extreme f(x)=-7(2x^3-125/32)^2+1

Lösung

extrempunkte

f (x) = - 7 (2 x^{3} - \frac{125}{32})^{2} + 1

Sattel (0, - \frac{108351}{1024}), Maximum (\frac{5}{4}, 1)

Schritte zur Lösung

f^{'} (x) = - 168 x^{5} + \frac{2625 x ^{2}}{8}

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{5}{4},

Absteigend

: \frac{5}{4} < x < \infty

Stecke

x = 0

- 7 (2 x^{3} - \frac{125}{32})^{2} + 1 : - \frac{108351}{1024}

Sattel (0, - \frac{108351}{1024})

Stecke

x = \frac{5}{4}

- 7 (2 x^{3} - \frac{125}{32})^{2} + 1 : 1

Maximum (\frac{5}{4}, 1)

Sattel (0, - \frac{108351}{1024}), Maximum (\frac{5}{4}, 1)

e x t re m e f (x, y) = \frac{1}{( x + y )}

e x t re m e f (x) = \frac{1}{1 - x ^{2}}

e x t re m e f (x) = \frac{x ^{2}}{x + 1}, - 2 \leq x \leq 2

e x t re m e f (x) = \frac{- x - 8}{10 x ^{2} - 40 x + 27}

e x t re m e f (x, y) = 4 x^{2} + 9 y^{2} - 2 x - 18 x y^{2}