extreme f(x)=(x-9)^3(x+5)^2

Lösung

extrempunkte

f (x) = (x - 9)^{3} (x + 5)^{2}

Maximum (- 5, 0), Minimum (\frac{3}{5}, - \frac{58084992}{3125}), Sattel (9, 0)

Schritte zur Lösung

f^{'} (x) = (x - 9)^{2} (x + 5) (5 x - 3)

Intervalle finden:

Ansteigend

: - \infty < x < - 5,

Absteigend

: - 5 < x < \frac{3}{5},

Ansteigend

: \frac{3}{5} < x < 9,

Ansteigend

: 9 < x < \infty

Stecke

x = - 5

(x - 9)^{3} (x + 5)^{2} : 0

Maximum (- 5, 0)

Stecke

x = \frac{3}{5}

(x - 9)^{3} (x + 5)^{2} : - \frac{58084992}{3125}

Minimum (\frac{3}{5}, - \frac{58084992}{3125})

Stecke

x = 9

(x - 9)^{3} (x + 5)^{2} : 0

Sattel (9, 0)

Maximum (- 5, 0), Minimum (\frac{3}{5}, - \frac{58084992}{3125}), Sattel (9, 0)

e x t re m e f (x) = x^{3} + x^{2} - x + 3

e x t re m e sin (4 x), 0 \leq x \leq \frac{π}{2}

e x t re m e f (x) = \frac{x ^{2} - 4 x + 4}{x - 6}

e x t re m e f (x) = x^{3} - 588 x

e x t re m e f (x) = 23 n - 166 n^{2}