extreme y=(2(x+3))/(x^2+x-2)

Lösung

extrempunkte

y = \frac{2 ( x + 3 )}{x ^{2} + x - 2}

Minimum (- 5, - \frac{2}{9}), Maximum (- 1, - 2)

Schritte zur Lösung

f^{'} (x) = \frac{2 ( - x ^{2} - 6 x - 5 )}{( x ^{2} + x - 2 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 5,

Ansteigend

: - 5 < x < - 2,

Ansteigend

: - 2 < x < - 1,

Absteigend

: - 1 < x < 1,

Absteigend

: 1 < x < \infty

Stecke

x = - 5

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

Minimum (- 5, - \frac{2}{9})

Stecke

x = - 1

\frac{2 ( x + 3 )}{x ^{2} + x - 2} : - 2

Maximum (- 1, - 2)

Minimum (- 5, - \frac{2}{9}), Maximum (- 1, - 2)

e x t re m e f (x, y) = ln (x) + y^{3}

e x t re m e f (x) = 2 x^{3} - 33 x^{2} + 180 x + 6

e x t re m e f (x) = (x^{4} - 32 x^{2} - 4)

e x t re m e f (x) = 5 x^{3} e^{- x}, - 1 \leq x \leq 4

e x t re m e f (x, y) = x^{2} + 2 x + y^{2} - 8 y + 8