{19 + 8}{3},

Absteigend

: \frac{19 + 8}{3} < x < \infty

Stecke

x = \frac{8 - 19}{3}

- x^{3} + 8 x^{2} - 15 x : - 5 (8 - 19) + \frac{1024 - 173 19}{27}

Minimum (\frac{8 - 19}{3}, - 5 (8 - 19) + \frac{1024 - 173 19}{27})

Stecke

x = \frac{8 + 19}{3}

- x^{3} + 8 x^{2} - 15 x : - 5 (8 + 19) + \frac{1024 + 173 19}{27}

Maximum (\frac{8 + 19}{3}, - 5 (8 + 19) + \frac{1024 + 173 19}{27})

Minimum (\frac{8 - 19}{3}, - 5 (8 - 19) + \frac{1024 - 173 19}{27}), Maximum (\frac{8 + 19}{3}, - 5 (8 + 19) + \frac{1024 + 173 19}{27})

r an g e f (x) = (x - 2)^{2} + 3

l in e (- 2, - 6), (4, 6)

l in e y = 8 x - 7

d o main \frac{- 6 x + 23}{7 x - 16}

d o main h (x) = \frac{2}{x ^{2} - 4}